det2.tex 41.8 KB
 Jim Hefferon committed Dec 05, 2011 1 % Chapter 4, Section 2 _Linear Algebra_ Jim Hefferon  Jim Hefferon committed Nov 11, 2013 2 % http://joshua.smcvt.edu/linearalgebra  Jim Hefferon committed Dec 05, 2011 3 4 5 % 2001-Jun-12 \section{Geometry of Determinants} The prior section develops the determinant algebraically, by  Jim Hefferon committed Nov 11, 2013 6 considering formulas satisfying certain conditions.  Jim Hefferon committed Dec 05, 2011 7 This section complements that with a geometric approach.  Jim Hefferon committed Nov 11, 2013 8 Beyond its intuitive appeal, an advantage of this approach is that while  Jim Hefferon committed Nov 11, 2013 9 we have so far only considered whether or not a determinant is zero,  Jim Hefferon committed Jan 13, 2012 10 here we shall give a meaning to the value of the determinant.  Jim Hefferon committed Nov 11, 2013 11 12 (The prior section treats the determinant as a function of the rows but this section focuses on columns.)  Jim Hefferon committed Dec 05, 2011 13 14 15 16 17 18 19 20 21 22 23 24  \subsection{Determinants as Size Functions} This parallelogram picture  Jim Hefferon committed Jan 13, 2012 25 is familiar from the construction of the sum of the two vectors.  Jim Hefferon committed Dec 05, 2011 26 27 28 \begin{center} \includegraphics{ch4.30} \end{center}  Jim Hefferon committed Jan 13, 2012 29   Jim Hefferon committed Jun 12, 2012 30 31 \begin{definition} \label{df:Box} %<*df:Box>  Jim Hefferon committed Jan 13, 2012 32 33 34 35 36 37 38 In $\Re^n$ the \definend{box}\index{box} (or \definend{parallelepiped}\index{parallelepiped}) formed by $$\sequence{\vec{v}_1,\dots,\vec{v}_n}$$ is the set $$\set{t_1\vec{v}_1+\dots+t_n\vec{v}_n  Jim Hefferon committed Nov 11, 2013 39  \suchthat t_1,\ldots,t_n\in \closedinterval{0}{1}}$$.  Jim Hefferon committed Jun 12, 2012 40 %  Jim Hefferon committed Jan 13, 2012 41 42 \end{definition}  Jim Hefferon committed Dec 23, 2013 43 \noindent Thus the parallelogram above is the box formed by  Jim Hefferon committed Jan 13, 2012 44 $\sequence{\binom{x_1}{y_1},\binom{x_2}{y_2}}$.  Jim Hefferon committed Nov 11, 2013 45 A three-space box is shown in \nearbyexample{ex:VolParPiped}.  Jim Hefferon committed Jan 13, 2012 46   Jim Hefferon committed Nov 12, 2013 47 48 We can find the area of the above box by drawing an enclosing rectangle and subtracting away areas not in the box.  Jim Hefferon committed Dec 05, 2011 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 \begin{center} \parbox{1.5in}{\hbox{}\hfil\includegraphics{ch4.31}\hfil\hbox{}} \quad \parbox{3.0in}{ \hbox{}\hfil $\begin{array}{l} \text{area of parallelogram} \\ \hbox{}\quad \hbox{} =\text{area of rectangle} -\text{area of$A$}-\text{area of$B$} \\ \hbox{}\qquad \hbox{} -\cdots-\text{area of$F$} \\ \hbox{}\quad \hbox{} =(x_1+x_2)(y_1+y_2)-x_2y_1-x_1y_1/2 \\ \hbox{}\qquad \hbox{} -x_2y_2/2-x_2y_2/2-x_1y_1/2-x_2y_1 \\ \hbox{}\quad \hbox{} =x_1y_2-x_2y_1 \end{array}$ \hfil\hbox{}} \end{center}  Jim Hefferon committed Nov 11, 2013 70 That the area equals the value of the determinant  Jim Hefferon committed Dec 05, 2011 71 \begin{equation*}  Jim Hefferon committed Jan 14, 2012 72  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 73 74  x_1 &x_2 \\ y_1 &y_2  Jim Hefferon committed Jan 14, 2012 75  \end{vmat}  Jim Hefferon committed Dec 05, 2011 76 77 78  =x_1y_2-x_2y_1 \end{equation*} is no coincidence.  Jim Hefferon committed Nov 11, 2013 79 80 81 The definition of determinants contains four properties that we know lead to a unique function for each dimension~$n$. We shall argue that these properties  Jim Hefferon committed Jan 13, 2012 82 make good postulates for a function  Jim Hefferon committed Nov 11, 2013 83 that measure the size of boxes in $n$-space.\index{size}  Jim Hefferon committed Dec 05, 2011 84   Jim Hefferon committed Jan 13, 2012 85 86 For instance, a function that measures the size of the box should have the property that multiplying one of the box-defining vectors by  Jim Hefferon committed Nov 11, 2013 87 a scalar  Jim Hefferon committed Jan 13, 2012 88 will multiply the size by that scalar.  Jim Hefferon committed Dec 05, 2011 89 90 91 92 93 \begin{center} \includegraphics{ch4.32} \qquad \includegraphics{ch4.33} \end{center}  Jim Hefferon committed Nov 11, 2013 94 95 96 Shown here is $k=1.4$. On the right the rescaled region is in solid lines with the original region shaded for comparison.  Jim Hefferon committed Jan 13, 2012 97 98 99 100 101 % The region formed by $k\vec{v}$ and~$\vec{w}$ % is bigger by a factor of $$k$$ % than the shaded region enclosed by $\vec{v}$ and~$\vec{w}$. % That is, % $$\size (k\vec{v},\vec{w})=k\cdot\size (\vec{v},\vec{w})$$ and  Jim Hefferon committed Nov 11, 2013 102 103  That is, we can reasonably expect that  Jim Hefferon committed Dec 05, 2011 104 $\size (\dots,k\vec{v},\dots)=k\cdot\size (\dots,\vec{v},\dots)$.  Jim Hefferon committed Nov 11, 2013 105 Of course, this condition is one of those in  Jim Hefferon committed Jan 13, 2012 106 the definition of determinants.  Jim Hefferon committed Dec 05, 2011 107   Jim Hefferon committed Jan 13, 2012 108 Another property of determinants that should apply to any  Jim Hefferon committed Nov 11, 2013 109 110 function measuring the size of a box is that it is unaffected by row combinations.  Jim Hefferon committed Dec 05, 2011 111 Here are before-combining and  Jim Hefferon committed Nov 12, 2013 112 after-combining boxes (the scalar shown is $k=-0.35$).  Jim Hefferon committed Dec 05, 2011 113 114 115 116 117 \begin{center} \includegraphics{ch4.34} \qquad \includegraphics{ch4.35} \end{center}  Jim Hefferon committed Nov 11, 2013 118 The box formed by $v$ and  Jim Hefferon committed Nov 12, 2013 119 120 121 $k\vec{v}+\vec{w}$ slants differently than the original one but the two have the same base and the same height, and hence the same area.  Jim Hefferon committed Nov 11, 2013 122 123 % (As before, the figure on the right has the % original region in shade for comparison.)  Jim Hefferon committed Nov 12, 2013 124 So we expect that size is not affected by a shear operation  Jim Hefferon committed Dec 05, 2011 125 $\size (\dots,\vec{v},\dots,\vec{w},\dots)  Jim Hefferon committed Nov 11, 2013 126 =\size (\dots,\vec{v},\dots,k\vec{v}+\vec{w},\dots)$.  