map2.tex 107 KB
 Jim Hefferon committed Dec 05, 2011 1 2 3 4 5 % Chapter 3, Section 2 _Linear Algebra_ Jim Hefferon % http://joshua.smcvt.edu/linalg.html % 2001-Jun-11 \section{Homomorphisms} The definition of isomorphism has two conditions.  Jim Hefferon committed Dec 31, 2011 6 7 In this section we will consider the second one. We will study maps that  Jim Hefferon committed Jan 02, 2012 8 9 are required only to preserve structure, maps that are not also required to be correspondences.  Jim Hefferon committed Dec 05, 2011 10   Jim Hefferon committed Dec 31, 2011 11 12 13 14 Experience shows that these maps are tremendously useful. For one thing we shall see in the second subsection below that while isomorphisms describe how spaces are the same,  Jim Hefferon committed Jan 02, 2012 15 we can think of these maps as describe how spaces are alike.  Jim Hefferon committed Dec 05, 2011 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75  \subsection{Definition} \begin{definition} \label{def:Homo} A function between vector spaces $$\map{h}{V}{W}$$ that preserves\index{preserves structure}\index{structure! preservation} the operations of addition \begin{center} if $$\vec{v}_1,\vec{v}_2\in V$$ then $$h(\vec{v}_1+\vec{v}_2)=h(\vec{v}_1)+h(\vec{v}_2)$$ \end{center} and scalar multiplication \begin{center} if $$\vec{v}\in V$$ and $$r\in\Re$$ then $$h(r\cdot\vec{v})=r\cdot h(\vec{v})$$ \end{center} is a \definend{homomorphism}\index{homomorphism}% \index{function!structure preserving!see{homomorphism}}% \index{vector space!homomorphism}\index{vector space!map} or \definend{linear map}\index{linear map!see{homomorphism}}. \end{definition} \begin{example} \label{ex:RThreeHomoRTwoFirst} The projection\index{projection} map $$\map{\pi}{\Re^3}{\Re^2}$$ \begin{equation*} \colvec{x \\ y \\ z} \mapsunder{\pi} \colvec{x \\ y} \end{equation*} is a homomorphism. It preserves addition \begin{equation*} \pi(\colvec{x_1 \\ y_1 \\ z_1}\!+\!\colvec{x_2 \\ y_2 \\ z_2}) = \pi(\colvec{x_1+x_2 \\ y_1+y_2 \\ z_1+z_2}) = \colvec{x_1+x_2 \\ y_1+y_2} = \pi(\colvec{x_1 \\ y_1 \\ z_1}) + \pi(\colvec{x_2 \\ y_2 \\ z_2}) \end{equation*} and scalar multiplication. \begin{equation*} \pi(r\cdot\colvec{x_1 \\ y_1 \\ z_1}) = \pi(\colvec{rx_1 \\ ry_1 \\ rz_1}) = \colvec{rx_1 \\ ry_1} = r\cdot\pi(\colvec{x_1 \\ y_1 \\ z_1}) \end{equation*}  Jim Hefferon committed Jan 02, 2012 76 This is not an isomorphism since it is not one-to-one.  Jim Hefferon committed Jan 01, 2012 77 For instance, both $\zero$ and $\vec{e}_3$ in $\Re^3$ map to  Jim Hefferon committed Dec 05, 2011 78 79 80 81 82 the zero vector in $\Re^2$. \end{example} \begin{example} \label{exam:TwoMapsHomoNotIso} Of course, the domain and codomain  Jim Hefferon committed Dec 31, 2011 83 can be other than spaces of column vectors.  Jim Hefferon committed Dec 05, 2011 84 85 86 87 88 89 90 91 92 Both of these are homomorphisms; the verifications are straightforward. \begin{enumerate} \item $$\map{f_1}{\polyspace_2}{\polyspace_3}$$ given by \begin{equation*} a_0+a_1x+a_2x^2 \;\mapsto\; a_0x+(a_1/2)x^2+(a_2/3)x^3 \end{equation*} \item $$\map{f_2}{M_{\nbyn{2}}}{\Re}$$ given by \begin{equation*}  Jim Hefferon committed Dec 31, 2011 93  \begin{mat}  Jim Hefferon committed Dec 05, 2011 94 95  a &b \\ c &d  Jim Hefferon committed Dec 31, 2011 96  \end{mat}  Jim Hefferon committed Dec 05, 2011 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117  \mapsto a+d \end{equation*} \end{enumerate} \end{example} \begin{example} Between any two spaces there is a \definend{zero homomorphism},% \index{zero homomorphism}\index{homomorphism!zero}\index{function!zero} mapping every vector in the domain to the zero vector in the codomain. \end{example} \begin{example} These two suggest why we use the term linear map'. \begin{enumerate} \item The map $$\map{g}{\Re^3}{\Re}$$ given by \begin{equation*} \colvec{x \\ y \\ z} \mapsunder{g} 3x+2y-4.5z \end{equation*}  Jim Hefferon committed Dec 31, 2011 118  is linear, that is, is a homomorphism.  Jim Hefferon committed Dec 05, 2011 119 120 121 122 123 124  In contrast, the map $$\map{\hat{g}}{\Re^3}{\Re}$$ given by \begin{equation*} \colvec{x \\ y \\ z} \mapsunder{\hat{g}} 3x+2y-4.5z+1 \end{equation*}  Jim Hefferon committed Dec 31, 2011 125  is not.  Jim Hefferon committed Dec 05, 2011 126  \begin{equation*}  Jim Hefferon committed Dec 31, 2011 127  \hat{g}(\colvec[r]{0 \\ 0 \\ 0}+\colvec[r]{1 \\ 0 \\ 0})=4  Jim Hefferon committed Jan 02, 2012 128  \qquad  Jim Hefferon committed Dec 31, 2011 129 130  \hat{g}(\colvec[r]{0 \\ 0 \\ 0}) +\hat{g}(\colvec[r]{1 \\ 0 \\ 0})=5  Jim Hefferon committed Dec 05, 2011 131  \end{equation*}  Jim Hefferon committed Jan 02, 2012 132 133  To show that a map is not linear we need only produce a single linear combination that the map does not preserve.  Jim Hefferon committed Dec 05, 2011 134 135 136 137 138 139 140  \item The first of these two maps $$\map{t_1,t_2}{\Re^3}{\Re^2}$$ is linear while the second is not. \begin{equation*} \colvec{x \\ y \\ z} \mapsunder{t_1} \colvec{5x-2y \\ x+y}  Jim Hefferon committed Jan 01, 2012 141  \qquad  Jim Hefferon committed Dec 05, 2011 142 143 144  \colvec{x \\ y \\ z} \mapsunder{t_2} \colvec{5x-2y \\ xy}  Jim Hefferon committed Dec 31, 2011 145  \end{equation*}  Jim Hefferon committed Jan 01, 2012 146 147  Finding a linear combination that the second map does not preserve is easy.  Jim Hefferon committed Dec 05, 2011 148 \end{enumerate}  Jim Hefferon committed Dec 31, 2011 149 150 The homomorphisms have coordinate functions that are linear combinations of the arguments.  Jim Hefferon committed Jan 02, 2012 151 % See also \nearbyexercise{exer:GrpahNotALine}.  Jim Hefferon committed Dec 05, 2011 152 153 \end{example}  Jim Hefferon committed Dec 31, 2011 154 155 Any isomorphism is a homomorphism, since an isomorphism is a  Jim Hefferon committed Dec 05, 2011 156 homomorphism that is also a correspondence.  Jim Hefferon committed Dec 31, 2011 157 158 So one way to think of homomorphism' is as a generalization of isomorphism'  Jim Hefferon committed Dec 05, 2011 159 160 motivated by the observation that many of the properties of isomorphisms have only to do with the map's structure preservation property  Jim Hefferon committed Dec 31, 2011 161 and not to do with being a correspondence.  Jim Hefferon committed Jan 02, 2012 162 163 The next two results are examples of that thinking. The proof for each given in the prior section  Jim Hefferon committed Dec 31, 2011 164 does not use one-to-one-ness or onto-ness  Jim Hefferon committed Jan 01, 2012 165 and therefore applies here.  Jim Hefferon committed Dec 05, 2011 166 167 168 169 170 171  \begin{lemma} \label{le:HomoSendsZeroToZero} A homomorphism sends a zero vector to a zero vector. \end{lemma} \begin{lemma} \label{le:HomoPreserveLinCombo}  Jim Hefferon committed Jan 02, 2012 172 173 174 175 For any map $$\map{f}{V}{W}$$ between vector spaces, the following are equivalent. \begin{tfae}  Jim Hefferon committed Dec 31, 2011 176 177  \item $f$ is a homomorphism  Jim Hefferon committed Dec 05, 2011 178 179 180 181 182 183 184 185 186  \item $f(c_1\cdot\vec{v}_1+c_2\cdot\vec{v}_2) =c_1\cdot f(\vec{v}_1)+c_2\cdot f(\vec{v}_2)$ for any $$c_1,c_2\in\Re$$ and $$\vec{v}_1,\vec{v}_2\in V$$ \item $f(c_1\cdot\vec{v}_1+\dots+c_n\cdot\vec{v}_n) =c_1\cdot f(\vec{v}_1)+\dots+c_n\cdot f(\vec{v}_n)$ for any $$c_1,\dots,c_n\in\Re$$ and $$\vec{v}_1,\ldots,\vec{v}_n\in V$$  Jim Hefferon committed Jan 02, 2012 187 \end{tfae}  Jim Hefferon committed Dec 05, 2011 188 189 190 \end{lemma} \begin{example}  Jim Hefferon committed Dec 31, 2011 191 The function $$\map{f}{\Re^2}{\Re^4}$$ given by  Jim Hefferon committed Dec 05, 2011 192 193 194 195 196 \begin{equation*} \colvec{x \\ y} \mapsunder{f} \colvec{x/2 \\ 0 \\ x+y \\ 3y} \end{equation*}  Jim Hefferon committed Dec 31, 2011 197 is linear since it satisfies item~(2).  Jim Hefferon committed Dec 05, 2011 198 199 200 201 202 203 204 205 206 207 208 \begin{equation*} \colvec{r_1(x_1/2)+r_2(x_2/2) \\ 0 \\ r_1(x_1+y_1)+r_2(x_2+y_2) \\ r_1(3y_1)+r_2(3y_2)} = r_1\colvec{x_1/2 \\ 0 \\ x_1+y_1 \\ 3y_1} + r_2\colvec{x_2/2 \\ 0 \\ x_2+y_2 \\ 3y_2} \end{equation*} \end{example} However,  Jim Hefferon committed Jan 01, 2012 209 some of the things that we have seen for isomorphisms fail to hold for  Jim Hefferon committed Dec 05, 2011 210 homomorphisms in general.  Jim Hefferon committed Jan 01, 2012 211 212 213 One example is the proof of Lemma~I.\ref{lem:IsoImpliesSameDim}, which shows that an isomorphism between spaces gives a correspondence between their bases.  Jim Hefferon committed Dec 05, 2011 214 Homomorphisms do not give any such correspondence;  Jim Hefferon committed Dec 31, 2011 215 216 \nearbyexample{ex:RThreeHomoRTwoFirst} shows this and another example is the zero map between two nontrivial spaces.  Jim Hefferon committed Dec 05, 2011 217 218 219 220 Instead, for homomorphisms a weaker but still very useful result holds. \begin{theorem} \label{th:HomoDetActOnBasis}  Jim Hefferon committed Jan 01, 2012 221 A homomorphism is determined by its action on a basis:~if  Jim Hefferon committed Dec 05, 2011 222 223 $$\sequence{\vec{\beta}_1,\dots,\vec{\beta}_n}$$ is a basis of a vector space $$V$$ and  Jim Hefferon committed Dec 31, 2011 224 225 226 227 $$\vec{w}_1,\dots,\vec{w}_n$$ are elements of a vector space $$W$$ (perhaps not distinct elements) then there exists a homomorphism from $$V$$ to $$W$$ sending each $$\vec{\beta}_i$$ to $$\vec{w}_i$$, and that homomorphism is unique.  