Commit dcc4189f by Jim Hefferon

### spell checks for the third chapter

parent 5526a5ff
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 ... ... @@ -11167,7 +11167,7 @@ \end{aligned} \end{multline*} (An alternate proof is to simply note that this is a property of differentiation that is familar from calculus.) property of differentiation that is familiar from calculus.) These two maps are not inverses as this composition does not act as the identity map on ... ... @@ -11687,7 +11687,7 @@ \colvec{f_1(\vec{v}) \\ f_2(\vec{v})} \end{equation*} They are linear because they are the composition of linear functions, and the fact that the compoistion of linear functions is linear and the fact that the composition of linear functions is linear was part of the proof that isomorphism is an equivalence relation (alternatively, the check that they are linear is straightforward). ... ... @@ -12757,7 +12757,7 @@ \colvec[r]{2 \\ 0}-\colvec[r]{-1 \\ 0}=\colvec[r]{3 \\ 0} \end{equation*} A more systemmatic way to find the image of $\vec{e}_2$ is to A more systematic way to find the image of $\vec{e}_2$ is to use the given information to represent the transformation, and then use that representation to determine the image. Taking this for a basis, ... ... @@ -13344,7 +13344,7 @@ \end{equation*} gives the additional information (beyond that there is at least one solution) that there are infinitely many solutions. Parametizing gives $c_2=-1+c_3$ and $c_1=1$, and so taking $c_3$ to Parametrizing gives $c_2=-1+c_3$ and $c_1=1$, and so taking $c_3$ to be zero gives a particular solution of $c_1=1$, $c_2=-1$, and $c_3=0$ (which is, of course, the observation made at the start). \end{exparts} ... ... @@ -13368,7 +13368,7 @@ \colvec[r]{0 \\ 0 \\ 1} \!\mapsto\colvec[r]{3 \\ 4} \end{equation*} So, for this first one, we are asking whether thare are scalars such that So, for this first one, we are asking whether there are scalars such that \begin{equation*} c_1\colvec[r]{1 \\ 0}+c_2\colvec[r]{1 \\ 1} +c_3\colvec[r]{3 \\ 4}=\colvec[r]{1 \\ 3} ... ... @@ -13529,7 +13529,7 @@ \end{ans} \begin{ans}{Three.III.2.16} Let the matrix be $G$, and suppose that it rperesents $\map{g}{V}{W}$ Let the matrix be $G$, and suppose that it represents $\map{g}{V}{W}$ with respect to bases $B$ and $D$. Because $G$ has two columns, $V$ is two-dimensional. Because $G$ has two rows, $W$ is two-dimensional. ... ... @@ -13579,7 +13579,7 @@ \end{ans} \begin{ans}{Three.III.2.18} Recall that the represention map Recall that the representation map \begin{equation*} V\mapsunder{\text{Rep}_{B}}\Re^n \end{equation*} ... ... @@ -13710,7 +13710,7 @@ to its dot product with $\vec{x}$ is linear (this is a matrix-vector product and so \nearbytheorem{th:MatIsLinMap} applies). Thus the map under consideration $h_{\vec{x}}$ is linear because it is the composistion of two linear maps. it is the composition of two linear maps. \begin{equation*} \vec{v}\mapsto \rep{\vec{v}}{B} \mapsto \vec{x}\cdot\rep{\vec{v}}{B} ... ... @@ -13898,7 +13898,7 @@ h_{1,j}\vec{\delta}_1+\dots+h_{i,j}\vec{\delta}_i +\dots+h_{m,j}\vec{\delta}_m \end{equation*} and with respcet to $B,2\cdot D$ it also represents and with respect to $B,2\cdot D$ it also represents $$\map{h_2}{V}{W}$$ sending \begin{equation*} \vec{\beta}_j\mapsto ... ... @@ -16421,7 +16421,7 @@ The proof tells us what how the bases change. We start by swapping the first and second rows of the representation with respect to $B$ to get a representation with resepect to a new basis $B_1$. with respect to a new basis $B_1$. \begin{equation*} \rep{1-x+3x^2-x^3}{B_1}= \colvec[r]{1 \\ 0 \\ 1 \\ 2}_{B_1} ... ... @@ -16631,7 +16631,7 @@ \colvec[r]{1 \\ 1}=1\cdot\colvec[r]{1 \\ 0} +1\cdot\colvec[r]{0 \\ 1} \end{equation*} give the other nonsinguar matrix. give the other nonsingular matrix. \begin{equation*} \rep{\identity}{\hat{B},B}=\begin{mat}[r] 0 &1 \\ ... ... @@ -17301,7 +17301,7 @@ Suppose that $$\vec{v}\in\Re^n$$ with $$n>1$$. If $$\vec{v}\neq\zero$$ then we consider the line $$\ell=\set{c\vec{v}\suchthat c\in\Re}$$ and if $$\vec{v}=\zero$$ we take $$\ell$$ to be any (nondegenerate) line at all we take $$\ell$$ to be any (non-degenerate) line at all (actually, we needn't distinguish between these two cases\Dash see the prior exercise). Let $$v_1,\dots,v_n$$ be the components of $$\vec{v}$$; ... ... @@ -17329,7 +17329,7 @@ The dimension $$n=0$$ case is the trivial vector space, here there is only one vector and so it cannot be expressed as the projection of a different vector. In the dimension $n=1$ case there is only one (nondegenerate) line, In the dimension $n=1$ case there is only one (non-degenerate) line, and every vector is in it, hence every vector is the projection only of itself. ... ... @@ -17670,7 +17670,7 @@ \end{ans} \begin{ans}{Three.VI.2.12} We can paramatrize the given space can in this way. We can parametrize the given space can in this way. \begin{equation*} \set{\colvec{x \\ y \\ z} \suchthat x=y-z} =\set{\colvec[r]{1 \\ 1 \\ 0}\cdot y+\colvec[r]{-1 \\ 0 \\ 1}\cdot z ... ... @@ -17865,7 +17865,7 @@ meets the vertical dashed line $\vec{v}-(1\cdot\vec{e}_1+2\cdot\vec{e}_2)$; this is what first item of this question proved. The Pythagorean theorem then gives that the hypoteneuse\Dash the The Pythagorean theorem then gives that the hypotenuse\Dash the segment from $\vec{v}$ to any other vector\Dash is longer than the vertical dashed line. ... ... @@ -19644,7 +19644,7 @@ perform the row operations and, if needed, column operations to reduce it to a partial-identity matrix. We will then translate that into a factorization $H=PBQ$. Subsitituting into the general matrix Substituting into the general matrix \begin{equation*} \rep{r_\theta}{\stdbasis_2,\stdbasis_2} \begin{mat}
 ... ... @@ -158,7 +158,7 @@ For contrast the next picture shows the effect of the map represented by $C_{2,1}(1)$. Here vectors are affected according to their second component: $\binom{x}{y}$ slides horozontally by twice $y$. $\binom{x}{y}$ slides horizontally by twice $y$. \begin{center} \includegraphics{ch3.57} \end{center} ... ... @@ -174,7 +174,7 @@ $H=T_nT_{n-1}\cdots T_jBT_{j-1}\cdots T_1$, and so, in some sense, we have an understanding of the action of any matrix $H$. We will illustrate the usefullness of our understanding in two ways. We will illustrate the usefulness of our understanding in two ways. The first is that we will use it to prove something about linear maps. Recall that under a linear map, the image of a subspace is a subspace and thus the linear transformation $h$ represented by $H$ maps lines ... ... @@ -192,7 +192,7 @@ Therefore their composition also preserves lines. % Thus, by understanding its components we can understand arbitrary square % matrices $H$, in the sense that we can prove things about them. The second way that we will illustrate the usefullness of The second way that we will illustrate the usefulness of our understanding is to apply it to Calculus. Below is a picture of the action of the one-variable real function $$y(x)=x^2+x$$. ... ... @@ -430,7 +430,7 @@ is appealing both for its simplicity and for its usefulness. perform the row operations and, if needed, column operations to reduce it to a partial-identity matrix. We will then translate that into a factorization $H=PBQ$. Subsitituting into the general matrix Substituting into the general matrix \begin{equation*} \rep{r_\theta}{\stdbasis_2,\stdbasis_2} \begin{mat} ... ...