Jim Hefferon committed Nov 12, 2013 127 Again, this is a determinant condition.  Jim Hefferon committed Dec 05, 2011 128   Jim Hefferon committed Nov 12, 2013 129 We expect that the box formed by unit vectors has unit size  Jim Hefferon committed Dec 05, 2011 130 131 132 \begin{center} \includegraphics{ch4.36} \end{center}  Jim Hefferon committed Nov 11, 2013 133 and we naturally extend that to any $n$-space  Jim Hefferon committed Jan 13, 2012 134 $\size(\vec{e}_1,\dots,\vec{e}_n)=1$.  Jim Hefferon committed Jan 14, 2012 135 % Again, that is a determinant property.  Jim Hefferon committed Dec 05, 2011 136   Jim Hefferon committed Nov 12, 2013 137 138 Condition~(2) of the definition of determinant is redundant, as remarked following the definition.  Jim Hefferon committed Nov 11, 2013 139 140 141 We know from the prior section that for each~$n$ the determinant exists and is unique so we know that these postulates for size functions are consistent and  Jim Hefferon committed Jan 14, 2012 142 that we do not need any more postulates.  Jim Hefferon committed Nov 12, 2013 143 Therefore, we are justified in  Jim Hefferon committed Nov 11, 2013 144 interpreting $$\det(\vec{v}_1,\dots,\vec{v}_n)$$ as giving the  Jim Hefferon committed Dec 05, 2011 145 size of the box formed by the vectors.  Jim Hefferon committed Jan 13, 2012 146 147 148 % (\textit{Comment.} % An even more basic approach, which also leads to the definition % below, is in \cite{Weston59}.)  Jim Hefferon committed Dec 05, 2011 149   Jim Hefferon committed Jun 12, 2012 150 \begin{remark} \label{re:PropertyTwoGivesSign}  Jim Hefferon committed Nov 12, 2013 151 Although condition~(2) is redundant it raises an important point.  Jim Hefferon committed Dec 05, 2011 152 Consider these two.  Jim Hefferon committed Jan 14, 2012 153 \begin{center} \small  Jim Hefferon committed Jan 13, 2012 154  \begin{tabular}{c@{\hspace*{8em}}c}  Jim Hefferon committed Dec 05, 2011 155  \includegraphics{ch4.37}  Jim Hefferon committed Jan 13, 2012 156  &\includegraphics{ch4.38} \\[.25ex]  Jim Hefferon committed Jan 14, 2012 157  \ $\begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 158 159  4 &1 \\ 2 &3  Jim Hefferon committed Jan 14, 2012 160 161  \end{vmat}=10$ &\ $\begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 162 163  1 &4 \\ 3 &2  Jim Hefferon committed Jan 14, 2012 164  \end{vmat}=-10$  Jim Hefferon committed Dec 05, 2011 165 166  \end{tabular} \end{center}  Jim Hefferon committed Nov 11, 2013 167 168 169 170 171 172 Swapping the columns changes the sign. On the left, starting with $\vec{u}$ and following the arc inside the angle to $\vec{v}$ (that is, going counterclockwise), we get a positive size. On the right, starting at $\vec{v}$ and going to~$\vec{u}$, and so following the clockwise arc, gives a negative size.  Jim Hefferon committed Dec 05, 2011 173 174 175 The sign returned by the size function reflects the \definend{orientation}\index{box!orientation}\index{orientation} or \definend{sense}\index{box!sense}\index{sense} of the box.  Jim Hefferon committed Jan 13, 2012 176 177 (We see the same thing if we picture the effect of scalar multiplication by a negative scalar.)  Jim Hefferon committed Nov 11, 2013 178 179 180 % Although it is both interesting and important, we don't need the idea of % orientation for the development below and so we will pass it by. % (See \nearbyexercise{exer:BasisOrient}.)  Jim Hefferon committed Dec 05, 2011 181 182 \end{remark}  Jim Hefferon committed Jun 13, 2012 183 184 \begin{definition} \label{df:Volume} %<*df:Volume>  Jim Hefferon committed Dec 05, 2011 185 186 The \definend{volume}\index{volume}\index{box!volume} of a box is the absolute value of the determinant of  Jim Hefferon committed Jan 13, 2012 187 a matrix with those vectors as columns.  Jim Hefferon committed Jun 13, 2012 188 %  Jim Hefferon committed Dec 05, 2011 189 190 \end{definition}  Jim Hefferon committed Jan 13, 2012 191 \begin{example} \label{ex:VolParPiped}  Jim Hefferon committed Jan 14, 2012 192 By the formula that takes the area of the  Jim Hefferon committed Jan 13, 2012 193 194 base times the height, the volume of this parallelepiped is $12$.  Jim Hefferon committed Jan 14, 2012 195 That agrees with the determinant.  Jim Hefferon committed Jan 13, 2012 196 \begin{center}  Jim Hefferon committed Nov 14, 2013 197 198  \parbox{2in}{\hbox{}\hfil\includegraphics{asy/ppiped.pdf}\hfil\hbox{}} % \parbox{2in}{\hbox{}\hfil\includegraphics{ch4.39}\hfil\hbox{}}  Jim Hefferon committed Jan 13, 2012 199  \quad  Jim Hefferon committed Jan 14, 2012 200  $\begin{vmat}[r]  Jim Hefferon committed Jan 13, 2012 201 202 203  2 &0 &-1\\ 0 &3 &0 \\ 2 &1 &1  Jim Hefferon committed Jan 14, 2012 204  \end{vmat}=12$  Jim Hefferon committed Jan 13, 2012 205 206 207 \end{center} We can also compute the volume as the absolute value of this determinant.  Jim Hefferon committed Dec 05, 2011 208 \begin{equation*}  Jim Hefferon committed Jan 14, 2012 209  \begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 210 211 212  0 &2 &0 \\ 3 &0 &3 \\ 1 &2 &1  Jim Hefferon committed Jan 14, 2012 213  \end{vmat}=-12  Jim Hefferon committed Dec 05, 2011 214 215 216 \end{equation*} \end{example}  Jim Hefferon committed Nov 11, 2013 217 218 219 % The next result describes some of the geometry of the linear % functions that act on % $$\Re^n$$.  Jim Hefferon committed Dec 05, 2011 220   Jim Hefferon committed Jun 13, 2012 221 \begin{theorem}\label{th:MatChVolByDetMat}  Jim Hefferon committed Dec 05, 2011 222 \index{size}\index{transformation!size change}  Jim Hefferon committed Jun 13, 2012 223 %<*th:MatChVolByDetMat>  Jim Hefferon committed Dec 05, 2011 224 A transformation $$\map{t}{\Re^n}{\Re^n}$$ changes the size of all boxes  Jim Hefferon committed Dec 09, 2013 225 by the same factor, namely, the size of the image of a box  Jim Hefferon committed Dec 05, 2011 226 227 228 $\deter{t(S)}$ is $\deter{T}$ times the size of the box $\deter{S}$, where $T$ is the matrix representing $t$ with respect to the standard basis.  Jim Hefferon committed Jan 13, 2012 229   Jim Hefferon committed Nov 11, 2013 230 That is, the determinant of a product is the  Jim Hefferon committed Dec 05, 2011 231 product of the determinants $\deter{TS}=\deter{T}\cdot\deter{S}$.  