Jim Hefferon committed Dec 05, 2011 228 229 230 231 \end{theorem} \begin{proof} We will define the map by  Jim Hefferon committed Dec 31, 2011 232 associating $\vec{\beta}_i$ with $\vec{w}_i$  Jim Hefferon committed Dec 05, 2011 233 and then extending linearly to all of the domain.  Jim Hefferon committed Jan 01, 2012 234 That is, given the input $\vec{v}$, we find its coordinates with  Jim Hefferon committed Dec 31, 2011 235 236 respect to the basis $$\vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n$$ and define  Jim Hefferon committed Jan 01, 2012 237 the associated output by using the same $c_i$ coordinates  Jim Hefferon committed Dec 05, 2011 238 $h(\vec{v})=c_1\vec{w}_1+\dots+c_n\vec{w}_n$.  Jim Hefferon committed Dec 31, 2011 239 This is a well-defined function because, with respect to the basis,  Jim Hefferon committed Dec 05, 2011 240 241 242 243 244 245 246 247 248 249 250 251 252 253 the representation of each domain vector $$\vec{v}$$ is unique. This map is a homomorphism since it preserves linear combinations; where $$\vec{v_1}=c_1\vec{\beta}_1+\cdots+c_n\vec{\beta}_n$$ and $$\vec{v_2}=d_1\vec{\beta}_1+\cdots+d_n\vec{\beta}_n$$, we have this. \begin{align*} h(r_1\vec{v}_1+r_2\vec{v}_2) &=h((r_1c_1+r_2d_1)\vec{\beta}_1+\dots+(r_1c_n+r_2d_n)\vec{\beta}_n) \\ &=(r_1c_1+r_2d_1)\vec{w}_1+\dots+(r_1c_n+r_2d_n)\vec{w}_n \\ &=r_1h(\vec{v}_1)+r_2h(\vec{v}_2) \end{align*} And, this map is unique since if $$\map{\hat{h}}{V}{W}$$  Jim Hefferon committed Jan 01, 2012 254 is another homomorphism satisfying that $$\hat{h}(\vec{\beta}_i)=\vec{w}_i$$  Jim Hefferon committed Jan 02, 2012 255 for each $$i$$  Jim Hefferon committed Dec 05, 2011 256 then $$h$$ and $$\hat{h}$$ agree on all of the vectors in the domain.  Jim Hefferon committed Jan 01, 2012 257 \begin{multline*}  Jim Hefferon committed Dec 05, 2011 258  \hat{h}(\vec{v})  Jim Hefferon committed Jan 01, 2012 259 260 261 262 263  =\hat{h}(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) =c_1\hat{h}(\vec{\beta}_1)+\dots+c_n\hat{h}(\vec{\beta}_n) \\ =c_1\vec{w}_1+\dots+c_n\vec{w}_n =h(\vec{v}) \end{multline*}  Jim Hefferon committed Dec 05, 2011 264 265 266 267 Thus, $h$ and $\hat{h}$ are the same map. \end{proof} \begin{example}  Jim Hefferon committed Jan 01, 2012 268 If we specify a map $$\map{h}{\Re^2}{\Re^2}$$  Jim Hefferon committed Dec 05, 2011 269 270 that acts on the standard basis $\stdbasis_2$ in this way \begin{equation*}  Jim Hefferon committed Dec 31, 2011 271 272 273  h(\colvec[r]{1 \\ 0})=\colvec[r]{-1 \\ 1} \qquad h(\colvec[r]{0 \\ 1})=\colvec[r]{-4 \\ 4}  Jim Hefferon committed Dec 05, 2011 274 \end{equation*}  Jim Hefferon committed Dec 31, 2011 275 then we have also specified the action of $h$ on any other member of the domain.  Jim Hefferon committed Dec 05, 2011 276 277 278 For instance, the value of $h$ on this argument \begin{equation*}  Jim Hefferon committed Dec 31, 2011 279 280 281  h(\colvec[r]{3 \\ -2})=h(3\cdot \colvec[r]{1 \\ 0}-2\cdot \colvec[r]{0 \\ 1}) =3\cdot h(\colvec[r]{1 \\ 0})-2\cdot h(\colvec[r]{0 \\ 1}) =\colvec[r]{5 \\ -5}  Jim Hefferon committed Dec 05, 2011 282 283 284 285 \end{equation*} is a direct consequence of the value of $h$ on the basis vectors. \end{example}  Jim Hefferon committed Jan 01, 2012 286 287 288 289 290 291 So we can construct a homomorphism by selecting a basis for the domain and specifying where the map sends those basis vectors. The prior lemma shows that we can always extend the action on the map linearly to the entire domain. Later in this chapter we shall develop a convenient scheme for computations  Jim Hefferon committed Dec 31, 2011 292 293 like this one, using matrices.  Jim Hefferon committed Dec 05, 2011 294 295 296 297 298 299 300 301 302 303 Just as the isomorphisms of a space with itself are useful and interesting, so too are the homomorphisms of a space with itself. \begin{definition} A linear map from a space into itself $$\map{t}{V}{V}$$ is a {\em linear transformation}\index{linear transformation!see{transformation}}. \end{definition} \begin{remark} In this book we use linear transformation' only in the case where  Jim Hefferon committed Dec 31, 2011 304 305 the codomain equals the domain but it is often used instead as a synonym for homomorphism'.  Jim Hefferon committed Dec 05, 2011 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 \end{remark} \begin{example} The map on $\Re^2$ that projects all vectors down to the $x$-axis \begin{equation*} \colvec{x \\ y}\mapsto\colvec{x \\ 0} \end{equation*} is a linear transformation. \end{example} \begin{example} The derivative map $$\map{d/dx}{\polyspace_n}{\polyspace_n}$$ \begin{equation*} a_0+a_1x+\cdots+a_nx^n \mapsunder{d/dx} a_1+2a_2x+3a_3x^2+\cdots+na_nx^{n-1} \end{equation*}  Jim Hefferon committed Jan 01, 2012 323 is a linear transformation as this result from calculus shows:  Jim Hefferon committed Dec 05, 2011 324 325 326 327 $$d(c_1f+c_2g)/dx=c_1\,(df/dx)+c_2\,(dg/dx)$$. \end{example} \begin{example} \label{ex:MatTransMapLinear}  Jim Hefferon committed Dec 31, 2011 328 The matrix transpose operation  Jim Hefferon committed Dec 05, 2011 329 \begin{equation*}  Jim Hefferon committed Dec 31, 2011 330  \begin{mat}  Jim Hefferon committed Dec 05, 2011 331 332  a &b \\ c &d  Jim Hefferon committed Dec 31, 2011 333  \end{mat}  Jim Hefferon committed Dec 05, 2011 334  \;\mapsto\;  Jim Hefferon committed Dec 31, 2011 335  \begin{mat}  Jim Hefferon committed Dec 05, 2011 336 337  a &c \\ b &d  Jim Hefferon committed Dec 31, 2011 338  \end{mat}  Jim Hefferon committed Dec 05, 2011 339 340 \end{equation*} is a linear transformation of $$\matspace_{\nbyn{2}}$$.  Jim Hefferon committed Jan 01, 2012 341 (Transpose is one-to-one and onto and so in fact it is  Jim Hefferon committed Dec 31, 2011 342 an automorphism.)  Jim Hefferon committed Dec 05, 2011 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 \end{example} We finish this subsection about maps by recalling that we can linearly combine maps. For instance, for these maps from $$\Re^2$$ to itself \begin{equation*} \colvec{x \\ y} \mapsunder{f} \colvec{2x \\ 3x-2y} \quad\text{and}\quad \colvec{x \\ y} \mapsunder{g} \colvec{0 \\ 5x} \end{equation*} the linear combination $$5f-2g$$ is also a map from $R^2$ to itself. \begin{equation*} \colvec{x \\ y} \mapsunder{5f-2g} \colvec{10x \\ 5x-10y} \end{equation*} \begin{lemma} \label{le:SpLinFcns} For vector spaces $$V$$ and $$W$$, the set of linear functions from $$V$$ to $$W$$ is itself a vector space, a subspace of the space of all functions from $$V$$ to $$W$$. \end{lemma}  Jim Hefferon committed Jan 01, 2012 372 373 374 \noindent We denote the space of linear maps with $$\linmaps{V}{W}$$\index{linear maps!space of}.  Jim Hefferon committed Dec 05, 2011 375 376 377 378 379 \begin{proof} This set is non-empty because it contains the zero homomorphism. So to show that it is a subspace we need only check that it is closed under linear combinations. Let $$\map{f,g}{V}{W}$$ be linear.  Jim Hefferon committed Jan 02, 2012 380 Then the sum of the two is linear  Jim Hefferon committed Dec 05, 2011 381 382 383 384 385 386 387 388 \begin{align*} (f+g)(c_1\vec{v}_1+c_2\vec{v}_2) &=f(c_1\vec{v}_1+c_2\vec{v}_2) + g(c_1\vec{v}_1+c_2\vec{v}_2) \\ &=c_1f(\vec{v}_1)+c_2f(\vec{v}_2) +c_1g(\vec{v}_1)+c_2g(\vec{v}_2) \\ &=c_1\bigl(f+g\bigr)(\vec{v}_1)+c_2\bigl(f+g\bigr)(\vec{v}_2) \end{align*}  Jim Hefferon committed Jan 02, 2012 389 and any scalar multiple of a map is also linear.  Jim Hefferon committed Dec 05, 2011 390 391 392 393 394 395 396 397 \begin{align*} (r\cdot f)(c_1\vec{v}_1+c_2\vec{v}_2) &=r(c_1f(\vec{v}_1)+c_2f(\vec{v}_2)) \\ &=c_1(r\cdot f)(\vec{v}_1)+c_2(r\cdot f)(\vec{v}_2) \end{align*} Hence $$\linmaps{V}{W}$$ is a subspace. \end{proof}  Jim Hefferon committed Dec 31, 2011 398 399 400 We started this section by defining homomorphisms as a generalization of isomorphisms, isolating the structure preservation property.  Jim Hefferon committed Jan 01, 2012 401 402 Some of the properties of isomorphisms carried over unchanged while we adapted others.  Jim Hefferon committed Dec 31, 2011 403   Jim Hefferon committed Jan 02, 2012 404 However, if we thereby get an impression that  Jim Hefferon committed Dec 31, 2011 405 the idea of  Jim Hefferon committed Jan 01, 2012 406 407 homomorphism' is in some way secondary to that of isomorphism' then that is mistaken.  Jim Hefferon committed Dec 31, 2011 408 409 In the rest of this chapter we shall work mostly with homomorphisms. This is  Jim Hefferon committed Dec 05, 2011 410 partly because any statement made about homomorphisms is automatically true  Jim Hefferon committed Dec 31, 2011 411 412 about isomorphisms but more because, while the isomorphism concept is more natural,  Jim Hefferon committed Dec 05, 2011 413 experience shows that the homomorphism concept  Jim Hefferon committed Dec 31, 2011 414 is more fruitful and more central to further progress.  