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 ... ... @@ -616,7 +616,7 @@ record was 1954-May-06. \textit{(This illustrates that there are data sets for which a linear model is not right, and that the line of best fit doesn't in that case have any predictive value.)} In a highway resturant a trucker told me that his boss often sends In a highway restaurant a trucker told me that his boss often sends him by a roundabout route, using more gas but paying lower bridge tolls. He said that New York state sets the bridge ... ...
 ... ... @@ -2178,7 +2178,7 @@ classes, the reduced echelon form matrices. In this section we have followed that outline, except that the appropriate notion of same-ness except that the appropriate notion of sameness here is vector space isomorphism. First we defined isomorphism, saw some examples, and established some properties. ... ...
 ... ... @@ -600,7 +600,7 @@ is more fruitful and more central to further progress. \end{aligned} \end{multline*} (An alternate proof is to simply note that this is a property of differentiation that is familar from calculus.) property of differentiation that is familiar from calculus.) These two maps are not inverses as this composition does not act as the identity map on ... ... @@ -1295,7 +1295,7 @@ is more fruitful and more central to further progress. \colvec{f_1(\vec{v}) \\ f_2(\vec{v})} \end{equation*} They are linear because they are the composition of linear functions, and the fact that the compoistion of linear functions is linear and the fact that the composition of linear functions is linear was part of the proof that isomorphism is an equivalence relation (alternatively, the check that they are linear is straightforward). ... ... @@ -1486,7 +1486,7 @@ We lose that the domain corresponds perfectly to the range. What we retain, as the examples below illustrate, is that a homomorphism describes how the domain is like'' or analgous to'' the range. the domain is like'' or analogous to'' the range. \begin{example} \label{ex:RThreeHomoRTwo} %\label{exPicProj} We think of $\Re^3$ as like $\Re^2$ except that vectors have an extra ... ... @@ -1802,7 +1802,7 @@ Equality holds if and only if the nullity of the map is $0$. We know that an isomorphism exists between two spaces if and only if the dimension of the range equals the dimension of the domain. We have now seen that for a homomorphism to exist a nexessary condition is that We have now seen that for a homomorphism to exist a necessary condition is that the dimension of the range must be less than or equal to the dimension of the domain. For instance, there is no homomorphism ... ...