Jim Hefferon committed Jun 13, 2012 232 %  Jim Hefferon committed Dec 05, 2011 233 234 \end{theorem}  Jim Hefferon committed Jan 13, 2012 235 The two sentences say the same thing, first in map terms and then  Jim Hefferon committed Dec 05, 2011 236 in matrix terms.  Jim Hefferon committed Jan 13, 2012 237 This is because  Jim Hefferon committed Dec 05, 2011 238 $\deter{t(S)}=\deter{TS}$, as both give the size of the box that is the  Jim Hefferon committed Nov 11, 2013 239 240 image of the unit box $\stdbasis_n$ under the composition $\composed{t}{s}$, where the maps are represented with respect to the standard basis.  Jim Hefferon committed Nov 12, 2013 241 We will prove the second sentence.  Jim Hefferon committed Dec 05, 2011 242   Jim Hefferon committed Nov 12, 2013 243 \begin{proof}  Jim Hefferon committed Jun 13, 2012 244 %<*pf:MatChVolByDetMat0>  Jim Hefferon committed Nov 15, 2014 245 First consider the case that $T$ is singular and thus  Jim Hefferon committed Nov 12, 2013 246 does not have an inverse.  Jim Hefferon committed Jan 13, 2012 247 248 Observe that if $$TS$$ is invertible then there is an $M$ such that $$(TS)M=I$$, so  Jim Hefferon committed Nov 12, 2013 249 $$T(SM)=I$$, and so $$T$$ is invertible.  Jim Hefferon committed Dec 09, 2013 250 The contrapositive of that observation is that  Jim Hefferon committed Jan 13, 2012 251 252 if $$T$$ is not invertible then neither is $$TS$$ \Dash if $\deter{T}=0$ then $\deter{TS}=0$.  Jim Hefferon committed Jun 13, 2012 253 %  Jim Hefferon committed Dec 05, 2011 254   Jim Hefferon committed Jun 13, 2012 255 %<*pf:MatChVolByDetMat1>  Jim Hefferon committed Nov 12, 2013 256 257 Now consider the case that $T$ is nonsingular. Any nonsingular matrix factors into a product  Jim Hefferon committed Jan 14, 2012 258 of elementary matrices $T=E_1E_2\cdots E_r$.  Jim Hefferon committed Jan 13, 2012 259 To finish this argument  Jim Hefferon committed Jan 14, 2012 260 261 262 we will verify that $$\deter{ES}=\deter{E}\cdot\deter{S}$$ for all matrices~$S$ and elementary matrices~$E$.  Jim Hefferon committed Jan 13, 2012 263 The result will then follow because  Jim Hefferon committed Dec 05, 2011 264 265 $\deter{TS}=\deter{E_1\cdots E_rS}=\deter{E_1}\cdots\deter{E_r}\cdot\deter{S} =\deter{E_1\cdots E_r}\cdot\deter{S}=\deter{T}\cdot\deter{S}$.  Jim Hefferon committed Jun 13, 2012 266 %  Jim Hefferon committed Dec 05, 2011 267   Jim Hefferon committed Jun 13, 2012 268 %<*pf:MatChVolByDetMat2>  Jim Hefferon committed Nov 11, 2013 269 There are three types of elementary matrix.  Jim Hefferon committed Jan 14, 2012 270 271 We will cover the $M_i(k)$ case; the $P_{i,j}$ and $C_{i,j}(k)$ checks are similar.  Jim Hefferon committed Nov 11, 2013 272 273 The matrix $M_i(k)S$ equals $S$ except that row~$i$ is multiplied by $k$. The third condition of determinant functions  Jim Hefferon committed Dec 05, 2011 274 then gives that $\deter{M_i(k)S}=k\cdot\deter{S}$.  Jim Hefferon committed Nov 11, 2013 275 But $\deter{M_i(k)}=k$, again by the third condition because  Jim Hefferon committed Dec 05, 2011 276 $M_i(k)$ is derived from the identity by multiplication of row~$i$ by  Jim Hefferon committed Jan 14, 2012 277 278 $k$. Thus $$\deter{ES}=\deter{E}\cdot\deter{S}$$ holds for  Jim Hefferon committed Dec 05, 2011 279 $E=M_i(k)$.  Jim Hefferon committed Jun 13, 2012 280 %  Jim Hefferon committed Dec 05, 2011 281 282 283 284 285 286 287 \end{proof} \begin{example} Application of the map $t$ represented with respect to the standard bases by \begin{equation*}  Jim Hefferon committed Jan 13, 2012 288  \begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 289 290  1 &1 \\ -2 &0  Jim Hefferon committed Jan 13, 2012 291  \end{mat}  Jim Hefferon committed Dec 05, 2011 292 293 294 295 296 \end{equation*} will double sizes of boxes, e.g., from this \begin{center} \parbox{1.5in}{\hbox{}\hfil\includegraphics{ch4.40}\hfil\hbox{}} \quad  Jim Hefferon committed Jan 14, 2012 297  $\begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 298 299  2 &1 \\ 1 &2  Jim Hefferon committed Jan 14, 2012 300  \end{vmat}=3$  Jim Hefferon committed Dec 05, 2011 301 302 303 304 305 \end{center} to this \begin{center} \parbox{1.5in}{\hbox{}\hfil\includegraphics{ch4.41}\hfil\hbox{}} \quad  Jim Hefferon committed Jan 14, 2012 306  $\begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 307 308  3 &3 \\ -4 &-2  Jim Hefferon committed Jan 14, 2012 309  \end{vmat}=6$  Jim Hefferon committed Dec 05, 2011 310 311 312 313 \end{center} \end{example}  Jim Hefferon committed Nov 12, 2013 314 315 316 317 318 % Recall that determinants are not additive homomorphisms, that % $\det(A+B)$ need not equal $\det(A)+\det(B)$. % In contrast, the above theorem says that determinants are % multiplicative homomorphisms: % $\det(AB)$ equals $\det(A)\cdot \det(B)$.  Jim Hefferon committed Sep 21, 2013 319   Jim Hefferon committed Jun 13, 2012 320 321 \begin{corollary} \label{co:DeterminantOfInverseIsInverseOfDeterminant} %<*co:DeterminantOfInverseIsInverseOfDeterminant>  Jim Hefferon committed Dec 05, 2011 322 323 If a matrix is invertible then the determinant of its inverse is the inverse of its determinant $\deter{T^{-1}}=1/\deter{T}$.  Jim Hefferon committed Jun 13, 2012 324 %  Jim Hefferon committed Dec 05, 2011 325 326 327 \end{corollary} \begin{proof}  Jim Hefferon committed Jun 13, 2012 328 %<*pf:DeterminantOfInverseIsInverseOfDeterminant>  Jim Hefferon committed Dec 05, 2011 329 $1=\deter{I}=\deter{TT^{-1}}=\deter{T}\cdot\deter{T^{-1}}$  Jim Hefferon committed Jun 13, 2012 330 %  Jim Hefferon committed Dec 05, 2011 331 332 333 334 335 \end{proof} \begin{exercises} \item  Jim Hefferon committed Jan 14, 2012 336  Find the volume of the region defined by the vectors.  