Commit 9593797d by Jim Hefferon

### computational exercises, including one of similarity

parent 88d60eaa
 ... ... @@ -25464,6 +25464,61 @@ octave:6> gplot z \end{ans} \begin{ans}{Four.II.1.12} \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{ans} \begin{ans}{Four.II.1.13} Express each transformation with respect to the standard bases and find the determinant. \begin{exparts*} ... ... @@ -25473,17 +25528,17 @@ octave:6> gplot z \end{exparts*} \end{ans} \begin{ans}{Four.II.1.13} \begin{ans}{Four.II.1.14} The starting area is $$6$$ and the matrix changes sizes by $$-14$$. Thus the area of the image is $$84$$. \end{ans} \begin{ans}{Four.II.1.14} \begin{ans}{Four.II.1.15} By a factor of $$21/2$$. \end{ans} \begin{ans}{Four.II.1.15} \begin{ans}{Four.II.1.16} For a box we take a sequence of vectors (as described in the remark, the order of the vectors matters), while for a span we take a set of vectors. ... ... @@ -25494,7 +25549,7 @@ octave:6> gplot z span the coefficients are free to range over all of $\Re$. \end{ans} \begin{ans}{Four.II.1.16} \begin{ans}{Four.II.1.17} We have drawn that picture to mislead. 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 ... ... @@ -25516,19 +25571,19 @@ octave:6> gplot z which has an area of $4$. \end{ans} \begin{ans}{Four.II.1.17} \begin{ans}{Four.II.1.18} Yes to both. For instance, the first is $$\deter{TS}=\deter{T}\cdot\deter{S}= \deter{S}\cdot\deter{T}=\deter{ST}$$. \end{ans} \begin{ans}{Four.II.1.18} \begin{ans}{Four.II.1.19} % 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{ans} \begin{ans}{Four.II.1.19} \begin{ans}{Four.II.1.20} \begin{exparts} \partsitem If it is defined then it is $$(3^2)\cdot (2)\cdot (2^{-2})\cdot (3)$$. ... ... @@ -25536,14 +25591,14 @@ octave:6> gplot z \end{exparts} \end{ans} \begin{ans}{Four.II.1.20} \begin{ans}{Four.II.1.21} $$\begin{vmat} \cos\theta &-\sin\theta \\ \sin\theta &\cos\theta \end{vmat}=1$$ \end{ans} \begin{ans}{Four.II.1.21} \begin{ans}{Four.II.1.22} No, for instance the determinant of \begin{equation*} T=\begin{mat}[r] ... ... @@ -25555,11 +25610,11 @@ octave:6> gplot z has length $$2$$. \end{ans} \begin{ans}{Four.II.1.22} \begin{ans}{Four.II.1.23} It is zero. \end{ans} \begin{ans}{Four.II.1.23} \begin{ans}{Four.II.1.24} Two of the three sides of the triangle are formed by these vectors. \begin{equation*} \colvec[r]{2 \\ 2 \\ 2}-\colvec[r]{1 \\ 2 \\ 1}=\colvec[r]{1 \\ 0 \\ 1} ... ... @@ -25610,7 +25665,7 @@ octave:6> gplot z \end{equation*} \end{ans} \begin{ans}{Four.II.1.24} \begin{ans}{Four.II.1.25} \begin{exparts} \partsitem Because the image of a linearly dependent set is linearly dependent, ... ... @@ -25661,7 +25716,7 @@ octave:6> gplot z \end{exparts} \end{ans} \begin{ans}{Four.II.1.25} \begin{ans}{Four.II.1.26} Any permutation matrix has the property that the transpose of the matrix is its inverse. ... ... @@ -25678,19 +25733,19 @@ octave:6> gplot z \end{equation*} \end{ans} \begin{ans}{Four.II.1.26} \begin{ans}{Four.II.1.27} Where the sides of the box are $$c$$ times longer, the box has $$c^3$$ times as many cubic units of volume. \end{ans} \begin{ans}{Four.II.1.27} \begin{ans}{Four.II.1.28} 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{ans} \begin{ans}{Four.II.1.28} \begin{ans}{Four.II.1.29} \begin{exparts} \partsitem The new basis is the old basis rotated by $$\pi/4$$. \partsitem ... ... @@ -25738,7 +25793,7 @@ octave:6> gplot z \end{exparts} \end{ans} \begin{ans}{Four.II.1.29} \begin{ans}{Four.II.1.30} 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 $\deter{T}$. ... ... @@ -25832,7 +25887,7 @@ octave:6> gplot z \end{equation*} \end{ans} \begin{ans}{Four.II.1.30} \begin{ans}{Four.II.1.31} \begin{exparts} \partsitem An algebraic check is easy. \begin{equation*} ... ... @@ -27231,9 +27286,103 @@ octave:6> gplot z \end{ans} \begin{ans}{Five.II.1.7} Gauss's