 ... ... @@ -1214,7 +1214,7 @@ for any matrix there is an associated linear map. \colvec[r]{2 \\ 0}-\colvec[r]{-1 \\ 0}=\colvec[r]{3 \\ 0} \end{equation*} A more systemmatic way to find the image of $\vec{e}_2$ is to A more systematic way to find the image of $\vec{e}_2$ is to use the given information to represent the transformation, and then use that representation to determine the image. Taking this for a basis, ... ... @@ -1988,7 +1988,7 @@ but we do not have particular spaces or bases in mind then we often take the domain and codomain to be $\Re^n$ and $\Re^m$ and use the standard bases. This is convienent because with the standard bases This is convenient because with the standard bases vector representation is transparent\Dash the representation of $\vec{v}$ is $\vec{v}$. (In this case the ... ... @@ -2081,7 +2081,7 @@ that superspace (because any basis for the rangespace is a linearly independent subset of the codomain whose size is equal to the dimension of the codomain, and thus so this basis for the reagespace must also be basis for the rangespace must also be a basis for the codomain). For the other half, ... ... @@ -2257,7 +2257,7 @@ And, we shall see how to find the matrix that represents a map's inverse. \end{equation*} gives the additional information (beyond that there is at least one solution) that there are infinitely many solutions. Parametizing gives $c_2=-1+c_3$ and $c_1=1$, and so taking $c_3$ to Parametrizing gives $c_2=-1+c_3$ and $c_1=1$, and so taking $c_3$ to be zero gives a particular solution of $c_1=1$, $c_2=-1$, and $c_3=0$ (which is, of course, the observation made at the start). \end{exparts} ... ... @@ -2293,7 +2293,7 @@ And, we shall see how to find the matrix that represents a map's inverse. \colvec[r]{0 \\ 0 \\ 1} \!\mapsto\colvec[r]{3 \\ 4} \end{equation*} So, for this first one, we are asking whether thare are scalars such that So, for this first one, we are asking whether there are scalars such that \begin{equation*} c_1\colvec[r]{1 \\ 0}+c_2\colvec[r]{1 \\ 1} +c_3\colvec[r]{3 \\ 4}=\colvec[r]{1 \\ 3} ... ... @@ -2509,7 +2509,7 @@ And, we shall see how to find the matrix that represents a map's inverse. domain. \end{exparts} \begin{answer} Let the matrix be $G$, and suppose that it rperesents $\map{g}{V}{W}$ Let the matrix be $G$, and suppose that it represents $\map{g}{V}{W}$ with respect to bases $B$ and $D$. Because $G$ has two columns, $V$ is two-dimensional. Because $G$ has two rows, $W$ is two-dimensional. ... ... @@ -2574,7 +2574,7 @@ And, we shall see how to find the matrix that represents a map's inverse. respect to $$D$$. Show that map is a linear transformation of $$\Re^n$$. \begin{answer} Recall that the represention map Recall that the representation map \begin{equation*} V\mapsunder{\text{Rep}_{B}}\Re^n \end{equation*} ... ... @@ -2787,7 +2787,7 @@ And, we shall see how to find the matrix that represents a map's inverse. to its dot product with $\vec{x}$ is linear (this is a matrix-vector product and so \nearbytheorem{th:MatIsLinMap} applies). Thus the map under consideration $h_{\vec{x}}$ is linear because it is the composistion of two linear maps. it is the composition of two linear maps. \begin{equation*} \vec{v}\mapsto \rep{\vec{v}}{B} \mapsto \vec{x}\cdot\rep{\vec{v}}{B} ... ...
 ... ... @@ -400,7 +400,7 @@ no matter what domain and codomain bases we use. h_{1,j}\vec{\delta}_1+\dots+h_{i,j}\vec{\delta}_i +\dots+h_{m,j}\vec{\delta}_m \end{equation*} and with respcet to $B,2\cdot D$ it also represents and with respect to $B,2\cdot D$ it also represents $$\map{h_2}{V}{W}$$ sending \begin{equation*} \vec{\beta}_j\mapsto ... ... @@ -466,7 +466,7 @@ no matter what domain and codomain bases we use. \index{transpose!interaction with sum and scalar multiplication} of a matrix $M$ is another matrix, whose $i,j$ entry is the $j,i$ entry of $M$. Verifiy these identities. Verify these identities. \begin{exparts} \partsitem $$\trans{(G+H)}=\trans{G}+\trans{H}$$ \partsitem $$\trans{(r\cdot H)}=r\cdot\trans{H}$$ ... ... @@ -2466,7 +2466,7 @@ is square and has with all entries zero except for ones in the main diagonal. \end{definition} \begin{example} Here is the $$\nbyn{2}$$ identity matrix leaving its multiplicand unchaged Here is the $$\nbyn{2}$$ identity matrix leaving its multiplicand unchanged when it acts from the right. \begin{equation*} \begin{mat}[r] ... ... @@ -2916,7 +2916,7 @@ Until now we have taken the point of view that our primary objects of study are vector spaces and the maps between them, and have adopted matrices only for computational convenience. This subsection show that this isn't the whole story. Understanding matrices operations vy how the entries combine can Understanding matrices operations by how the entries combine can be useful also. In the rest of this book we shall continue to focus on maps as the primary objects but we will be pragmatic\Dash if the matrix point of view gives some ... ... @@ -3500,7 +3500,7 @@ clearer idea then we will go with it. \end{answer} \item Combine the two generalizations of the identity matrix, the one allowing entires to be other than ones, and the one allowing the the one allowing entries to be other than ones, and the one allowing the single one in each row and column to be off the diagonal. What is the action of this type of matrix? \begin{answer} ... ... @@ -5120,7 +5120,7 @@ elementary real number system can be interesting and useful. items. \end{exparts} When two things multiply to give zero despite that neither is zero, each is said to be a \definend{zero divisor}.\index{zero divison} said to be a \definend{zero divisor}.\index{zero division} Prove that no zero divisor is invertible. \begin{answer} For the answer to the items making up the first half, see ... ...