Jim Hefferon committed Dec 05, 2011 415 416 417 418 419 420  \begin{exercises} \recommended \item Decide if each $$\map{h}{\Re^3}{\Re^2}$$ is linear. \begin{exparts*} \partsitem $$h(\colvec{x \\ y \\ z})=\colvec{x \\ x+y+z}$$  Jim Hefferon committed Dec 31, 2011 421 422  \partsitem $$h(\colvec{x \\ y \\ z})=\colvec[r]{0 \\ 0}$$ \partsitem $$h(\colvec{x \\ y \\ z})=\colvec[r]{1 \\ 1}$$  Jim Hefferon committed Dec 05, 2011 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445  \partsitem $$h(\colvec{x \\ y \\ z})=\colvec{2x+y \\ 3y-4z}$$ \end{exparts*} \begin{answer} \begin{exparts} \partsitem Yes. The verification is straightforward. \begin{align*} h( c_1\cdot\colvec{x_1 \\ y_1 \\ z_1} +c_2\cdot\colvec{x_2 \\ y_2 \\ z_2} ) &=h( \colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2} ) \\ &=\colvec{c_1x_1+c_2x_2 \\ c_1x_1+c_2x_2+c_1y_1+c_2y_2+c_1z_1+c_2z_2} \\ &=c_1\cdot\colvec{x_1 \\ x_1+y_1+z_1} +c_2\cdot\colvec{x_2 \\ c_2+y_2+z_2} \\ &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2}) \end{align*} \partsitem Yes. The verification is easy. \begin{align*} h(c_1\cdot\colvec{x_1 \\ y_1 \\ z_1} +c_2\cdot\colvec{x_2 \\ y_2 \\ z_2}) &=h( \colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2} ) \\  Jim Hefferon committed Dec 31, 2011 446  &=\colvec[r]{0 \\ 0} \\  Jim Hefferon committed Dec 05, 2011 447 448 449 450 451 452  &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2}) \end{align*} \partsitem No. An example of an addition that is not respected is this. \begin{equation*}  Jim Hefferon committed Dec 31, 2011 453 454 455  h(\colvec[r]{0 \\ 0 \\ 0}+\colvec[r]{0 \\ 0 \\ 0}) =\colvec[r]{1 \\ 1} \neq h(\colvec[r]{0 \\ 0 \\ 0})+h(\colvec[r]{0 \\ 0 \\ 0})  Jim Hefferon committed Dec 05, 2011 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474  \end{equation*} \partsitem Yes. The verification is straightforward. \begin{align*} h( c_1\cdot\colvec{x_1 \\ y_1 \\ z_1} +c_2\cdot\colvec{x_2 \\ y_2 \\ z_2} ) &=h( \colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2} ) \\ &=\colvec{2(c_1x_1+c_2x_2)+(c_1y_1+c_2y_2) \\ 3(c_1y_1+c_2y_2)-4(c_1z_1+c_2z_2)} \\ &=c_1\cdot\colvec{2x_1+y_1 \\ 3y_1-4z_1} +c_2\cdot\colvec{2x_2+y_2 \\ 3y_2-4z_2} \\ &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2}) \end{align*} \end{exparts} \end{answer} \recommended \item Decide if each map $$\map{h}{\matspace_{\nbyn{2}}}{\Re}$$ is linear. \begin{exparts}  Jim Hefferon committed Dec 31, 2011 475 476 477 478  \partsitem $$h(\begin{mat} a &b \\ c &d \end{mat})=a+d$$ \partsitem $$h(\begin{mat} a &b \\ c &d \end{mat})=ad-bc$$ \partsitem $$h(\begin{mat} a &b \\ c &d \end{mat})=2a+3b+c-d$$ \partsitem $$h(\begin{mat} a &b \\ c &d \end{mat})=a^2+b^2$$  Jim Hefferon committed Dec 05, 2011 479 480  \end{exparts} \begin{answer}  Jim Hefferon committed Jan 01, 2012 481 482  For each, we must either check that the map preserves linear combinations or give an example of a linear combination that is  Jim Hefferon committed Dec 05, 2011 483 484 485 486 487  not. \begin{exparts*} \partsitem Yes. The check that it preserves combinations is routine. \begin{align*}  Jim Hefferon committed Dec 31, 2011 488  h(r_1\cdot\begin{mat}  Jim Hefferon committed Dec 05, 2011 489 490  a_1 &b_1 \\ c_1 &d_1  Jim Hefferon committed Dec 31, 2011 491 492  \end{mat} +r_2\cdot\begin{mat}  Jim Hefferon committed Dec 05, 2011 493 494  a_2 &b_2 \\ c_2 &d_2  Jim Hefferon committed Dec 31, 2011 495 496  \end{mat}) &=h(\begin{mat}  Jim Hefferon committed Dec 05, 2011 497 498  r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2  Jim Hefferon committed Dec 31, 2011 499  \end{mat}) \\  Jim Hefferon committed Dec 05, 2011 500 501  &=(r_1a_1+r_2a_2)+(r_1d_1+r_2d_2) \\ &=r_1(a_1+d_1)+r_2(a_2+d_2) \\  Jim Hefferon committed Dec 31, 2011 502  &=r_1\cdot h(\begin{mat}  Jim Hefferon committed Dec 05, 2011 503 504  a_1 &b_1 \\ c_1 &d_1  Jim Hefferon committed Dec 31, 2011 505 506  \end{mat}) +r_2\cdot h(\begin{mat}  Jim Hefferon committed Dec 05, 2011 507 508  a_2 &b_2 \\ c_2 &d_2  Jim Hefferon committed Dec 31, 2011 509  \end{mat})  Jim Hefferon committed Dec 05, 2011 510 511 512 513  \end{align*} \partsitem No. For instance, not preserved is multiplication by the scalar $2$. \begin{equation*}  Jim Hefferon committed Dec 31, 2011 514  h(2\cdot\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 515 516  1 &0 \\ 0 &1  Jim Hefferon committed Dec 31, 2011 517 518  \end{mat}) =h(\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 519 520  2 &0 \\ 0 &2  Jim Hefferon committed Dec 31, 2011 521  \end{mat})  Jim Hefferon committed Dec 05, 2011 522 523  =4 \quad\text{while}\quad  Jim Hefferon committed Dec 31, 2011 524  2\cdot h(\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 525 526  1 &0 \\ 0 &1  Jim Hefferon committed Dec 31, 2011 527  \end{mat})  Jim Hefferon committed Dec 05, 2011 528 529 530 531 532 533  =2\cdot 1=2 \end{equation*} \partsitem Yes. This is the check that it preserves combinations of two members of the domain. \begin{align*}  Jim Hefferon committed Dec 31, 2011 534 535 536  h(r_1\cdot\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} +r_2\cdot\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat}) &=h(\begin{mat}  Jim Hefferon committed Dec 05, 2011 537 538  r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2  Jim Hefferon committed Dec 31, 2011 539  \end{mat}) \\  Jim Hefferon committed Dec 05, 2011 540 541 542 543  &=2(r_1a_1+r_2a_2)+3(r_1b_1+r_2b_2) +(r_1c_1+r_2c_2)-(r_1d_1+r_2d_2) \\ &=r_1(2a_1+3b_1+c_1-d_1) +r_2(2a_2+3b_2+c_2-d_2) \\  Jim Hefferon committed Dec 31, 2011 544 545  &=r_1\cdot h(\begin{mat} a_1 &b_1 \\ c_1 &d_1 \end{mat} +r_2\cdot h(\begin{mat} a_2 &b_2 \\ c_2 &d_2 \end{mat})  Jim Hefferon committed Dec 05, 2011 546 547 548 549  \end{align*} \partsitem No. An example of a combination that is not preserved is this. \begin{equation*}  Jim Hefferon committed Dec 31, 2011 550  h(\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 551 552  1 &0 \\ 0 &0  Jim Hefferon committed Dec 31, 2011 553 554  \end{mat} +\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 555 556  1 &0 \\ 0 &0  Jim Hefferon committed Dec 31, 2011 557 558  \end{mat}) =h(\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 559 560  2 &0 \\ 0 &0  Jim Hefferon committed Dec 31, 2011 561  \end{mat})  Jim Hefferon committed Dec 05, 2011 562 563  =4 \quad\text{while}\quad  Jim Hefferon committed Dec 31, 2011 564  h(\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 565 566  1 &0 \\ 0 &0  Jim Hefferon committed Dec 31, 2011 567 568  \end{mat}) +h(\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 569 570  1 &0 \\ 0 &0  Jim Hefferon committed Dec 31, 2011 571  \end{mat})  Jim Hefferon committed Dec 05, 2011 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602  =1+1 =2 \end{equation*} \end{exparts*} \end{answer} \recommended \item Show that these two maps are homomorphisms. \begin{exparts} \partsitem $$\map{d/dx}{\polyspace_3}{\polyspace_2}$$ given by $$a_0+a_1x+a_2x^2+a_3x^3$$ maps to $$a_1+2a_2x+3a_3x^2$$ \partsitem $$\map{\int}{\polyspace_2}{\polyspace_3}$$ given by $$b_0+b_1x+b_2x^2$$ maps to $$b_0x+(b_1/2)x^2+(b_2/3)x^3$$ \end{exparts} Are these maps inverse to each other? \begin{answer} The check that each is a homomorphisms is routine. Here is the check for the differentiation map. \begin{multline*} \frac{d}{dx}(r\cdot (a_0+a_1x+a_2x^2+a_3x^3) +s\cdot (b_0+b_1x+b_2x^2+b_3x^3)) \\ \begin{aligned} &=\frac{d}{dx}((ra_0+sb_0)+(ra_1+sb_1)x+(ra_2+sb_2)x^2 +(ra_3+sb_3)x^3) \\ &=(ra_1+sb_1)+2(ra_2+sb_2)x+3(ra_3+sb_3)x^2 \\ &=r\cdot (a_1+2a_2x+3a_3x^2)+s\cdot (b_1+2b_2x+3b_3x^2) \\ &=r\cdot \frac{d}{dx}(a_0+a_1x+a_2x^2+a_3x^3) +s\cdot \frac{d}{dx} (b_0+b_1x+b_2x^2+b_3x^3) \end{aligned} \end{multline*} (An alternate proof is to simply note that this is a  Jim Hefferon committed Jan 09, 2012 603  property of differentiation that is familiar from calculus.)  Jim Hefferon committed Dec 05, 2011 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640  These two maps are not inverses as this composition does not act as the identity map on this element of the domain. \begin{equation*} 1\in\polyspace_3\;\mapsunder{d/dx}\; 0\in\polyspace_2\;\mapsunder{\int}\; 0\in\polyspace_3 \end{equation*} \end{answer} \item Is (perpendicular) projection from $$\Re^3$$ to the $$xz$$-plane a homomorphism? Projection to the $$yz$$-plane? To the $$x$$-axis? The $$y$$-axis? The $$z$$-axis? Projection to the origin? \begin{answer} Each of these projections is a homomorphism. Projection to the $xz$-plane and to the $yz$-plane are these maps. \begin{equation*} \colvec{x \\ y \\ z}\mapsto\colvec{x \\ 0 \\ z} \qquad \colvec{x \\ y \\ z}\mapsto\colvec{0 \\ y \\ z} \end{equation*} Projection to the $x$-axis, to the $y$-axis, and to the $z$-axis are these maps. \begin{equation*} \colvec{x \\ y \\ z}\mapsto\colvec{x \\ 0 \\ 0} \qquad \colvec{x \\ y \\ z}\mapsto\colvec{0 \\ y \\ 0} \qquad \colvec{x \\ y \\ z}\mapsto\colvec{0 \\ 0 \\ z} \end{equation*} And projection to the origin is this map. \begin{equation*}  Jim Hefferon committed Dec 31, 2011 641  \colvec{x \\ y \\ z}\mapsto\colvec[r]{0 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 642 643 644 645 646 647 648 649 650 651 652 653  \end{equation*} Verification that each is a homomorphism is straightforward. (The last one, of course, is the zero transformation on $\Re^3$.) \end{answer} \item Show that, while the maps from \nearbyexample{exam:TwoMapsHomoNotIso} preserve linear operations, they are not isomorphisms. \begin{answer} The first is not onto; for instance, there is no polynomial that is sent the constant polynomial $p(x)=1$. The second is not one-to-one; both of these members of the domain \begin{equation*}  Jim Hefferon committed Dec 31, 2011 654  \begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 655 656  1 &0 \\ 0 &0  Jim Hefferon committed Dec 31, 2011 657  \end{mat}  Jim Hefferon committed Dec 05, 2011 658  \quad\text{and}\quad  Jim Hefferon committed Dec 31, 2011 659  \begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 660 661  0 &0 \\ 0 &1  Jim Hefferon committed Dec 31, 2011 662  \end{mat}  Jim Hefferon committed Dec 05, 2011 663  \end{equation*}  Jim Hefferon committed Jan 01, 2012 664  map to the same member of the codomain, $1\in\Re$.  