Jim Hefferon committed Dec 05, 2011 337  \begin{exparts}  Jim Hefferon committed Jan 13, 2012 338 339 340 341 342 343 344  \partsitem $\sequence{\colvec[r]{1 \\ 3},\colvec[r]{-1 \\ 4}}$ \partsitem $\sequence{\colvec[r]{2 \\ 1 \\ 0},\colvec[r]{3 \\ -2 \\ 4}, \colvec[r]{8 \\ -3 \\ 8}}$ \partsitem $\sequence{\colvec[r]{1 \\ 2 \\ 0 \\ 1}, \colvec[r]{2 \\ 2 \\ 2 \\ 2}, \colvec[r]{-1 \\ 3 \\ 0 \\ 5}, \colvec[r]{0 \\ 1 \\ 0 \\ 7}}$  Jim Hefferon committed Dec 05, 2011 345 346 347 348 349 350 351 352 353 354 355 356  \end{exparts} \begin{answer} For each, find the determinant and take the absolute value. \begin{exparts*} \partsitem $7$ \partsitem $0$ \partsitem $58$ \end{exparts*} \end{answer} \recommended \item Is \begin{equation*}  Jim Hefferon committed Jan 13, 2012 357  \colvec[r]{4 \\ 1 \\ 2}  Jim Hefferon committed Dec 05, 2011 358 359 360  \end{equation*} inside of the box formed by these three? \begin{equation*}  Jim Hefferon committed Jan 13, 2012 361  \colvec[r]{3 \\ 3 \\ 1}  Jim Hefferon committed Dec 05, 2011 362  \quad  Jim Hefferon committed Jan 13, 2012 363  \colvec[r]{2 \\ 6 \\ 1}  Jim Hefferon committed Dec 05, 2011 364  \quad  Jim Hefferon committed Jan 13, 2012 365  \colvec[r]{1 \\ 0 \\ 5}  Jim Hefferon committed Dec 05, 2011 366 367 368 369  \end{equation*} \begin{answer} Solving \begin{equation*}  Jim Hefferon committed Jan 13, 2012 370 371 372 373  c_1\colvec[r]{3 \\ 3 \\ 1} +c_2\colvec[r]{2 \\ 6 \\ 1} +c_3\colvec[r]{1 \\ 0 \\ 5} =\colvec[r]{4 \\ 1 \\ 2}  Jim Hefferon committed Dec 05, 2011 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388  \end{equation*} gives the unique solution $$c_3=11/57$$, $$c_2=-40/57$$ and $$c_1=99/57$$. Because $$c_1>1$$, the vector is not in the box. \end{answer} \recommended \item Find the volume of this region. \begin{center} \includegraphics{ch4.42} \end{center} \begin{answer} Move the parallelepiped to start at the origin, so that it becomes the box formed by \begin{equation*} \sequence{  Jim Hefferon committed Jan 13, 2012 389 390  \colvec[r]{3 \\ 0}, \colvec[r]{2 \\ 1}  Jim Hefferon committed Dec 05, 2011 391 392 393 394 395  } \end{equation*} and now the absolute value of this determinant is easily computed as $3$. \begin{equation*}  Jim Hefferon committed Jan 14, 2012 396  \begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 397 398  3 &2 \\ 0 &1  Jim Hefferon committed Jan 14, 2012 399  \end{vmat}=3  Jim Hefferon committed Dec 05, 2011 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416  \end{equation*} \end{answer} \recommended \item Suppose that $$\deter{A}=3$$. By what factor do these change volumes? \begin{exparts*} \partsitem $$A$$ \partsitem $$A^2$$ \partsitem $$A^{-2}$$ \end{exparts*} \begin{answer} \begin{exparts*} \partsitem $$3$$ \partsitem $$9$$ \partsitem $1/9$ \end{exparts*} \end{answer}  Jim Hefferon committed Dec 11, 2014 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498  \recommended \item Consider the linear transformation of~$\Re^3$ represented with respect to the standard bases by this matrix. \begin{equation*} \begin{mat} 1 &0 &-1 \\ 3 &1 &1 \\ -1 &0 &3 \end{mat} \end{equation*} \begin{exparts} \partsitem Compute the determinant of the matrix. Does the transformation preserve orientation or reverse it? \partsitem Find the size of the box defined by these vectors. What is its orientation? \begin{equation*} \colvec{1 \\ -1 \\ 2} \quad \colvec{2 \\ 0 \\ -1} \quad \colvec{1 \\ 1 \\ 0} \end{equation*} \partsitem Find the images under $t$ of the vectors in the prior item and find the size of the box that they define. What is the orientation? \end{exparts} \begin{answer} \begin{exparts} \partsitem Gauss's Method \begin{equation*} \grstep[\rho_1+\rho_3]{-3\rho_1+\rho_2} \begin{mat} 1 &0 &-1 \\ 0 &1 &4 \\ 0 &0 &2 \end{mat} \end{equation*} gives the determinant as~$+2$. The sign is positive so the transformation preserves orientation. \partsitem The size of the box is the value of this determinant. \begin{equation*} \begin{vmat} 1 &2 &1 \\ -1 &0 &1 \\ 2 &-1 &0 \end{vmat} =+6 \end{equation*} The orientation is positive. \partsitem Since this transformation is represented by the given matrix with respect to the standard bases, and with respect to the standard basis the vectors represent themselves, to find the image of the vectors under the transformation just multiply them, from the left, by the matrix. \begin{equation*} \colvec{1 \\ -1 \\ 2}\mapsto\colvec{-1 \\ 4 \\ 5} \qquad \colvec{2 \\ 0 \\ -1}\mapsto\colvec{3 \\ 5 \\ -5} \qquad \colvec{1 \\ 1 \\ 0}\mapsto\colvec{1 \\ 4 \\ -1} \end{equation*} Then compute the size of the resulting box. \begin{equation*} \begin{vmat} -1 &3 &1 \\ 4 &5 &4 \\ 5 &-5 &-1 \end{vmat} =+12 \end{equation*} The starting box is positively oriented, the transformation preserves orientations (since the determinant of the matrix is positive), and the ending box is also positively oriented. \end{exparts} \end{answer} \item  Jim Hefferon committed Dec 05, 2011 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519  By what factor does each transformation change the size of boxes? \begin{exparts*} \partsitem $\colvec{x \\ y}\mapsto\colvec{2x \\ 3y}$ \partsitem $\colvec{x \\ y}\mapsto\colvec{3x-y \\ -2x+y}$ \partsitem $\colvec{x \\ y \\ z}\mapsto\colvec{x-y \\ x+y+z \\ y-2z}$ \end{exparts*} \begin{answer} Express each transformation with respect to the standard bases and find the determinant. \begin{exparts*} \partsitem $6$ \partsitem $-1$ \partsitem $-5$ \end{exparts*} \end{answer} \item What is the area of the image of the rectangle $$[2..4]\times [2..5]$$ under the action of this matrix? \begin{equation*}  Jim Hefferon committed Jan 13, 2012 520  \begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 521 522  2 &3 \\ 4 &-1  Jim Hefferon committed Jan 13, 2012 523  \end{mat}  Jim Hefferon committed Dec 05, 2011 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538  \end{equation*} \begin{answer} The starting area is $$6$$ and the matrix changes sizes by $$-14$$. Thus the area of the image is $$84$$. \end{answer} \item If $$\map{t}{\Re^3}{\Re^3}$$ changes volumes by a factor of $$7$$ and $$\map{s}{\Re^3}{\Re^3}$$ changes volumes by a factor of $$3/2$$ then by what factor will their composition changes volumes? \begin{answer} By a factor of $$21/2$$. \end{answer} \item In what way does the definition of a box differ from the  Jim Hefferon committed Jan 13, 2012 539  definition of a span?  