 ... ... @@ -90,7 +90,7 @@ map $$\map{\identity}{V}{V}$$ with respect to those bases. Left-multiplication by the change of basis matrix for $$B,D$$ converts a representation with respect to $$B$$ to one with respect to $$D$$. Conversly, if left-multiplication by a matrix changes bases Conversely, if left-multiplication by a matrix changes bases $M\cdot\rep{\vec{v}}{B}=\rep{\vec{v}}{D}$ then $M$ is a change of basis matrix. \end{lemma} ... ... @@ -178,7 +178,7 @@ to some ending basis. Because the matrix is nonsingular it will Gauss-Jordan reduce to the identity. If the matrix is the identity~$I$ then the statement is obvious. Otherwise there are elementatry reduction matrices such that Otherwise there are elementary reduction matrices such that $R_r\cdots R_1\cdot M=I$ with $r\geq 1$. Elementary matrices are invertible and their inverses are also elementary so multiplying both sides of that equation from the left ... ... @@ -608,7 +608,7 @@ the same space, and where the map is the identity map. \end{equation*} \end{answer} \item Conside the vector space of real-valued functions with basis Consider the vector space of real-valued functions with basis $$\sequence{\sin(x),\cos(x)}$$. Show that $$\sequence{2\sin(x)+\cos(x),3\cos(x)}$$ is also a basis for this space. ... ... @@ -789,7 +789,7 @@ the same space, and where the map is the identity map. \begin{exparts} \partsitem In $$\polyspace_3$$ with basis $$B=\sequence{1+x,1-x,x^2+x^3,x^2-x^3}$$ we have this represenatation. representation. \begin{equation*} \rep{1-x+3x^2-x^3}{B}= \colvec[r]{0 \\ 1 \\ 1 \\ 2}_B ... ... @@ -815,7 +815,7 @@ the same space, and where the map is the identity map. The proof tells us what how the bases change. We start by swapping the first and second rows of the representation with respect to $B$ to get a representation with resepect to a new basis $B_1$. with respect to a new basis $B_1$. \begin{equation*} \rep{1-x+3x^2-x^3}{B_1}= \colvec[r]{1 \\ 0 \\ 1 \\ 2}_{B_1} ... ... @@ -1184,7 +1184,7 @@ has been \definend{diagonalized}\index{matrix!diagonalized} when its representation is diagonal with respect to $B,B$, that is, with respect to equal starting and ending bases. In Chaper Five we shall see which maps and matrices are diagonalizable. In Chapter Five we shall see which maps and matrices are diagonalizable. In the rest of this subsection we consider the easier case where representations are with respect to $B,D$, which are possibly different starting and ending bases. ... ... @@ -1223,7 +1223,7 @@ the set of matrices into matrix equivalence classes. \end{center} We can get some insight into the classes by comparing matrix equivalence with row equivalence (rememeber that matrices are row equivalent when they can be reduced to each (remember that matrices are row equivalent when they can be reduced to each other by row operations). In $\hat{H}=PHQ$, the matrices $P$ and $Q$ are nonsingular and thus we can write each as a product of elementary reduction matrices ... ... @@ -1609,7 +1609,7 @@ this is a good classification of linear maps. \colvec[r]{1 \\ 1}=1\cdot\colvec[r]{1 \\ 0} +1\cdot\colvec[r]{0 \\ 1} \end{equation*} give the other nonsinguar matrix. give the other nonsingular matrix. \begin{equation*} \rep{\identity}{\hat{B},B}=\begin{mat}[r] 0 &1 \\ ... ...