Jim Hefferon committed Dec 05, 2011 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 680 681 682 683 684 685 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 705 706 707 708 709 710 711 712 713 714 715 716 717 718 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 748 749 750 751 752 753 754 755 756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792  \end{answer} \item Is an identity map a linear transformation? \begin{answer} Yes;~in any space $$\text{id}(c\cdot \vec{v}+d\cdot \vec{w}) = c\cdot \vec{v}+d\cdot \vec{w} = c\cdot\text{id}(\vec{v})+d\cdot\text{id}(\vec{w})$$. \end{answer} \recommended \item \label{exer:GrpahNotALine} Stating that a function is linear' is different than stating that its graph is a line. \begin{exparts} \partsitem The function $$\map{f_1}{\Re}{\Re}$$ given by $$f_1(x)=2x-1$$ has a graph that is a line. Show that it is not a linear function. \partsitem The function $$\map{f_2}{\Re^2}{\Re}$$ given by \begin{equation*} \colvec{x \\ y} \mapsto x+2y \end{equation*} does not have a graph that is a line. Show that it is a linear function. \end{exparts} \begin{answer} \begin{exparts} \partsitem This map does not preserve structure since $$f(1+1)=3$$, while $$f(1)+f(1)=2$$. \partsitem The check is routine. \begin{align*} f(r_1\cdot\colvec{x_1 \\ y_1}+r_2\cdot\colvec{x_2 \\ y_2}) &=f(\colvec{r_1x_1+r_2x_2 \\ r_1y_1+r_2y_2}) \\ &=(r_1x_1+r_2x_2)+2(r_1y_1+r_2y_2) \\ &=r_1\cdot (x_1+2y_1)+r_2\cdot (x_2+2y_2) \\ &=r_1\cdot f(\colvec{x_1 \\ y_1})+r_2\cdot f(\colvec{x_2 \\ y_2}) \end{align*} \end{exparts} \end{answer} \recommended \item Part of the definition of a linear function is that it respects addition. Does a linear function respect subtraction? \begin{answer} Yes. Where $$\map{h}{V}{W}$$ is linear, $$h(\vec{u}-\vec{v}) =h(\vec{u}+(-1)\cdot\vec{v}) =h(\vec{u})+(-1)\cdot h(\vec{v}) =h(\vec{u})-h(\vec{v})$$. \end{answer} \item Assume that $$h$$ is a linear transformation of $$V$$ and that $$\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$$ is a basis of $$V$$. Prove each statement. \begin{exparts} \partsitem If $$h(\vec{\beta}_i)=\zero$$ for each basis vector then $$h$$ is the zero map. \partsitem If $$h(\vec{\beta}_i)=\vec{\beta}_i$$ for each basis vector then $$h$$ is the identity map. \partsitem If there is a scalar $$r$$ such that $$h(\vec{\beta}_i)=r\cdot\vec{\beta}_i$$ for each basis vector then $$h(\vec{v})=r\cdot\vec{v}$$ for all vectors in $V$. \end{exparts} \begin{answer} \begin{exparts} \partsitem Let $$\vec{v}\in V$$ be represented with respect to the basis as $$\vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n$$. Then $$h(\vec{v})=h(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) =c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n) =c_1\cdot\zero+\dots+c_n\cdot\zero =\zero$$. \partsitem This argument is similar to the prior one. Let $$\vec{v}\in V$$ be represented with respect to the basis as $$\vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n$$. Then $$h(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) =c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n) =c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n =\vec{v}$$. \partsitem As above, only $$c_1h(\vec{\beta}_1)+\dots+c_nh(\vec{\beta}_n) =c_1r\vec{\beta}_1+\dots+c_nr\vec{\beta}_n =r(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n) =r\vec{v}$$. \end{exparts} \end{answer} \recommended \item Consider the vector space $$\Re^+$$ where vector addition and scalar multiplication are not the ones inherited from $\Re$ but rather are these: $$a+b$$ is the product of $$a$$ and $$b$$, and $$r\cdot a$$ is the $$r$$-th power of $$a$$. (This was shown to be a vector space in an earlier exercise.) Verify that the natural logarithm map $$\map{\ln}{\Re^+}{\Re}$$ is a homomorphism between these two spaces. Is it an isomorphism? \begin{answer} That it is a homomorphism follows from the familiar rules that the logarithm of a product is the sum of the logarithms $\ln(ab)=\ln(a)+\ln(b)$ and that the logarithm of a power is the multiple of the logarithm $\ln(a^r)=r\ln(a)$. This map is an isomorphism because it has an inverse, namely, the exponential map, so it is a correspondence, and therefore it is an isomorphism. \end{answer} \recommended \item Consider this transformation of $$\Re^2$$. \begin{equation*} \colvec{x \\ y} \mapsto \colvec{x/2 \\ y/3} \end{equation*} Find the image under this map of this ellipse. \begin{equation*} \set{\colvec{x \\ y} \suchthat (x^2/4)+(y^2/9)=1} \end{equation*} \begin{answer} Where $$\hat{x}=x/2$$ and $$\hat{y}=y/3$$, the image set is \begin{equation*} \set{\colvec{\hat{x} \\ \hat{y}} \suchthat \frac{\displaystyle (2\hat{x})^2}{\displaystyle 4} +\frac{\displaystyle (3\hat{y})^2}{\displaystyle 9}=1} =\set{\colvec{\hat{x} \\ \hat{y}} \suchthat \hat{x}^2+\hat{y}^2=1} \end{equation*} the unit circle in the $$\hat{x}\hat{y}$$-plane. \end{answer} \recommended \item Imagine a rope wound around the earth's equator so that it fits snugly (suppose that the earth is a sphere).  Jim Hefferon committed Jan 01, 2012 793  How much extra rope must we add to raise the circle to a constant  Jim Hefferon committed Dec 05, 2011 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809  six feet off the ground? \begin{answer} The circumference function $r\mapsto 2\pi r$ is linear. Thus we have $2\pi\cdot (r_{\text{earth}}+6)- 2\pi\cdot (r_{\text{earth}})=12\pi$. Observe that it takes the same amount of extra rope to raise the circle from tightly wound around a basketball to six feet above that basketball as it does to raise it from tightly wound around the earth to six feet above the earth. \end{answer} \recommended \item Verify that this map $$\map{h}{\Re^3}{\Re}$$ \begin{equation*} \colvec{x \\ y \\ z}\;\mapsto\;  Jim Hefferon committed Dec 31, 2011 810  \colvec{x \\ y \\ z}\dotprod\colvec[r]{3 \\ -1 \\ -1}=3x-y-z  Jim Hefferon committed Dec 05, 2011 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928  \end{equation*} is linear. Generalize. \begin{answer} Verifying that it is linear is routine. \begin{align*} h(c_1\cdot \colvec{x_1 \\ y_1 \\ z_1} +c_2\cdot \colvec{x_2 \\ y_2 \\ z_2}) &=h(\colvec{c_1x_1+c_2x_2 \\ c_1y_1+c_2y_2 \\ c_1z_1+c_2z_2}) \\ &=3(c_1x_1+c_2x_2)-(c_1y_1+c_2y_2)-(c_1z_1+c_2z_2) \\ &=c_1\cdot (3x_1-y_1-z_1)+c_2\cdot (3x_2-y_2-z_2) \\ &=c_1\cdot h(\colvec{x_1 \\ y_1 \\ z_1}) +c_2\cdot h(\colvec{x_2 \\ y_2 \\ z_2}) \end{align*} The natural guess at a generalization is that for any fixed $$\vec{k}\in\Re^3$$ the map $$\vec{v}\mapsto\vec{v}\dotprod\vec{k}$$ is linear. This statement is true. It follows from properties of the dot product we have seen earlier: $$(\vec{v}+\vec{u})\dotprod\vec{k}=\vec{v}\dotprod\vec{k}+ \vec{u}\dotprod\vec{k}$$ and $$(r\vec{v})\dotprod\vec{k}=r(\vec{v}\dotprod\vec{k})$$. (The natural guess at a generalization of this generalization, that the map from $$\Re^n$$ to $$\Re$$ whose action consists of taking the dot product of its argument with a fixed vector $$\vec{k}\in\Re^n$$ is linear, is also true.) \end{answer} \item \label{exer:HomoRONeMultByScalar} Show that every homomorphism from $$\Re^1$$ to $$\Re^1$$ acts via multiplication by a scalar. Conclude that every nontrivial linear transformation of $$\Re^1$$ is an isomorphism. Is that true for transformations of $$\Re^2$$? $$\Re^n$$? \begin{answer} Let $$\map{h}{\Re^1}{\Re^1}$$ be linear. A linear map is determined by its action on a basis, so fix the basis $$\sequence{1}$$ for $$\Re^1$$. For any $$r\in\Re^1$$ we have that $$h(r)=h(r\cdot 1)=r\cdot h(1)$$ and so $$h$$ acts on any argument $r$ by multiplying it by the constant $$h(1)$$. If $$h(1)$$ is not zero then the map is a correspondence\Dash its inverse is division by $$h(1)$$\Dash so any nontrivial transformation of $\Re^1$ is an isomorphism. This projection map is an example that shows that not every transformation of $$\Re^n$$ acts via multiplication by a constant when $$n>1$$, including when $n=2$. \begin{equation*} \colvec{x_1 \\ x_2 \\ \vdots \\ x_n} \mapsto\colvec{x_1 \\ 0 \\ \vdots \\ 0} \end{equation*} \end{answer} \item %(This will be used in \nearbyexercise{exer:Cosets} below.) \begin{exparts} \partsitem Show that for any scalars $$a_{1,1},\dots, a_{m,n}$$ this map $$\map{h}{\Re^n}{\Re^m}$$ is a homomorphism. \begin{equation*} \colvec{x_1 \\ \vdots \\ x_n} \mapsto \colvec{a_{1,1}x_1+\dots+a_{1,n}x_n \\ \vdots \\ a_{m,1}x_1+\cdots+a_{m,n}x_n} \end{equation*} \partsitem Show that for each $i$, the $$i$$-th derivative operator $d^i/dx^i$ is a linear transformation of $$\polyspace_n$$. Conclude that for any scalars $$c_k,\ldots, c_0$$ this map is a linear