Jim Hefferon committed Dec 05, 2011 540 541  \begin{answer} For a box we take a sequence of vectors (as described  Jim Hefferon committed Jan 13, 2012 542  in the remark, the order of the vectors matters),  Jim Hefferon committed Dec 05, 2011 543 544 545 546  while for a span we take a set of vectors. Also, for a box subset of $\Re^n$ there must be $n$ vectors; of course for a span there can be any number of vectors. Finally, for a box the coefficients $t_1$,~\ldots, $t_n$  Jim Hefferon committed Jan 13, 2012 547  are in the interval $[0..1]$, while for a  Jim Hefferon committed Dec 05, 2011 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566  span the coefficients are free to range over all of $\Re$. \end{answer} \recommended \item Why doesn't this picture contradict \nearbytheorem{th:MatChVolByDetMat}? \begin{center} \begin{tabular}{ccc} \includegraphics{ch4.43} &\raisebox{12pt}{$$\grstep{\bigl(\begin{smallmatrix} 2 &1 \\ 0 &1 \end{smallmatrix}\bigr)}$$} &\includegraphics{ch4.44} \\ area is $2$ &determinant is $2$ &area is $5$ \end{tabular} \end{center} \begin{answer}  Jim Hefferon committed Jan 13, 2012 567  We have drawn that picture to mislead.  Jim Hefferon committed Dec 05, 2011 568 569 570 571 572  The picture on the left is not the box formed by two vectors. If we slide it to the origin then it becomes the box formed by this sequence. \begin{equation*} \sequence{  Jim Hefferon committed Jan 13, 2012 573 574  \colvec[r]{0 \\ 1}, \colvec[r]{2 \\ 0}  Jim Hefferon committed Dec 05, 2011 575 576 577 578 579 580  } \end{equation*} Then the image under the action of the matrix is the box formed by this sequence. \begin{equation*} \sequence{  Jim Hefferon committed Jan 13, 2012 581 582  \colvec[r]{1 \\ 1}, \colvec[r]{4 \\ 0}  Jim Hefferon committed Dec 05, 2011 583 584 585 586 587 588 589 590 591 592 593 594  } \end{equation*} which has an area of $4$. \end{answer} \recommended \item Does $$\deter{TS}=\deter{ST}$$? $$\deter{T(SP)}=\deter{(TS)P}$$? \begin{answer} Yes to both. For instance, the first is $$\deter{TS}=\deter{T}\cdot\deter{S}= \deter{S}\cdot\deter{T}=\deter{ST}$$. \end{answer}  Jim Hefferon committed Aug 15, 2014 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611  \item Show that there are no $\nbyn{2}$ matrices $A$ and~$B$ satisfying these. % http://math.stackexchange.com/questions/827262/need-help-with-a-linear-algebra-proof \begin{equation*} AB=\begin{mat}[r] 1 &-1 \\ 2 &0 \end{mat} \quad BA=\begin{mat}[r] 2 &1 \\ 1 &1 \end{mat} \end{equation*} \begin{answer} % due to math.stackexchange.com user dgrasines517 Because $\deter{AB}=\deter{A}\cdot\deter{B}=\deter{BA}$ and these two matrices have different determinants. \end{answer}  Jim Hefferon committed Dec 05, 2011 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631  \item \begin{exparts} \partsitem Suppose that $$\deter{A}=3$$ and that $$\deter{B}=2$$. Find $$\deter{A^2\cdot \trans{B}\cdot B^{-2}\cdot \trans{A} }$$. \partsitem Assume that $$\deter{A}=0$$. Prove that $$\deter{6A^3+5A^2+2A}=0$$. \end{exparts} \begin{answer} \begin{exparts} \partsitem If it is defined then it is $$(3^2)\cdot (2)\cdot (2^{-2})\cdot (3)$$. \partsitem $$\deter{6A^3+5A^2+2A}=\deter{A}\cdot\deter{6A^2+5A+2I}$$. \end{exparts} \end{answer} \recommended \item Let $$T$$ be the matrix representing (with respect to the standard bases) the map that rotates plane vectors counterclockwise thru $$\theta$$ radians. By what factor does $$T$$ change sizes? \begin{answer}  Jim Hefferon committed Jan 14, 2012 632  $$\begin{vmat}  Jim Hefferon committed Dec 05, 2011 633 634  \cos\theta &-\sin\theta \\ \sin\theta &\cos\theta  Jim Hefferon committed Jan 14, 2012 635  \end{vmat}=1$$  Jim Hefferon committed Dec 05, 2011 636 637 638 639 640 641 642  \end{answer} \recommended \item Must a transformation $$\map{t}{\Re^2}{\Re^2}$$ that preserves areas also preserve lengths? \begin{answer} No, for instance the determinant of \begin{equation*}  Jim Hefferon committed Jan 13, 2012 643  T=\begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 644 645  2 &0 \\ 0 &1/2  Jim Hefferon committed Jan 13, 2012 646  \end{mat}  Jim Hefferon committed Dec 05, 2011 647 648 649 650 651 652 653 654 655 656 657 658 659 660 661 662 663 664 665  \end{equation*} is $$1$$ so it preserves areas, but the vector $$T\vec{e}_1$$ has length $$2$$. \end{answer} \recommended \item What is the volume of a parallelepiped in $$\Re^3$$ bounded by a linearly dependent set? \begin{answer} It is zero. \end{answer} \recommended \item Find the area of the triangle in $$\Re^3$$ with endpoints $$(1,2,1)$$, $$(3,-1,4)$$, and $$(2,2,2)$$. (Area, not volume. The triangle defines a plane\Dash what is the area of the triangle in that plane?) \begin{answer} Two of the three sides of the triangle are formed by these vectors. \begin{equation*}  Jim Hefferon committed Jan 13, 2012 666  \colvec[r]{2 \\ 2 \\ 2}-\colvec[r]{1 \\ 2 \\ 1}=\colvec[r]{1 \\ 0 \\ 1}  Jim Hefferon committed Dec 05, 2011 667  \qquad  Jim Hefferon committed Jan 13, 2012 668  \colvec[r]{3 \\ -1 \\ 4}-\colvec[r]{1 \\ 2 \\ 1}=\colvec[r]{2 \\ -3 \\ 3}  Jim Hefferon committed Dec 05, 2011 669 670 671 672 673  \end{equation*} One way to find the area of this triangle is to produce a length-one vector orthogonal to these two. From these two relations \begin{equation*}  Jim Hefferon committed Jan 13, 2012 674  \colvec[r]{1 \\ 0 \\ 1}  Jim Hefferon committed Dec 05, 2011 675  \cdot\colvec{x \\ y \\ z}  Jim Hefferon committed Jan 13, 2012 676  =\colvec[r]{0 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 677  \qquad  Jim Hefferon committed Jan 13, 2012 678  \colvec[r]{2 \\ -3 \\ 3}  Jim Hefferon committed Dec 05, 2011 679  \cdot\colvec{x \\ y \\ z}  Jim Hefferon committed Jan 13, 2012 680  =\colvec[r]{0 \\ 0 \\ 0}  Jim Hefferon committed Dec 05, 2011 681 682 683 684 685 686 687  \end{equation*} we get a system \begin{equation*} \begin{linsys}{3} x & & &+ &z &= &0 \\ 2x &- &3y &+ &3z &= &0 \end{linsys}  Jim Hefferon committed Dec 23, 2013 688  \grstep{-2\rho_1+\rho_2}  Jim Hefferon committed Dec 05, 2011 689 690 691 692 693 694 695  \begin{linsys}{3} x & & &+ &z &= &0 \\ & &-3y&+ &z &= &0 \end{linsys} \end{equation*} with this solution set. \begin{equation*}  Jim Hefferon committed Jan 13, 2012 696  \set{\colvec[r]{-1 \\ 1/3 \\ 1}z\suchthat z\in\Re},  Jim Hefferon committed Dec 05, 2011 697 698 699  \end{equation*} A solution of length one is this. \begin{equation*}  Jim Hefferon committed Jan 13, 2012 700  \frac{1}{\sqrt{19/9}}\colvec[r]{-1 \\ 1/3 \\ 1}  Jim Hefferon committed Dec 05, 2011 701 702 703 704  \end{equation*} Thus the area of the triangle is the absolute value of this determinant. \begin{equation*}  Jim Hefferon committed Jan 14, 2012 705  \begin{vmat}[r]  Jim Hefferon committed Dec 05, 2011 706 707 708  1 &2 &-3/\sqrt{19} \\ 0 &-3 &1/\sqrt{19} \\ 1 &3 &3/\sqrt{19}  