 ... ... @@ -22,7 +22,7 @@ is the $\vec{p}$ in the plane with the property that someone standing on $\vec{p}$ and looking directly up or down sees $\vec{v}$. In this section we will generalize this to other projections, both orthogonal and nonorthogonal. both orthogonal and non-orthogonal. ... ... @@ -202,7 +202,7 @@ This subsection has developed a natural projection map, orthogonal projection into a line. As suggested by the examples, we use it often in applications. The next subsection shows how the definition of orthogonal projection into a line gives us a way to calculate especially convienent bases projection into a line gives us a way to calculate especially convenient bases for vector spaces, again something that we often see in applications. The final subsection completely generalizes projection, orthogonal or not, into any subspace at all. ... ... @@ -317,7 +317,7 @@ into any subspace at all. \partsitem $\colvec[r]{1 \\ 2}$ \partsitem $\colvec[r]{0 \\ 4}$ \end{exparts*} Show that in general the projection tranformation is this. Show that in general the projection transformation is this. \begin{equation*} \colvec{x_1 \\ x_2} \mapsto ... ... @@ -467,7 +467,7 @@ into any subspace at all. Suppose that $$\vec{v}\in\Re^n$$ with $$n>1$$. If $$\vec{v}\neq\zero$$ then we consider the line $$\ell=\set{c\vec{v}\suchthat c\in\Re}$$ and if $$\vec{v}=\zero$$ we take $$\ell$$ to be any (nondegenerate) line at all we take $$\ell$$ to be any (non-degenerate) line at all (actually, we needn't distinguish between these two cases\Dash see the prior exercise). Let $$v_1,\dots,v_n$$ be the components of $$\vec{v}$$; ... ... @@ -495,7 +495,7 @@ into any subspace at all. The dimension $$n=0$$ case is the trivial vector space, here there is only one vector and so it cannot be expressed as the projection of a different vector. In the dimension $n=1$ case there is only one (nondegenerate) line, In the dimension $n=1$ case there is only one (non-degenerate) line, and every vector is in it, hence every vector is the projection only of itself. \end{answer} ... ... @@ -1244,7 +1244,7 @@ An example is in \nearbyexercise{exer:OrthoRepEasy}. Find an orthonormal basis for this subspace of $\Re^3$:~the plane $x-y+z=0$. \begin{answer} We can paramatrize the given space can in this way. We can parametrize the given space can in this way. \begin{equation*} \set{\colvec{x \\ y \\ z} \suchthat x=y-z} =\set{\colvec[r]{1 \\ 1 \\ 0}\cdot y+\colvec[r]{-1 \\ 0 \\ 1}\cdot z ... ... @@ -1465,7 +1465,7 @@ An example is in \nearbyexercise{exer:OrthoRepEasy}. meets the vertical dashed line $\vec{v}-(1\cdot\vec{e}_1+2\cdot\vec{e}_2)$; this is what first item of this question proved. The Pythagorean theorem then gives that the hypoteneuse\Dash the The Pythagorean theorem then gives that the hypotenuse\Dash the segment from $\vec{v}$ to any other vector\Dash is longer than the vertical dashed line. ... ...
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