transformation of that space. \begin{equation*} f\mapsto \frac{d^k}{dx^k}f+c_{k-1}\frac{d^{k-1}}{dx^{k-1}}f +\dots+ c_1\frac{d}{dx}f+c_0f \end{equation*} \end{exparts} \begin{answer} \begin{exparts} \partsitem Where $$c$$ and $$d$$ are scalars, we have this. \begin{align*} h(c\cdot \colvec{x_1 \\ \vdots \\ x_n} +d\cdot \colvec{y_1 \\ \vdots \\ y_n}) &=h(\colvec{cx_1+dy_1 \\ \vdots \\ cx_n+dy_n}) \\ &=\colvec{a_{1,1}(cx_1+dy_1)+\dots+a_{1,n}(cx_n+dy_n) \\ \vdots \\ a_{m,1}(cx_1+dy_1)+\dots+a_{m,n}(cx_n+dy_n)} \\ &=c\cdot\colvec{a_{1,1}x_1+\dots+a_{1,n}x_n \\ \vdots \\ a_{m,1}x_1+\dots+a_{m,n}x_n} +d\cdot\colvec{a_{1,1}y_1+\dots+a_{1,n}y_n \\ \vdots \\ a_{m,1}y_1+\dots+a_{m,n}y_n} \\ &=c\cdot h(\colvec{x_1 \\ \vdots \\ x_n}) +d\cdot h(\colvec{y_1 \\ \vdots \\ y_n}) \end{align*} \partsitem Each power $i$ of the derivative operator is linear because of these rules familiar from calculus. \begin{equation*} \frac{d^i}{dx^i}(\,f(x)+g(x)\,)=\frac{d^i}{dx^i}f(x) +\frac{d^i}{dx^i}g(x) \quad\text{and}\quad \frac{d^i}{dx^i}\,r\cdot f(x)=r\cdot\frac{d^i}{dx^i}f(x) \end{equation*} Thus the given map is a linear transformation of $$\polyspace_n$$ because any linear combination of linear maps is also a linear map. \end{exparts} \end{answer} \item \nearbylemma{le:SpLinFcns} shows that a sum of linear functions is linear and that a scalar multiple of a linear function is linear. Show also that a composition of linear functions is linear. \begin{answer} (This argument has already appeared, as part of the proof that isomorphism is an equivalence.) Let $\map{f}{U}{V}$ and $\map{g}{V}{W}$ be linear.  Jim Hefferon committed Jan 01, 2012 929  The composition preserves linear combinations  Jim Hefferon committed Dec 05, 2011 930 931 932 933 934 935 936 937  \begin{multline*} \composed{g}{f}(c_1\vec{u}_1+c_2\vec{u}_2) =g(\,f(c_1\vec{u}_1+c_2\vec{u}_2)\,) =g(\,c_1f(\vec{u}_1)+c_2f(\vec{u}_2)\,) \\ =c_1\cdot g(f(\vec{u}_1))+c_2\cdot g(f(\vec{u}_2)) =c_1\cdot \composed{g}{f}(\vec{u}_1) +c_2\cdot \composed{g}{f}(\vec{u}_2) \end{multline*}  Jim Hefferon committed Jan 01, 2012 938  where $\vec{u}_1,\vec{u}_2\in U$ and scalars $c_1,c_2$  Jim Hefferon committed Dec 05, 2011 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973  \end{answer} \recommended \item Where $$\map{f}{V}{W}$$ is linear, suppose that $$f(\vec{v}_1)=\vec{w}_1$$, \ldots, $$f(\vec{v}_n)=\vec{w}_n$$ for some vectors $$\vec{w}_1$$, \ldots, $$\vec{w}_n$$ from $$W$$. \begin{exparts} \partsitem If the set of $$\vec{w}\,$$'s is independent, must the set of $$\vec{v}\,$$'s also be independent? \partsitem If the set of $$\vec{v}\,$$'s is independent, must the set of $$\vec{w}\,$$'s also be independent? \partsitem If the set of $$\vec{w}\,$$'s spans $$W$$, must the set of $$\vec{v}\,$$'s span $$V$$? \partsitem If the set of $$\vec{v}\,$$'s spans $$V$$, must the set of $$\vec{w}\,$$'s span $$W$$? \end{exparts} \begin{answer} \begin{exparts} \partsitem Yes. The set of $\vec{w}\,$'s cannot be linearly independent if the set of $\vec{v}\,$'s is linearly dependent because any nontrivial relationship in the domain $$\zero_V=c_1\vec{v}_1+\dots+c_n\vec{v}_n$$ would give a nontrivial relationship in the range $$f(\zero_V)=\zero_W=f(c_1\vec{v}_1+\dots+c_n\vec{v}_n) =c_1f(\vec{v}_1)+\dots+c_nf(\vec{v}_n) =c_1\vec{w}+\dots+c_n\vec{w}_n$$. \partsitem Not necessarily. For instance, the transformation of $$\Re^2$$ given by \begin{equation*} \colvec{x \\ y} \mapsto \colvec{x+y \\ x+y} \end{equation*} sends this linearly independent set in the domain to a linearly dependent image. \begin{equation*}  Jim Hefferon committed Dec 31, 2011 974  \set{\vec{v}_1,\vec{v}_2}=\set{\colvec[r]{1 \\ 0},\colvec[r]{1 \\ 1}}  Jim Hefferon committed Dec 05, 2011 975  \;\mapsto\;  Jim Hefferon committed Dec 31, 2011 976  \set{\colvec[r]{1 \\ 1},\colvec[r]{2 \\ 2}}=\set{\vec{w}_1,\vec{w}_2}  Jim Hefferon committed Dec 05, 2011 977 978 979 980 981 982 983 984 985 986  \end{equation*} \partsitem Not necessarily. An example is the projection map $$\map{\pi}{\Re^3}{\Re^2}$$ \begin{equation*} \colvec{x \\ y \\ z}\mapsto\colvec{x \\ y} \end{equation*} and this set that does not span the domain but maps to a set that does span the codomain. \begin{equation*}  Jim Hefferon committed Dec 31, 2011 987 988  \set{\colvec[r]{1 \\ 0 \\ 0},\colvec[r]{0 \\ 1 \\ 0}} \mapsunder{\pi}\set{\colvec[r]{1 \\ 0},\colvec[r]{0 \\ 1}}  Jim Hefferon committed Dec 05, 2011 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008  \end{equation*} \partsitem Not necessarily. For instance, the injection map $\map{\iota}{\Re^2}{\Re^3}$ sends the standard basis $\stdbasis_2$ for the domain to a set that does not span the codomain. (\textit{Remark.} However, the set of $\vec{w}$'s does span the range. A proof is easy.) \end{exparts} \end{answer} \item Generalize \nearbyexample{ex:MatTransMapLinear} by proving that the matrix transpose map is linear. What is the domain and codomain? \begin{answer} Recall that the entry in row~$i$ and column~$j$ of the transpose of $M$ is the entry $m_{j,i}$ from row~$j$ and column~$i$ of $M$. Now, the check is routine. \begin{align*}  Jim Hefferon committed Dec 31, 2011 1009 1010  \trans{[r\cdot\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1011 1012  \cdots &a_{i,j} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1013 1014 1015  \end{mat} +s\cdot\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1016 1017  \cdots &b_{i,j} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1018 1019 1020  \end{mat}]} &=\trans{\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1021 1022  \cdots &ra_{i,j}+sb_{i,j} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1023 1024 1025  \end{mat}} \\ &=\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1026 1027  \cdots &ra_{j,i}+sb_{j,i} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1028 1029 1030  \end{mat} \\ &=r\cdot\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1031 1032  \cdots &a_{j,i} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1033 1034 1035  \end{mat} +s\cdot\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1036 1037  \cdots &b_{j,i} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1038 1039 1040  \end{mat} \\ &=r\cdot\trans{\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1041 1042  \cdots &a_{j,i} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1043 1044 1045  \end{mat} } +s\cdot\trans{\begin{mat} \ &\vdots \\  Jim Hefferon committed Dec 05, 2011 1046 1047  \cdots &b_{j,i} &\cdots \\ &\vdots  Jim Hefferon committed Dec 31, 2011 1048  \end{mat} }  Jim Hefferon committed Dec 05, 2011 1049 1050 1051 1052 1053 1054 1055  \end{align*} The domain is $$\matspace_{\nbym{m}{n}}$$ while the codomain is $$\matspace_{\nbym{n}{m}}$$. \end{answer} \item \begin{exparts} \partsitem Where $$\vec{u},\vec{v}\in \Re^n$$,  Jim Hefferon committed Jan 01, 2012 1056  by definition the line segment connecting them is the set  Jim Hefferon committed Dec 05, 2011 1057 1058 1059 1060 1061 1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 1091 1092 1093 1094  $$\ell=\set{t\cdot\vec{u}+(1-t)\cdot\vec{v}\suchthat t\in [0..1]}$$. Show that the image, under a homomorphism $h$, of the segment between $\vec{u}$ and $\vec{v}$ is the segment between $h(\vec{u})$ and $h(\vec{v})$. \partsitem A subset of $$\Re^n$$ is \definend{convex}\index{convex set} if, for any two points in that set, the line segment joining them lies entirely in that set. (The inside of a sphere is convex while the skin of a sphere is not.) Prove that linear maps from $$\Re^n$$ to $$\Re^m$$ preserve the property of set convexity. \end{exparts} \begin{answer} \begin{exparts} \partsitem For any homomorphism $$\map{h}{\Re^n}{\Re^m}$$ we have \begin{equation*} h(\ell) =\set{h(t\cdot\vec{u}+(1-t)\cdot\vec{v})\suchthat t\in [0..1]} =\set{t\cdot h(\vec{u})+(1-t)\cdot h(\vec{v})\suchthat t\in [0..1]} \end{equation*} which is the line segment from $h(\vec{u})$ to $h(\vec{v})$. \partsitem We must show that if a subset of the domain is convex then its image, as a subset of the range, is also convex. Suppose that $$C\subseteq \Re^n$$ is convex and consider its image $h(C)$. To show $h(C)$ is convex we must show that for any two of its members, $\vec{d}_1$ and $\vec{d}_2$, the line segment connecting them \begin{equation*} \ell=\set{t\cdot\vec{d}_1+(1-t)\cdot\vec{d}_2\suchthat t\in [0..1]} \end{equation*} is a subset of $h(C)$. Fix any member $\hat{t}\cdot\vec{d}_1+(1-\hat{t})\cdot\vec{d}_2$ of that line segment. Because the endpoints of $\ell$ are in the image of $C$, there are members of $C$ that map to them, say $h(\vec{c}_1)=\vec{d}_1$ and $h(\vec{c}_2)=\vec{d}_2$.  