Jim Hefferon committed Jan 14, 2012 709  \end{vmat}  Jim Hefferon committed Dec 05, 2011 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  =-12/\sqrt{19} \end{equation*} \end{answer} \recommended \item \label{exer:DetProdEqProdDetsFcn} An alternate proof of \nearbytheorem{th:MatChVolByDetMat} uses the definition of determinant functions. \begin{exparts} \partsitem Note that the vectors forming $S$ make a linearly dependent set if and only if $\deter{S}=0$, and check that the result holds in this case. \partsitem For the $\deter{S}\neq 0$ case, to show that $\deter{TS}/\deter{S}=\deter{T}$ for all transformations, consider the function $$\map{d}{\matspace_{\nbyn{n}}}{\Re}$$ given by $$T\mapsto \deter{TS}/\deter{S}$$. Show that $d$ has the first property of a determinant. \partsitem Show that $d$ has the remaining three properties of a determinant function. \partsitem Conclude that $\deter{TS}=\deter{T}\cdot\deter{S}$. \end{exparts} \begin{answer} \begin{exparts} \partsitem Because the image of a linearly dependent set is linearly dependent, if the vectors forming $S$ make a linearly dependent set, so that $\deter{S}=0$, then the vectors forming $t(S)$ make a linearly dependent set, so that $\deter{TS}=0$, and in this case the equation holds. \partsitem We must check that if $T\smash[b]{\grstep{k\rho_i+\rho_j}}\hat{T}$ then $d(T)=\deter{TS}/\deter{S}=\deter{\hat{T}S}/\deter{S}=d(\hat{T})$. We can do this by checking that combining rows first and then multiplying to get $$\hat{T}S$$ gives the same result as multiplying first to get $$TS$$ and then combining (because the determinant $$\deter{TS}$$ is unaffected by the combining rows so we'll then have that $$\deter{\hat{T}S}=\deter{TS}$$ and hence that $$d(\hat{T})=d(T)$$). This check runs:~after adding $$k$$ times row~$$i$$ of $$TS$$ to row~$j$ of $$TS$$, the $$j,p$$ entry is $$(kt_{i,1}+t_{j,1})s_{1,p}+\dots+(kt_{i,r}+t_{j,r})s_{r,p}$$, which is the $$j,p$$ entry of $$\hat{T}S$$. \partsitem For the second property, we need only check that swapping $T\smash[b]{\grstep{\rho_i\swap\rho_j}}\hat{T}$ and then multiplying to get $$\hat{T}S$$ gives the same result as multiplying $$T$$ by $$S$$ first and then swapping (because, as the determinant $$\deter{TS}$$ changes sign on the row swap, we'll then have $$\deter{\hat{T}S}=-\deter{TS}$$, and so $$d(\hat{T})=-d(T)$$).  Jim Hefferon committed Jan 13, 2012 761  This check runs just like the one for the first property.  Jim Hefferon committed Dec 05, 2011 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805  For the third property, we need only show that performing $T\smash[b]{\grstep{k\rho_i}}\hat{T}$ and then computing $$\hat{T}S$$ gives the same result as first computing $$TS$$ and then performing the scalar multiplication (as the determinant $$\deter{TS}$$ is rescaled by $$k$$, we'll have $$\deter{\hat{T}S}=k\deter{TS}$$ and so $$d(\hat{T})=k\,d(T)$$). Here too, the argument runs just as above. The fourth property, that if $T$ is $I$ then the result is $1$, is obvious. \partsitem Determinant functions are unique, so $$\deter{TS}/\deter{S}=d(T)=\deter{T}$$, and so $\deter{TS}=\deter{T}\deter{S}$. \end{exparts} \end{answer} % \item % Use the fact that % $$\deter{TS}=\deter{T}\,\deter{S}$$ % to prove that if $$\phi$$ and $$\sigma$$ are $$n$$-permutations % then $$\sgn(\phi\sigma)=\sgn(\phi)\sgn(\sigma)$$. % \cite{HoffmanKunze} % \begin{answer} % Take $$T=P_\phi$$ and $$S=P_\sigma$$. % Note that $$TS=P_{\phi\sigma}$$ and so % $$\deter{P_{\phi\sigma}}=\deter{P_\phi}\cdot\deter{P_\sigma}$$. % \end{answer} \item Give a non-identity matrix with the property that $$\trans{A}=A^{-1}$$. Show that if $$\trans{A}=A^{-1}$$ then $$\deter{A}=\pm 1$$. Does the converse hold? \begin{answer} Any permutation matrix has the property that the transpose of the matrix is its inverse. For the implication, we know that $$\deter{\trans{A}}=\deter{A}$$. Then $$1=\deter{A\cdot A^{-1}}=\deter{A\cdot\trans{A}} =\deter{A}\cdot\deter{\trans{A}}=\deter{A}^2$$. The converse does not hold; here is an example. \begin{equation*}  Jim Hefferon committed Jan 13, 2012 806  \begin{mat}[r]  Jim Hefferon committed Dec 05, 2011 807 808  3 &1 \\ 2 &1  Jim Hefferon committed Jan 13, 2012 809  \end{mat}  Jim Hefferon committed Dec 05, 2011 810 811 812 813 814 815 816 817 818 819 820 821 822  \end{equation*} \end{answer} \item The algebraic property of determinants that factoring a scalar out of a single row will multiply the determinant by that scalar shows that where $$H$$ is $$\nbyn{3}$$, the determinant of $$cH$$ is $$c^3$$ times the determinant of $$H$$. Explain this geometrically, that is, using \nearbytheorem{th:MatChVolByDetMat}. (The observation that increasing the linear size of a three-dimensional object by a factor of $c$ will increase its volume by a factor of  Jim Hefferon committed Mar 10, 2012 823  $c^3$ while only increasing its surface area by an amount proportional  Jim Hefferon committed Dec 05, 2011 824  to a factor of  Jim Hefferon committed Mar 10, 2012 825  $c^2$ is the \definend{Square-cube law}~\cite{Wikipedia}.)  Jim Hefferon committed Dec 05, 2011 826 827 828 829 830  \begin{answer} Where the sides of the box are $$c$$ times longer, the box has $$c^3$$ times as many cubic units of volume. \end{answer} \recommended \item  Jim Hefferon committed Jan 13, 2012 831  We say that matrices $H$ and $G$ are  Jim Hefferon committed Dec 05, 2011 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848  \definend{similar}\index{similar}\index{matrix!similar} if there is a nonsingular matrix $P$ such that $H=P^{-1}GP$ (we will study this relation in Chapter Five). Show that similar matrices have the same determinant. \begin{answer} If $$H=P^{-1}GP$$ then $$\deter{H}=\deter{P^{-1}}\deter{G}\deter{P} =\deter{P^{-1}}\deter{P}\deter{G}=\deter{P^{-1}P}\deter{G} =\deter{G}$$. \end{answer} \item \label{exer:BasisOrient} We usually represent vectors in $$\Re^2$$ with respect to the standard basis so vectors in the first quadrant have both coordinates positive. \begin{center} \parbox{.75in}{\hbox{}\hfil\includegraphics{ch4.45}\hfil\hbox{}} \qquad  Jim Hefferon committed Jan 13, 2012 849  $$\rep{\vec{v}}{\stdbasis_2}=\colvec[r]{+3 \\ +2}$$  Jim Hefferon committed Dec 05, 2011 850 851 852 853 854 855 856 857 858 859 860 861 862  \end{center} Moving counterclockwise around the origin, we cycle thru four regions: {\scriptsize \begin{equation*} \cdots \;\longrightarrow\colvec{+ \\ +} \;\longrightarrow\colvec{- \\ +} \;\longrightarrow\colvec{- \\ -} \;\longrightarrow\colvec{+ \\ -} \;\longrightarrow\cdots\,. \end{equation*} } Using this basis \begin{center}  Jim Hefferon committed Jan 13, 2012 863  $$B=\sequence{\colvec[r]{0 \\ 1},\colvec[r]{-1 \\ 0}}$$  Jim Hefferon committed Dec 05, 2011 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887  \qquad \parbox{.75in}{\hbox{}\hfil\includegraphics{ch4.46}\hfil\hbox{}} \end{center} gives the same counterclockwise cycle. We say these two bases have the same \emph{orientation}.\index{orientation} \begin{exparts} \partsitem Why do they give the same cycle? \partsitem What other configurations of unit vectors on the axes give the same cycle? \partsitem Find the determinants of the matrices formed from those (ordered) bases. \partsitem What other counterclockwise cycles are possible, and what are the associated determinants? \partsitem What happens in $$\Re^1$$? \partsitem What happens in $$\Re^3$$? \end{exparts} A fascinating general-audience discussion of orientations is in \cite{Gardner}. \begin{answer} \begin{exparts} \partsitem The new basis is the old basis rotated by $$\pi/4$$. \partsitem $ Jim Hefferon committed Jan 13, 2012 888 889  \sequence{\colvec[r]{-1 \\ 0}, \colvec[r]{0 \\ -1}}  Jim Hefferon committed Dec 05, 2011 890 $, $ Jim Hefferon committed Jan 13, 2012 891 892  \sequence{\colvec[r]{0 \\ -1}, \colvec[r]{1 \\ 0}}  Jim Hefferon committed Dec 05, 2011 893 894 $ \partsitem In each case the determinant is $$+1$$  Jim Hefferon committed Jan 13, 2012 895  (we say that these bases  Jim Hefferon committed Dec 05, 2011 896 897 898 899 900 901 902 903 904 905 906 907  have \definend{positive orientation}). \partsitem Because only one sign can change at a time, the only other cycle possible is \begin{equation*} \cdots \;\longrightarrow\colvec{+ \\ +} \;\longrightarrow\colvec{+ \\ -} \;\longrightarrow\colvec{- \\ -} \;\longrightarrow\colvec{- \\ +} \;\longrightarrow\cdots\,. \end{equation*} Here each associated determinant is $$-1$$  Jim Hefferon committed Jan 13, 2012 908  (we say that such bases have a \definend{negative orientation}).  Jim Hefferon committed Dec 05, 2011 909 910 911 912 913 914 915 916 917 918 919 920 921 922  \partsitem There is one positively oriented basis $$\sequence{(1)}$$ and one negatively oriented basis $$\sequence{(-1)}$$. \partsitem There are $$48$$ bases ($$6$$ half-axis choices are possible for the first unit vector, $$4$$ for the second, and $$2$$ for the last). Half are positively oriented like the standard basis on the left below, and half are negatively oriented like the one on the right \begin{center} \includegraphics{ch4.47} \hspace*{4em} \includegraphics{ch4.48} \end{center} In $$\Re^3$$ positive orientation is sometimes called  Jim Hefferon committed Jan 13, 2012 923 924 925  right hand orientation' because if a person places their right hand with their fingers curling  Jim Hefferon committed Dec 05, 2011 926 927 928 929 930 931 932 933 934 935 936 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988  from $$\vec{e}_1$$ to $$\vec{e}_2$$ then the thumb will point with $$\vec{e}_3$$. \end{exparts} \end{answer} % \item % A region of $$\Re^n$$ is \definend{convex}\index{convex region} % if for any two points % connecting that region, the line segment joining them lies entirely % inside the region (the inside of a sphere is convex, while the skin of a % sphere or a horseshoe is not). % Prove that boxes are convex. % \begin{answer} % Let $$\vec{p}=p_1\vec{v}_1+\dots +p_n\vec{v}_n$$ and % $$\vec{q}=q_1\vec{v}_1+\dots +q_n\vec{v}_n$$ be two vectors from a box % so that $$p_1,\dots,\,p_n,q_1,\dots,\,q_n\in [0..1]$$. % The line segment between them is this. % \begin{equation*} % \set{t\cdot\vec{p}+(1-t)\cdot\vec{q} % \suchthat t\in [0..1]} % =\set{(tp_1+(1-t)q_1)\cdot\vec{v}_1+\dots+(tp_n+(1-t)q_n)\cdot\vec{v}_n % \suchthat t\in [0..1] } % \end{equation*} % Showing that each member of that set is in the box is routine. % \end{answer} \item \label{exer:DetProdEqProdDetsPerms} \textit{This question uses material from the optional Determinant Functions Exist subsection.} Prove \nearbytheorem{th:MatChVolByDetMat} by using the permutation expansion formula for the determinant. \begin{answer} We will compare $$\det(\vec{s}_1,\dots,\vec{s}_n)$$ with $$\det(t(\vec{s}_1),\dots,t(\vec{s}_n))$$ to show that the second differs from the first by a factor of $\deter{T}$. We represent the $$\vec{s}\,$$'s with respect to the standard bases \begin{equation*} \rep{\vec{s}_i}{\stdbasis_n}= \colvec{s_{1,i} \\ s_{2,i} \\ \vdots \\ s_{n,i}} \end{equation*} and then we represent the map application with matrix-vector multiplication \begin{align*} \rep{\,t(\vec{s}_i)\,}{\stdbasis_n} &=\generalmatrix{t}{n}{n} \colvec{s_{1,j} \\ s_{2,j} \\ \vdots \\ s_{n,j}} \\ &=s_{1,j}\colvec{t_{1,1} \\ t_{2,1} \\ \vdots \\ t_{n,1}} +s_{2,j}\colvec{t_{1,2} \\ t_{2,2} \\ \vdots \\ t_{n,2}} +\dots +s_{n,j}\colvec{t_{1,n} \\ t_{2,n} \\ \vdots \\ t_{n,n}} \\ &=s_{1,j}\vec{t}_1+s_{2,j}\vec{t}_2+\dots+s_{n,j}\vec{t}_n \end{align*} where $$\vec{t}_i$$ is column~$i$ of $$T$$. Then $\det(t(\vec{s}_1),\,\dots,\,t(\vec{s}_n))$ equals $\det(s_{1,1}\vec{t}_1\!+\!s_{2,1}\vec{t}_2\! +\!\dots\!+\!s_{n,1}\vec{t}_n,\, \dots,\, s_{1,n}\vec{t}_1\!+\!s_{2,n}\vec{t}_2 \!+\!\dots\!