Jim Hefferon committed Jan 01, 2012 1095  Now, where $\hat{t}$ is the scalar that we fixed in the first  Jim Hefferon committed Dec 05, 2011 1096 1097 1098 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 1123 1124 1125 1126 1127 1128 1129 1130 1131 1132 1133 1134 1135 1136 1137 1138 1139 1140 1141 1142 1143 1144 1145 1146 1147 1148 1149 1150 1151 1152 1153 1154 1155 1156 1157 1158 1159 1160 1161 1162 1163 1164 1165 1166 1167 1168 1169 1170 1171 1172 1173 1174 1175 1176 1177 1178 1179 1180 1181 1182 1183 1184 1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 1198 1199 1200 1201 1202 1203 1204 1205 1206 1207 1208 1209 1210 1211 1212 1213 1214 1215 1216 1217 1218 1219 1220 1221 1222 1223 1224 1225 1226 1227 1228 1229 1230 1231 1232 1233 1234 1235 1236 1237 1238 1239  sentence of this paragraph, observe that $h(\hat{t}\cdot\vec{c}_1+(1-\hat{t})\cdot\vec{c}_2) =\hat{t}\cdot h(\vec{c}_1)+(1-\hat{t})\cdot h(\vec{c}_2) =\hat{t}\cdot\vec{d}_1+(1-\hat{t})\cdot\vec{d}_2$ Thus, any member of $\ell$ is a member of $h(C)$, and so $h(C)$ is convex. \end{exparts} \end{answer} \recommended \item \label{exer:HomosPresLinStruc} Let $$\map{h}{\Re^n}{\Re^m}$$ be a homomorphism. \begin{exparts} \partsitem Show that the image under $$h$$ of a line in $$\Re^n$$ is a (possibly degenerate) line in $$\Re^m$$. \partsitem What happens to a $$k$$-dimensional linear surface? \end{exparts} \begin{answer} \begin{exparts} \partsitem For $$\vec{v}_0,\vec{v}_1\in\Re^n$$, the line through $$\vec{v}_0$$ with direction $$\vec{v}_1$$ is the set $\set{\vec{v}_0+t\cdot \vec{v}_1\suchthat t\in\Re}$. The image under $h$ of that line $\set{h(\vec{v}_0+t\cdot \vec{v}_1)\suchthat t\in\Re} =\set{h(\vec{v}_0)+t\cdot h(\vec{v}_1)\suchthat t\in\Re}$ is the line through $h(\vec{v}_0)$ with direction $h(\vec{v}_1)$. If $$h(\vec{v}_1)$$ is the zero vector then this line is degenerate. \partsitem A $$k$$-dimensional linear surface in $$\Re^n$$ maps to a (possibly degenerate) $$k$$-dimensional linear surface in $$\Re^m$$. The proof is just like that the one for the line. \end{exparts} \end{answer} \item Prove that the restriction of a homomorphism to a subspace of its domain is another homomorphism. \begin{answer} Suppose that $$\map{h}{V}{W}$$ is a homomorphism and suppose that $$S$$ is a subspace of $$V$$. Consider the map $$\map{\hat{h}}{S}{W}$$ defined by $$\hat{h}(\vec{s})=h(\vec{s})$$. (The only difference between $\hat{h}$ and $h$ is the difference in domain.) Then this new map is linear: $$\hat{h}(c_1\cdot\vec{s}_1+c_2\cdot\vec{s}_2)= h(c_1\vec{s}_1+c_2\vec{s}_2)=c_1h(\vec{s}_1)+c_2h(\vec{s}_2)= c_1\cdot\hat{h}(\vec{s}_1)+c_2\cdot\hat{h}(\vec{s}_2)$$. \end{answer} \item Assume that $$\map{h}{V}{W}$$ is linear. \begin{exparts} \partsitem Show that the \definend{rangespace} of this map $$\set{h(\vec{v})\suchthat \vec{v}\in V}$$ is a subspace of the codomain $$W$$. \partsitem Show that the \definend{nullspace} of this map $$\set{\vec{v}\in V\suchthat h(\vec{v})=\zero_W}$$ is a subspace of the domain $$V$$. \partsitem Show that if $$U$$ is a subspace of the domain $$V$$ then its image $$\set{h(\vec{u})\suchthat \vec{u}\in U}$$ is a subspace of the codomain $$W$$. This generalizes the first item. \partsitem Generalize the second item. \end{exparts} \begin{answer} This will appear as a lemma in the next subsection. \begin{exparts} \partsitem The range is nonempty because $$V$$ is nonempty. To finish we need to show that it is closed under combinations. A combination of range vectors has the form, where $$\vec{v}_1,\dots,\vec{v}_n\in V$$, \begin{equation*} c_1\cdot h(\vec{v}_1)+\dots+c_n\cdot h(\vec{v}_n) = h(c_1\vec{v}_1)+\dots+h(c_n\vec{v}_n) = h(c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n), \end{equation*} which is itself in the range as $$c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n$$ is a member of domain $$V$$. Therefore the range is a subspace. \partsitem The nullspace is nonempty since it contains $\zero_V$, as $$\zero_V$$ maps to $$\zero_W$$. It is closed under linear combinations because, where $$\vec{v}_1,\dots,\vec{v}_n\in V$$ are elements of the inverse image set $$\set{\vec{v}\in V\suchthat h(\vec{v})=\zero_W}$$, for $$c_1,\ldots,c_n\in\Re$$ \begin{equation*} \zero_W=c_1\cdot h(\vec{v}_1)+\dots+c_n\cdot h(\vec{v}_n) =h(c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n) \end{equation*} and so $$c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n$$ is also in the inverse image of $$\zero_W$$. \partsitem This image of $$U$$ nonempty because $$U$$ is nonempty. For closure under combinations, where $$\vec{u}_1,\ldots,\vec{u}_n\in U$$, \begin{equation*} c_1\cdot h(\vec{u}_1)+\dots+c_n\cdot h(\vec{u}_n) = h(c_1\cdot \vec{u}_1)+\dots+h(c_n\cdot \vec{u}_n) = h(c_1\cdot \vec{u}_1+\dots+c_n\cdot \vec{u}_n) \end{equation*} which is itself in $$h(U)$$ as $$c_1\cdot \vec{u}_1+\dots+c_n\cdot \vec{u}_n$$ is in $$U$$. Thus this set is a subspace. \partsitem The natural generalization is that the inverse image of a subspace of is a subspace. Suppose that $$X$$ is a subspace of $$W$$. Note that $$\zero_W\in X$$ so the set $$\set{\vec{v}\in V \suchthat h(\vec{v})\in X}$$ is not empty. To show that this set is closed under combinations, let $$\vec{v}_1,\dots,\vec{v}_n$$ be elements of $$V$$ such that $$h(\vec{v}_1)=\vec{x}_1$$, \ldots, $$h(\vec{v}_n)=\vec{x}_n$$ and note that \begin{equation*} h(c_1\cdot \vec{v}_1+\dots+c_n\cdot \vec{v}_n) =c_1\cdot h(\vec{v}_1)+\dots+c_n\cdot h(\vec{v}_n) =c_1\cdot \vec{x}_1+\dots+c_n\cdot \vec{x}_n \end{equation*} so a linear combination of elements of $$h^{-1}(X)$$ is also in $$h^{-1}(X)$$. \end{exparts} \end{answer} \item Consider the set of isomorphisms from a vector space to itself. Is this a subspace of the space $$\linmaps{V}{V}$$ of homomorphisms from the space to itself? \begin{answer} No; the set of isomorphisms does not contain the zero map (unless the space is trivial). \end{answer} \item Does \nearbytheorem{th:HomoDetActOnBasis} need that $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$ is a basis? That is, can we still get a well-defined and unique homomorphism if we drop either the condition that the set of $\vec{\beta}$'s be linearly independent, or the condition that it span the domain? \begin{answer} If $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$ doesn't span the space then the map needn't be unique. For instance, if we try to define a map from $\Re^2$ to itself by  Jim Hefferon committed Jan 01, 2012 1240  specifying only that $\vec{e}_1$ maps to itself, then  Jim Hefferon committed Dec 05, 2011 1241 1242 1243 1244 1245 1246 1247 1248 1249 1250 1251 1252 1253 1254 1255 1256 1257 1258 1259 1260 1261 1262 1263 1264 1265 1266 1267 1268 1269 1270 1271 1272 1273 1274 1275 1276 1277 1278 1279 1280 1281 1282 1283 1284 1285 1286 1287 1288 1289 1290 1291 1292 1293 1294 1295 1296 1297  there is more than one homomorphism possible; both the identity map and the projection map onto the first component fit this condition. If we drop the condition that $\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$ is linearly independent then we risk an inconsistent specification (i.e, there could be no such map). An example is if we consider $\sequence{\vec{e}_2,\vec{e}_1,2\vec{e}_1}$, and try to define a map from $\Re^2$ to itself that sends $\vec{e}_2$ to itself, and sends both $\vec{e}_1$ and $2\vec{e}_1$ to $\vec{e}_1$. No homomorphism can satisfy these three conditions. \end{answer} \item Let $$V$$ be a vector space and assume that the maps $$\map{f_1,f_2}{V}{\Re^1}$$ are linear. \begin{exparts} \partsitem Define a map $$\map{F}{V}{\Re^2}$$ whose component functions are the given linear ones. \begin{equation*} \vec{v}\mapsto\colvec{f_1(\vec{v}) \\ f_2(\vec{v})} \end{equation*} Show that $$F$$ is linear. \partsitem Does the converse hold\Dash is any linear map from $$V$$ to $$\Re^2$$ made up of two linear component maps to $$\Re^1$$? \partsitem Generalize. \end{exparts} \begin{answer} \begin{exparts} \partsitem Briefly, the check of linearity is this. \begin{equation*} F(r_1\cdot \vec{v}_1+r_2\cdot \vec{v}_2) =\colvec{f_1(r_1\vec{v}_1+r_2\vec{v}_2) \\ f_2(r_1\vec{v}_1+r_2\vec{v}_2)} =r_1\colvec{f_1(\vec{v}_1) \\ f_2(\vec{v}_1)} +r_2\colvec{f_1(\vec{v}_2) \\ f_2(\vec{v}_2)} =r_1\cdot F(\vec{v}_1)+r_2\cdot F(\vec{v}_2) \end{equation*} \partsitem Yes. Let $$\map{\pi_1}{\Re^2}{\Re^1}$$ and $$\map{\pi_2}{\Re^2}{\Re^1}$$ be the projections \begin{equation*} \colvec{x \\ y}\mapsunder{\pi_1} x \quad\text{and}\quad \colvec{x \\ y}\mapsunder{\pi_2} y \end{equation*} onto the two axes. Now, where $$f_1(\vec{v})=\pi_1(F(\vec{v}))$$ and $$f_2(\vec{v})=\pi_2(F(\vec{v}))$$ we have the desired component functions. \begin{equation*} F(\vec{v})= \colvec{f_1(\vec{v}) \\ f_2(\vec{v})} \end{equation*} They are linear because they are the composition of linear functions,  Jim Hefferon committed Jan 09, 2012 1298  and the fact that the composition of linear functions is linear  Jim Hefferon committed Jan 01, 2012 1299  was part of the proof that isomorphism is an equivalence  Jim Hefferon committed Dec 05, 2011 1300 1301 1302 1303 1304 1305 1306 1307 1308 1309 1310 1311 1312 1313 1314 1315 1316 1317 1318 1319 1320 1321 1322 1323 1324 1325  relation (alternatively, the check that they are linear is straightforward). \partsitem In general, a map from a vector space $$V$$ to an $$\Re^n$$ is linear if and only if each of the component functions is linear. The verification is as in the prior item. \end{exparts} \end{answer} \end{exercises} \subsection{Rangespace and Nullspace} Isomorphisms and homomorphisms both preserve structure.  Jim Hefferon committed Jan 02, 2012 1326 The difference is that homomorphisms are subject to fewer restrictions  Jim Hefferon committed Jan 01, 2012 1327 because they needn't be onto and  Jim Hefferon committed Dec 05, 2011 1328 needn't be one-to-one.  Jim Hefferon committed Jan 02, 2012 1329 1330 We will examine what can happen with homomorphisms that cannot happen to isomorphisms.  