+\!s_{n,n}\vec{t}_n)$. As in the derivation of the permutation expansion formula, we apply multilinearity, first splitting along the sum in the first argument  Jim Hefferon committed Dec 23, 2013 989  \begin{multline*}  Jim Hefferon committed Dec 05, 2011 990 991  \det(s_{1,1}\vec{t}_1,\, \dots,\,  Jim Hefferon committed Dec 23, 2013 992  s_{1,n}\vec{t}_1+s_{2,n}\vec{t}_2+\dots+s_{n,n}\vec{t}_n) \\  Jim Hefferon committed Dec 05, 2011 993 994 995 996  +\cdots{} +\det(s_{n,1}\vec{t}_n,\, \ldots,\, s_{1,n}\vec{t}_1+s_{2,n}\vec{t}_2+\dots+s_{n,n}\vec{t}_n)  Jim Hefferon committed Dec 23, 2013 997  \end{multline*}  Jim Hefferon committed Dec 05, 2011 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038  and then splitting each of those $n$ summands along the sums in the second arguments, etc. We end with, as in the derivation of the permutation expansion, $$n^n$$ summand determinants, each of the form $\det(s_{i_1,1}\vec{t}_{i_1},s_{i_2,2}\vec{t}_{i_2}, \,\dots,\, s_{i_n,n}\vec{t}_{i_n})$. Factor out each of the $s_{i,j}$'s $= s_{i_1,1}s_{i_2,2}\dots s_{i_n,n} \cdot\det(\vec{t}_{i_1},\vec{t}_{i_2}, \,\dots,\, \vec{t}_{i_n})$. As in the permutation expansion derivation, whenever two of the indices in $i_1$, \ldots, $i_n$ are equal then the determinant has two equal arguments, and evaluates to $0$. So we need only consider the cases where $i_1$, \ldots, $i_n$ form a permutation of the numbers $1$, \ldots, $n$. We thus have \begin{equation*} \det(t(\vec{s}_1),\dots,t(\vec{s}_n))= \sum_{\text{permutations\ } \phi} s_{\phi(1),1}\dots s_{\phi(n),n} \det(\vec{t}_{\phi(1)},\dots,\vec{t}_{\phi(n)}). \end{equation*} Swap the columns in $\det(\vec{t}_{\phi(1)},\ldots,\vec{t}_{\phi(n)})$ to get the matrix $$T$$ back, which changes the sign by a factor of $\sgn{\phi}$, and then factor out the determinant of $T$. \begin{equation*} =\sum_\phi s_{\phi(1),1}\dots s_{\phi(n),n} \det(\vec{t}_1,\dots,\vec{t}_n)\cdot\sgn{\phi} =\det(T)\sum_\phi s_{\phi(1),1}\dots s_{\phi(n),n}\cdot\sgn{\phi}. \end{equation*} As in the proof that the determinant of a matrix equals the determinant  Jim Hefferon committed Jan 13, 2012 1039  of its transpose, we commute the $s$'s to list them by ascending  Jim Hefferon committed Dec 05, 2011 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058 1059 1060 1061 1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075  row number instead of by ascending column number (and we substitute $\sgn(\phi^{-1})$ for $\sgn(\phi)$). \begin{equation*} =\det(T)\sum_\phi s_{1,\phi^{-1}(1)}\dots s_{n,\phi^{-1}(n)}\cdot\sgn{\phi^{-1}} =\det(T)\det(\vec{s}_1,\vec{s}_2,\dots,\vec{s}_n) \end{equation*} \end{answer} % \item % Suppose that $$\map{f}{\matspace_{\nbyn{n}}}{\Re}$$ is a non-constant % function with the property that $$f(GH)=f(G)f(H)$$. % \begin{exparts} % \partsitem Show that $$f$$ sends the identity to $$1$$. % \partsitem Show that $$f$$ maps the elementary matrix % $$C_{i,j}(k)$$ to $$1$$ % (this matrix results from performing $$k\rho_i+\rho_j$$ to the % identity). % \partsitem Show that % $$f$$ maps a row swap matrix to $$+1$$ or $$-1$$. % \end{exparts} % \begin{answer} % \begin{exparts} % \partsitem We have $$f(IH)=f(I)f(H)$$ and $$f(IH)=f(H)$$. % \partsitem % \partsitem A row swap matrix has the property that when % done twice it equals % the identity. % But $$f(R)f(R)=f(RR)=f(I)=1$$ implies that $$f(R)=\pm 1$$. % \end{exparts} % \end{answer} \recommended \item \begin{exparts} \partsitem Show that this gives the equation of a line in $$\Re^2$$ thru $$(x_2,y_2)$$ and $$(x_3,y_3)$$. \begin{equation*}  Jim Hefferon committed Jan 14, 2012 1076  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 1077 1078 1079  x &x_2 &x_3 \\ y &y_2 &y_3 \\ 1 &1 &1  Jim Hefferon committed Jan 14, 2012 1080  \end{vmat}=0  Jim Hefferon committed Dec 05, 2011 1081 1082 1083 1084 1085 1086  \end{equation*} \partsitem \cite{Monthly55p249} Prove that the area of a triangle with vertices $$(x_1,y_1)$$, $$(x_2,y_2)$$, and $$(x_3,y_3)$$ is \begin{equation*} \frac{1}{2}  Jim Hefferon committed Jan 14, 2012 1087  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 1088 1089 1090  x_1 &x_2 &x_3 \\ y_1 &y_2 &y_3 \\ 1 &1 &1  Jim Hefferon committed Jan 14, 2012 1091  \end{vmat}.  Jim Hefferon committed Dec 05, 2011 1092 1093 1094 1095 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  \end{equation*} \partsitem \cite{MathMag73p286} Prove that the area of a triangle with vertices at $$(x_1,y_1)$$, $$(x_2,y_2)$$, and $$(x_3,y_3)$$ whose coordinates are integers has an area of $$N$$ or $$N/2$$ for some positive integer $$N$$. \end{exparts} \begin{answer} \begin{exparts} \partsitem An algebraic check is easy. \begin{equation*} 0 =xy_2+x_2y_3+x_3y-x_3y_2-xy_3-x_2y =x\cdot (y_2-y_3)+y\cdot (x_3-x_2)+x_2y_3-x_3y_2 \end{equation*} simplifies to the familiar form \begin{equation*} y=x\cdot (x_3-x_2)/(y_3-y_2)+(x_2y_3-x_3y_2)/(y_3-y_2) \end{equation*} (the $y_3-y_2=0$ case is easily handled). For geometric insight, this picture shows that the box formed by the three vectors. Note that all three vectors end in the $z=1$ plane. Below the two vectors on the right is the line through $(x_2,y_2)$ and $(x_3,y_3)$. \begin{center} \includegraphics{ch4.49} \end{center} The box will have a nonzero volume unless the triangle formed by the ends of the three is degenerate. That only happens (assuming that $(x_2,y_3)\neq (x_3,y_3)$) if $(x,y)$ lies on the line through the other two. \partsitem \answerasgiven %  Jim Hefferon committed Jan 13, 2012 1127 1128  We find the altitude through $(x_1,y_1)$ of a triangle with vertices $(x_1,y_1)$ $(x_2,y_2)$ and $(x_3,y_3)$ in the usual  Jim Hefferon committed Dec 05, 2011 1129 1130 1131  way from the normal form of the above: \begin{equation*} \frac{1}{\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}}  Jim Hefferon committed Jan 14, 2012 1132  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 1133 1134 1135  x_1 &x_2 &x_3 \\ y_1 &y_2 &y_3 \\ 1 &1 &1  Jim Hefferon committed Jan 14, 2012 1136  \end{vmat}.  Jim Hefferon committed Dec 05, 2011 1137 1138 1139 1140  \end{equation*} Another step shows the area of the triangle to be \begin{equation*} \frac{1}{2}  Jim Hefferon committed Jan 14, 2012 1141  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 1142 1143 1144  x_1 &x_2 &x_3 \\ y_1 &y_2 &y_3 \\ 1 &1 &1  Jim Hefferon committed Jan 14, 2012 1145  \end{vmat}.  Jim Hefferon committed Dec 05, 2011 1146 1147  \end{equation*} This exposition reveals the \textit{modus operandi} more clearly  Jim Hefferon committed Jan 13, 2012 1148  than the usual proof of showing a collection of terms to be identical  Jim Hefferon committed Dec 05, 2011 1149 1150 1151 1152 1153  with the determinant. \partsitem \answerasgiven % Let \begin{equation*} D=  Jim Hefferon committed Jan 14, 2012 1154  \begin{vmat}  Jim Hefferon committed Dec 05, 2011 1155 1156 1157  x_1 &x_2 &x_3 \\ y_1 &y_2 &y_3 \\ 1 &1 &1  Jim Hefferon committed Jan 14, 2012 1158  \end{vmat}  Jim Hefferon committed Dec 05, 2011 1159 1160 1161 1162 1163 1164  \end{equation*} then the area of the triangle is $(1/2)\deter{D}$. Now if the coordinates are all integers, then $D$ is an integer. \end{exparts} \end{answer} \end{exercises}`