Jim Hefferon committed Dec 05, 2011 1331   Jim Hefferon committed Dec 31, 2011 1332 1333 We first consider the effect of not requiring that a homomorphism be  Jim Hefferon committed Dec 05, 2011 1334 onto its codomain.  Jim Hefferon committed Dec 31, 2011 1335 1336 1337 Of course, each homomorphism is onto some set, namely its range. For example, the injection map $$\map{\iota}{\Re^2}{\Re^3}$$  Jim Hefferon committed Dec 05, 2011 1338 1339 1340 \begin{equation*} \colvec{x \\ y} \mapsto \colvec{x \\ y \\ 0} \end{equation*}  Jim Hefferon committed Jan 01, 2012 1341 1342 is a homomorphism that is not onto.  Jim Hefferon committed Dec 31, 2011 1343 But, $\iota$ is onto the $xy$-plane subset of $$\Re^3$$.  Jim Hefferon committed Dec 05, 2011 1344 1345 1346 1347 1348 1349 1350 1351 1352 1353 1354  \begin{lemma} \label{le:RangeIsSubSp} Under a homomorphism, the image of any subspace of the domain is a subspace of the codomain. In particular, the image of the entire space, the range of the homomorphism, is a subspace of the codomain. \end{lemma} \begin{proof} Let $\map{h}{V}{W}$ be linear and let $S$ be a subspace of the domain $V$.  Jim Hefferon committed Jan 01, 2012 1355 1356 1357 The image $h(S)$ is a subset of the codomain $W$, which is nonempty because $S$ is nonempty. Thus, to show that $h(S)$ is a subspace of $W$  Jim Hefferon committed Dec 05, 2011 1358 1359 1360 1361 1362 1363 1364 1365 1366 1367 1368 1369 1370 1371 1372 1373 1374 1375 1376 1377 1378 1379 1380 we need only show that it is closed under linear combinations of two vectors. If $h(\vec{s}_1)$ and $h(\vec{s}_2)$ are members of $h(S)$ then $c_1\cdot h(\vec{s}_1)+c_2\cdot h(\vec{s}_2) = h(c_1\cdot \vec{s}_1)+h(c_2\cdot \vec{s}_2) = h(c_1\cdot \vec{s}_1+c_2\cdot \vec{s}_2)$ is also a member of $h(S)$ because it is the image of $$c_1\cdot \vec{s}_1+c_2\cdot \vec{s}_2$$ from $$S$$. \end{proof} \begin{definition} The \definend{rangespace}\index{rangespace}\index{homomorphism!rangespace} of a homomorphism $$\map{h}{V}{W}$$ is \begin{equation*} \rangespace{h}=\set{h(\vec{v})\suchthat \vec{v}\in V} \end{equation*} sometimes denoted $$h(V)$$. The dimension of the rangespace is the map's \definend{rank}.\index{rank!of a homomorphism} \end{definition} \noindent  Jim Hefferon committed Jan 01, 2012 1381 1382 We shall soon see the connection between the rank of a map and the rank of a matrix.  Jim Hefferon committed Dec 05, 2011 1383 1384  \begin{example} \label{ex:DerivMapRnge}  Jim Hefferon committed Jan 01, 2012 1385 For the derivative map  Jim Hefferon committed Dec 05, 2011 1386 1387 $$\map{d/dx}{\polyspace_3}{\polyspace_3}$$ given by $$a_0+a_1x+a_2x^2+a_3x^3 \mapsto a_1+2a_2x+3a_3x^2$$  Jim Hefferon committed Jan 01, 2012 1388 the rangespace  Jim Hefferon committed Dec 05, 2011 1389 1390 $$\rangespace{d/dx}$$ is the set of quadratic polynomials $$\set{r+sx+tx^2\suchthat r,s,t\in\Re }$$.  Jim Hefferon committed Jan 01, 2012 1391 Thus, this map's rank is~$$3$$.  Jim Hefferon committed Dec 05, 2011 1392 1393 1394 1395 1396 \end{example} \begin{example} \label{ex:MatToPolyRnge} With this homomorphism $$\map{h}{M_{\nbyn{2}}}{\polyspace_3}$$ \begin{equation*}  Jim Hefferon committed Dec 31, 2011 1397  \begin{mat}  Jim Hefferon committed Dec 05, 2011 1398 1399  a &b \\ c &d  Jim Hefferon committed Dec 31, 2011 1400  \end{mat}  Jim Hefferon committed Dec 05, 2011 1401  \mapsto  Jim Hefferon committed Dec 31, 2011 1402  (a+b+2d)+cx^2+cx^3  Jim Hefferon committed Dec 05, 2011 1403 1404 1405 1406 1407 \end{equation*} an image vector in the range can have any constant term, must have an $x$ coefficient of zero, and must have the same coefficient of $x^2$ as of $x^3$. That is, the rangespace is  Jim Hefferon committed Dec 31, 2011 1408 1409 $$\rangespace{h}=\set{r+sx^2+sx^3\suchthat r,s\in\Re}$$ and so the rank is~$$2$$.  Jim Hefferon committed Dec 05, 2011 1410 1411 1412 1413 1414 1415 1416 1417 1418 1419 1420 1421 1422 1423 1424 1425 1426 1427 1428 \end{example} The prior result shows that, in passing from the definition of isomorphism to the more general definition of homomorphism, omitting the onto' requirement doesn't make an essential difference. Any homomorphism is onto its rangespace. However, omitting the one-to-one' condition does make a difference. A homomorphism may have many elements of the domain that map to one element of the codomain. Below is a bean'' sketch of a many-to-one map between sets.\appendrefs{many-to-one maps}\spacefactor=1000 % It shows three elements of the codomain that are each the image of many members of the domain. \begin{center} \includegraphics{ch3.5} % bean to bean; many to one \end{center} Recall that for any function $\map{h}{V}{W}$,  Jim Hefferon committed Jan 01, 2012 1429 the set of elements of $V$ that map to $$\vec{w}\in W$$  Jim Hefferon committed Dec 05, 2011 1430 1431 1432 is the \definend{inverse image\/}\index{inverse image}% \index{function! inverse image} $h^{-1}(\vec{w})=\set{\vec{v}\in V\suchthat h(\vec{v})=\vec{w}}$.  Jim Hefferon committed Dec 31, 2011 1433 Above, the left bean shows three inverse image sets.  Jim Hefferon committed Dec 05, 2011 1434 1435 1436 1437 1438 1439 1440 1441 1442 1443  \begin{example} Consider the projection\index{projection} $$\map{\pi}{\Re^3}{\Re^2}$$ \begin{equation*} \colvec{x \\ y \\ z} \mapsunder{\pi} \colvec{x \\ y} \end{equation*} which is a homomorphism that is many-to-one.  Jim Hefferon committed Dec 31, 2011 1444 An inverse image set is a vertical line of vectors  Jim Hefferon committed Dec 05, 2011 1445 1446 1447 1448 in the domain. \begin{center} \includegraphics{ch3.11} \end{center}  Jim Hefferon committed Dec 31, 2011 1449 1450 1451 1452 One example is this. \begin{equation*} \pi^{-1}(\colvec[r]{1 \\ 3})=\set{\colvec[r]{1 \\ 3 \\ z}\suchthat z\in\Re} \end{equation*}  Jim Hefferon committed Dec 05, 2011 1453 1454 1455 1456 1457 1458 1459 1460 1461 \end{example} \begin{example} \label{ex:RTwoHomoREasyOneMap} This homomorphism $\map{h}{\Re^2}{\Re^1}$ \begin{equation*} \colvec{x \\ y} \mapsunder{h} x+y \end{equation*}  Jim Hefferon committed Dec 31, 2011 1462 1463 is also many-to-one. For a fixed $w\in\Re^1$,  Jim Hefferon committed Dec 05, 2011 1464 1465 1466 1467 1468 1469 1470 the inverse image $h^{-1}(w)$ \begin{center} \includegraphics{ch3.12} \end{center} is the set of plane vectors whose components add to $w$. \end{example}  Jim Hefferon committed Dec 31, 2011 1471 1472 1473 1474 1475 1476 1477 1478 1479 % The above examples have only to do with the % fact that we are considering functions, % specifically, many-to-one functions. % They show the inverse images % as sets of vectors that are % related to the image vector $\vec{w}$. % But these are more than just arbitrary functions, they are % homomorphisms; what do the two preservation % conditions say about the relationships?  Jim Hefferon committed Dec 05, 2011 1480 1481  In generalizing from isomorphisms to homomorphisms by  Jim Hefferon committed Jan 01, 2012 1482 dropping the one-to-one condition,  Jim Hefferon committed Dec 31, 2011 1483 1484 1485 we lose the property that we've stated intuitively as that the domain is the same'' as the range. We lose that the domain  Jim Hefferon committed Jan 01, 2012 1486 1487 corresponds perfectly to the range. What we retain, as the examples below illustrate,  Jim Hefferon committed Dec 31, 2011 1488 is that a homomorphism describes how  Jim Hefferon committed Jan 09, 2012 1489 the domain is like'' or analogous to'' the range.  Jim Hefferon committed Dec 05, 2011 1490 1491  \begin{example} \label{ex:RThreeHomoRTwo} %\label{exPicProj}  Jim Hefferon committed Dec 31, 2011 1492 We think of $\Re^3$ as like $\Re^2$ except that vectors have an extra  Jim Hefferon committed Dec 05, 2011 1493 1494 component. That is, we think of the vector with components $x$, $y$, and~$z$  Jim Hefferon committed Dec 31, 2011 1495 as somehow like the vector with components $x$ and~$y$.  Jim Hefferon committed Dec 05, 2011 1496 1497 1498 1499 In defining the projection map $\pi$, we make precise which members of the domain we are thinking of as related to which members of the codomain.  Jim Hefferon committed Dec 31, 2011 1500 To understanding how the  Jim Hefferon committed Dec 05, 2011 1501 preservation conditions in the definition of homomorphism  Jim Hefferon committed Jan 02, 2012 1502 show that the domain elements are like the codomain elements,  Jim Hefferon committed Jan 01, 2012 1503 1504 we start by picturing $\Re^2$ as the $xy$-plane inside of $\Re^3$. (Of course, $\Re^2$ is not the  Jim Hefferon committed Jan 02, 2012 1505 $xy$~plane inside of $\Re^3$ since the $xy$~plane  Jim Hefferon committed Jan 01, 2012 1506 1507 1508 is a set of three-tall vectors with a third component of zero, but there is a natural correspondence.)  Jim Hefferon committed Dec 31, 2011 1509 Then  Jim Hefferon committed Jan 02, 2012 1510 1511 the preservation of addition property says that vectors in $$\Re^3$$ act like their shadows in the plane.  Jim Hefferon committed Dec 05, 2011 1512 1513 1514 1515 1516 1517 1518 1519 1520 1521 1522 1523 1524 \begin{center} \small \begin{tabular}{@{}c@{}c@{}c@{}c@{}c@{}} \includegraphics{ch3.1} &&\includegraphics{ch3.2} &&\includegraphics{ch3.3} \\[1.5ex] {\small $\colvec{x_1 \\ y_1 \\ z_1}$ above $\colvec{x_1 \\ y_1}$} &{\small \ plus\ } &{\small $\colvec{x_2 \\ y_2 \\ z_2}$ above $\colvec{x_2 \\ y_2}$} &{\small \ equals\ } &{\small $\colvec{x_1+y_1 \\ y_1+y_2 \\ z_1+z_2}$ above $\colvec{x_1+x_2 \\ y_1+y_2}$} \end{tabular} \end{center}  Jim Hefferon committed Jan 02, 2012 1525 1526 1527 1528 \noindent Thinking of $\pi(\vec{v})$ as the shadow'' of $\vec{v}$ in the plane gives this restatement: the sum of the shadows $\pi(\vec{v}_1)+\pi(\vec{v}_2)$ equals  Jim Hefferon committed Jan 01, 2012 1529 1530 the shadow of the sum $\pi(\vec{v}_1+\vec{v}_2)$.  Jim Hefferon committed Dec 31, 2011 1531 Preservation of scalar multiplication is similar.  Jim Hefferon committed Dec 05, 2011 1532   Jim Hefferon committed Jan 02, 2012 1533 Redrawing by showing the codomain $\Re^2$ on the right  Jim Hefferon committed Dec 31, 2011 1534 gives a picture that is uglier but is more faithful to the  Jim Hefferon committed Dec 05, 2011 1535 1536 1537 1538 bean'' sketch. \begin{center} \small \includegraphics{ch3.4} \end{center}  Jim Hefferon committed Jan 02, 2012 1539 1540 1541 1542 Again, the domain vectors that map to $\vec{w}_1$ lie in a vertical line; the picture shows one in gray. Call any member of this inverse image $\pi^{-1}(\vec{w}_1)$  Jim Hefferon committed Jan 01, 2012 1543 a $\vec{w}_1$~vector.''  Jim Hefferon committed Dec 05, 2011 1544 Similarly, there is a vertical line of $\vec{w}_2$~vectors'' and  Jim Hefferon committed Jan 01, 2012 1545 a vertical line of $\vec{w}_1+\vec{w}_2$~vectors.''  Jim Hefferon committed Jan 02, 2012 1546 Now, saying that $\pi$ is a homomorphism is recognizing that  Jim Hefferon committed Dec 05, 2011 1547 1548 1549 if $\pi(\vec{v}_1)=\vec{w}_1$ and $\pi(\vec{v}_2)=\vec{w}_2$ then $\pi(\vec{v}_1+\vec{v}_2)=\pi(\vec{v}_1)+\pi(\vec{v}_2) =\vec{w}_1+\vec{w}_2$.  Jim Hefferon committed Jan 02, 2012 1550 That is, the classes add:~any  Jim Hefferon committed Dec 31, 2011 1551 $$\vec{w}_1$$~vector plus any  Jim Hefferon committed Dec 05, 2011 1552 $$\vec{w}_2$$~vector  Jim Hefferon committed Dec 31, 2011 1553 1554 equals a $$\vec{w}_1+\vec{w}_2$$~vector. Scalar multiplication is similar.  Jim Hefferon committed Dec 05, 2011 1555   Jim Hefferon committed Dec 31, 2011 1556 So although $\Re^3$ and $\Re^2$ are not isomorphic  Jim Hefferon committed Dec 05, 2011 1557 1558 1559 1560 1561 1562 $\pi$ describes a way in which they are alike:~vectors in $\Re^3$ add as do the associated vectors in $\Re^2$\Dash vectors add as their shadows add. \end{example} \begin{example} \label{ex:RTwoHomoRHardOne}  Jim Hefferon committed Dec 31, 2011 1563 A homomorphism can express  Jim Hefferon committed Dec 05, 2011 1564 1565 an analogy between spaces that is more subtle than the prior one. For the map  Jim Hefferon committed Jan 01, 2012 1566 from \nearbyexample{ex:RTwoHomoREasyOneMap}  Jim Hefferon committed Dec 05, 2011 1567 1568 1569 1570 1571 1572 1573 1574 \begin{equation*} \colvec{x \\ y} \mapsunder{h} x+y \end{equation*} fix two numbers $w_1, w_2$ in the range $$\Re$$. A $\vec{v}_1$ that maps to $w_1$ has components that add to $w_1$,  Jim Hefferon committed Jan 01, 2012 1575 so the inverse image $h^{-1}(w_1)$ is the set of vectors  Jim Hefferon committed Dec 05, 2011 1576 with endpoint on the diagonal line $x+y=w_1$.  Jim Hefferon committed Jan 02, 2012 1577 Think of these as $w_1$ vectors.''  Jim Hefferon committed Jan 01, 2012 1578 1579 Similarly we have $w_2$ vectors'' and $w_1+w_2$ vectors.'' The addition preservation property says this.  Jim Hefferon committed Dec 05, 2011 1580 1581 1582 1583 1584 1585 1586 1587 1588 \begin{center} \small \begin{tabular}{@{}ccccc@{}} \includegraphics{ch3.6} &&\includegraphics{ch3.7} &&\includegraphics{ch3.8} \\[1.5ex] {\small a $w_1$ vector''} &{\small plus} &{\small a $w_2$ vector''} &{\small equals}  Jim Hefferon committed Dec 31, 2011 1589  &{\small a $w_1+w_2$ vector''}  Jim Hefferon committed Dec 05, 2011 1590 1591  \end{tabular} \end{center}  Jim Hefferon committed Jan 01, 2012 1592 Restated, if we add a  Jim Hefferon committed Jan 02, 2012 1593 1594 $w_1$~vector to a $w_2$~vector then $h$ maps the result to a $w_1+w_2$ vector.  Jim Hefferon committed Jan 01, 2012 1595 Briefly, the sum of the images is the image of the sum.  Jim Hefferon committed Jan 02, 2012 1596 Even more briefly, $$h(\vec{v}_1)+h(\vec{v}_2)=h(\vec{v}_1+\vec{v}_2)$$.  Jim Hefferon committed Dec 31, 2011 1597 1598 % (The preservation of scalar multiplication condition has a % similar restatement.)  Jim Hefferon committed Dec 05, 2011 1599 1600 1601 1602 1603 1604 1605 1606 1607 1608 1609 1610 1611 1612 1613 1614 1615 1616 1617 1618 1619 1620 1621 1622 1623 1624 \end{example} \begin{example} \label{ex:PicRThreeToRTwo} The inverse images can be structures other than lines. For the linear map $$\map{h}{\Re^3}{\Re^2}$$ \begin{equation*} \colvec{x \\ y \\ z} \mapsto \colvec{x \\ x} \end{equation*} the inverse image sets are planes $x=0$, $x=1$, etc., perpendicular to the $$x$$-axis. \begin{center} \small \includegraphics{ch3.9} \end{center} \end{example} We won't describe how every homomorphism that we will use is an analogy because the formal sense that we make of alike in that~\ldots'' is a homomorphism exists such that~\ldots'. Nonetheless, the idea that a homomorphism between two spaces expresses how the domain's vectors fall into classes that act like the range's vectors is a good way to view homomorphisms. Another reason that we won't treat all of the homomorphisms that  Jim Hefferon committed Dec 31, 2011 1625 1626 we see as above is that many vector spaces are hard to draw, e.g., a space of polynomials.  Jim Hefferon committed Jan 01, 2012 1627 But there is nothing wrong with leveraging those spaces  Jim Hefferon committed Dec 31, 2011 1628 that we can draw.  Jim Hefferon committed Jan 01, 2012 1629 We derive two insights from the three examples  Jim Hefferon committed Dec 05, 2011 1630 1631 1632 \ref{ex:RThreeHomoRTwo}, \ref{ex:RTwoHomoRHardOne}, and~\ref{ex:PicRThreeToRTwo}.  Jim Hefferon committed Jan 01, 2012 1633 1634 The first insight is that in all three examples the inverse image of the range's zero vector is a line or plane  Jim Hefferon committed Jan 02, 2012 1635 through the origin, a subspace of the domain.  Jim Hefferon committed Dec 05, 2011 1636 1637 1638 1639 1640 1641 1642 1643 1644 1645  \begin{lemma} \label{le:NullspIsSubSp} For any homomorphism, the inverse image of a subspace of the range is a subspace of the domain. In particular, the inverse image of the trivial subspace of the range is a subspace of the domain. \end{lemma} \begin{proof} Let $\map{h}{V}{W}$ be a homomorphism  Jim Hefferon committed Dec 31, 2011 1646 1647 1648 and let $S$ be a subspace of the rangespace of $h$. Consider the inverse image $h^{-1}(S)=\set{\vec{v}\in V\suchthat h(\vec{v})\in S}$.  Jim Hefferon committed Dec 05, 2011 1649 It is nonempty because it contains $\zero_V$, since  Jim Hefferon committed Dec 31, 2011 1650 1651 $$h(\zero_V)=\zero_W$$ and $$\zero_W$$ is an element $S$, as $S$ is a subspace.  Jim Hefferon committed Jan 02, 2012 1652 1653 To finish we show that it is closed under linear combinations. Let $$\vec{v}_1$$ and $$\vec{v}_2$$ be two elements of $h^{-1}(S)$.  Jim Hefferon committed Dec 31, 2011 1654 1655 Then $h(\vec{v}_1)$ and $h(\vec{v}_2)$ are elements of $S$.  Jim Hefferon committed Jan 02, 2012 1656 1657 1658 1659 That implies that $c_1\vec{v}_1+c_2\vec{v}_2$ is an element of the inverse image $h^{-1}(S)$ because  Jim Hefferon committed Dec 05, 2011 1660 1661 $h(c_1\vec{v}_1+c_2\vec{v}_2) =c_1h(\vec{v}_1)+c_2h(\vec{v}_2)$  Jim Hefferon committed Dec 31, 2011 1662 is a member of $S$.  Jim Hefferon committed Dec 05, 2011 1663 1664 1665 1666 1667 \end{proof} \begin{definition} The \definend{nullspace}\index{homomorphism!nullspace}\index{nullspace} or \definend{kernel}\index{kernel} of a linear map  Jim Hefferon committed Dec 31, 2011 1668 $$\map{h}{V}{W}$$ is the inverse image of $0_W$.  Jim Hefferon committed Dec 05, 2011 1669 \begin{equation*}  Jim Hefferon committed Dec 31, 2011 1670  \nullspace{h}=h^{-1}(\zero_W)=\set{\vec{v}\in V\suchthat h(\vec{v})=\zero_W}  Jim Hefferon committed Dec 05, 2011 1671 1672 1673 1674 1675 1676 1677 1678 1679 1680 1681 \end{equation*} The dimension of the nullspace is the map's {\em nullity}\index{nullity}\index{homomorphism!nullity}. \end{definition} \begin{center} \includegraphics{ch3.10} \end{center} \begin{example} The map from \nearbyexample{ex:DerivMapRnge} has this nullspace  Jim Hefferon committed Dec 31, 2011 1682 1683 $$\nullspace{d/dx}=\set{a_0+0x+0x^2+0x^3\suchthat a_0\in\Re}$$ so its nullity is $1$.  Jim Hefferon committed Dec 05, 2011 1684 1685 1686 1687 \end{example} \begin{example} The map from \nearbyexample{ex:MatToPolyRnge}  Jim Hefferon committed Jan 01, 2012 1688 has this nullspace and nullity $2$.  Jim Hefferon committed Dec 05, 2011 1689 \begin{equation*}  Jim Hefferon committed Dec 31, 2011 1690  \nullspace{h}=\set{\begin{mat}  Jim Hefferon committed Dec 05, 2011 1691 1692  a &b \\ 0 &-(a+b)/2  Jim Hefferon committed Dec 31, 2011 1693  \end{mat}\suchthat a,b\in\Re}  Jim Hefferon committed Dec 05, 2011 1694 1695 1696 1697 \end{equation*} \end{example} Now for the second insight from the above pictures.  Jim Hefferon committed Jan 01, 2012 1698 1699 In \nearbyexample{ex:RThreeHomoRTwo} each of the vertical lines squashes down `