Method shows that the first matrix represents maps of rank two while the second matrix represents maps of rank three. \begin{exparts} \partsitem \begin{equation*} \begin{CD} \C^3_{\wrt{B}} @>t>T> \C^3_{\wrt{B}} \\ @V{\scriptstyle\identity} VV @V{\scriptstyle\identity} VV \\ \C^3_{\wrt{D}} @>t>\hat{T}> \C^3_{\wrt{D}} \end{CD} \end{equation*} \partsitem For each element of the starting basis~$B$ find the effect of the transformation \begin{equation*} \colvec{1 \\ 2 \\ 3}\mapsunder{t}\colvec{-2 \\ 3 \\ 4} \qquad \colvec{0 \\ 1 \\ 0}\mapsunder{t}\colvec{0 \\ 0 \\ 2} \qquad \colvec{0 \\ 0 \\ 1}\mapsunder{t}\colvec{-1 \\ 1 \\ 0} \end{equation*} and represented those outputs with respect to the ending basis~$B$ \begin{equation*} \rep{\colvec{-2 \\ 3 \\ 4}}{B}=\colvec{-2 \\ 7 \\ 10} \qquad \rep{\colvec{0 \\ 0 \\ 2}}{B}=\colvec{0 \\ 0 \\ 2} \qquad \rep{\colvec{-1 \\ 1 \\ 0}}{B}=\colvec{-1 \\ 3 \\ 3} \end{equation*} to get the matrix. \begin{equation*} T=\rep{t}{B,B}= \begin{mat} -2 &0 &-1 \\ 7 &0 &3 \\ 10 &2 &3 \end{mat} \end{equation*} \partsitem Find the effect of the transformation on the elements of~$D$ \begin{equation*} \colvec{1 \\ 0 \\ 0}\mapsunder{t}\colvec{1 \\ 0 \\ 0} \qquad \colvec{1 \\ 1 \\ 0}\mapsunder{t}\colvec{1 \\ 0 \\ 2} \qquad \colvec{1 \\ 0 \\ 1}\mapsunder{t}\colvec{0 \\ 1 \\ 0} \end{equation*} and represented those with respect to the ending basis~$D$ \begin{equation*} \rep{\colvec{1 \\ 0 \\ 0}}{D}=\colvec{1 \\ 0 \\ 0} \qquad \rep{\colvec{1 \\ 0 \\ 2}}{D}=\colvec{-1 \\ 0 \\ 2} \qquad \rep{\colvec{0 \\ 1 \\ 0}}{D}=\colvec{-1 \\ 1 \\ 0} \end{equation*} to get the matrix. \begin{equation*} \hat{T}=\rep{t}{D,D}= \begin{mat} 1 &-1 &-1 \\ 0 &0 &1 \\ 0 &2 &0 \end{mat} \end{equation*} \partsitem To go down on the right we need $\rep{\identity}{B,D}$ so we first compute the effect of the identity map on each element of~$D$, which is no effect, and then represent the results with respect to~$B$. \begin{equation*} \rep{\colvec{1 \\ 2 \\ 3}}{D}=\colvec{-4 \\ 2 \\ 3} \qquad \rep{\colvec{0 \\ 1 \\ 0}}{D}=\colvec{-1 \\ 1 \\ 0} \qquad \rep{\colvec{0 \\ 0 \\ 1}}{D}=\colvec{-1 \\ 0 \\ 1} \end{equation*} So this is~$P$. \begin{equation*} P= \begin{mat} -4 &-1 &-1 \\ 2 &1 &0 \\ 3 &0 &1 \end{mat} \end{equation*} For the other matrix~$\rep{\identity}{D,B}$ we can either find it directly, as we just have with~$P$, or we can do the usual calculation of a matrix inverse. \begin{equation*} P^{-1}= \begin{mat} 1 &1 &1 \\ -2 &-1 &-2 \\ -3 &-3 &-2 \end{mat} \end{equation*} \end{exparts} \end{ans} \begin{ans}{Five.II.1.8} ... ... @@ -27435,6 +27584,12 @@ octave:6> gplot z \end{ans} \begin{ans}{Five.II.1.10} Gauss's Method shows that the first matrix represents maps of rank two while the second matrix represents maps of rank three. \end{ans} \begin{ans}{Five.II.1.11} The only representation of a zero map is a zero matrix, no matter what the pair of bases $\rep{z}{B,D}=Z$, and so in particular for any single basis $B$ we have $\rep{z}{B,B}=Z$. ... ... @@ -27447,18 +27602,18 @@ octave:6> gplot z respect to some $B,D$.) \end{ans} \begin{ans}{Five.II.1.11} \begin{ans}{Five.II.1.12} No. If $$A=PBP^{-1}$$ then $$A^2=(PBP^{-1})(PBP^{-1})=PB^2P^{-1}$$. \end{ans} \begin{ans}{Five.II.1.12} \begin{ans}{Five.II.1.13} Matrix similarity is a special case of matrix equivalence (if matrices are similar then they are matrix equivalent) and matrix equivalence preserves nonsingularity. \end{ans} \begin{ans}{Five.II.1.13} \begin{ans}{Five.II.1.14} A matrix is similar to itself; take $$P$$ to be the identity matrix:~$P=IPI^{-1}=IPI$. ... ... @@ -27474,7 +27629,7 @@ octave:6> gplot z is similar to $$U$$. \end{ans} \begin{ans}{Five.II.1.14} \begin{ans}{Five.II.1.15} Let $f_x$ and $f_y$ be the reflection maps (sometimes called `flip's). For any bases $$B$$ and $$D$$, the matrices $$\rep{f_x}{B,B}$$ and ... ... @@ -27545,7 +27700,7 @@ octave:6> gplot z similar. \end{ans} \begin{ans}{Five.II.1.15} \begin{ans}{Five.II.1.16} We must show that if two matrices are similar then they have the same determinant and the same rank. Both determinant and rank are properties of matrices that ... ... @@ -27574,7 +27729,7 @@ octave:6> gplot z The argument for rank is much the same. \end{ans} \begin{ans}{Five.II.1.16} \begin{ans}{Five.II.1.17} The matrix equivalence class containing all $$\nbyn{n}$$ rank zero matrices contains only a single matrix, the zero matrix. Therefore it has as a subset only one similarity class. ... ... @@ -27590,7 +27745,7 @@ octave:6> gplot z infinitely many similarity classes. \end{ans} \begin{ans}{Five.II.1.17} \begin{ans}{Five.II.1.18} Yes, these are similar \begin{equation*} \begin{mat}[r] ... ... @@ -27609,7 +27764,7 @@ octave:6> gplot z $D=\sequence{\vec{\beta}_2,\vec{\beta}_1}$. \end{ans} \begin{ans}{Five.II.1.18} \begin{ans}{Five.II.1.19} The $$k$$-th powers are similar because, where each matrix represents the map $t$, the $k$-th powers represent $$t^k$$, the composition of $k$-many $t$'s. ... ... @@ -27628,7 +27783,7 @@ octave:6> gplot z Other negative powers are now given by the first paragraph. \end{ans} \begin{ans}{Five.II.1.19} \begin{ans}{Five.II.1.20} In conceptual terms, both represent $$p(t)$$ for some transformation $$t$$. In computational terms, we have this. ... ... @@ -27641,7 +27796,7 @@ octave:6> gplot z \end{align*} \end{ans} \begin{ans}{Five.II.1.20} \begin{ans}{Five.II.1.21} There are two equivalence classes, (i)~the class of rank~zero matrices, of which there is one: $\mathscr{C}_1=\set{(0)}$, ... ... @@ -27666,7 +27821,7 @@ octave:6> gplot z $(k)$ for $k\neq0$. \end{ans} \begin{ans}{Five.II.1.21} \begin{ans}{Five.II.1.22} No. Here is an example that has two pairs, each of two similar matrices: \begin{equation*} ... ... @@ -27745,7 +27900,7 @@ octave:6> gplot z since the zero matrix is similar only to itself. \end{ans} \begin{ans}{Five.II.1.22} \begin{ans}{Five.II.1.23} If $$N=P(T-\lambda I)P^{-1}$$ then $$N=PTP^{-1}-P(\lambda I)P^{-1}$$. The diagonal matrix $$\lambda I$$ commutes with anything, so
 ... ... @@ -414,7 +414,88 @@ $1=\deter{I}=\deter{TT^{-1}}=\deter{T}\cdot\deter{T^{-1}}$ \partsitem $1/9$ \end{exparts*} \end{answer} \recommended \item \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 By what factor does each transformation change the size of boxes? \begin{exparts*} ... ...
 ... ... @@ -358,6 +358,9 @@ the \definend{standard basis}\index{standard basis}\index{basis!standard}% \index{standard basis!complex number scalars} for $$\C^n$$ as a vector space over $\C$ and again denote it $$\stdbasis_n$$. Another example is that $\polyspace_n$ will be the vector space of degree~$n$ polynomials with coefficients that are complex. ... ...
 ... ... @@ -41,65 +41,12 @@ In matrix terms, \subsection{Definition and Examples} \begin{definition} \label{df:Similar} %<*df:Similar> The matrices $$T$$ and $\hat{T}$ are \definend{similar}\index{matrix!similarity}% \index{equivalence relation!matrix similarity}\index{similar matrices} if there is a nonsingular $$P$$ such that $\hat{T}=PTP^{-1}$. % \end{definition} \noindent Since nonsingular matrices are square, $T$ and $\hat{T}$ must be square and of the same size. \nearbyexercise{exer:SimIsEquivRel} checks that similarity is an equivalence relation. \begin{example} Calculation with these two \begin{equation*} P= \begin{mat}[r] 2 &1 \\ 1 &1 \end{mat} \qquad T= \begin{mat}[r] 2 &-3 \\ 1 &-1 \end{mat} \end{equation*} gives that $T$ is similar to this matrix. \begin{equation*} \hat{T}= \begin{mat}[r] 12 &-19 \\ 7 &-11 \end{mat} \end{equation*} \end{example} \begin{example} \label{ex:OnlyZeroSimToZero} %<*ex:OnlyZeroSimToZero> The only matrix similar to the zero matrix is itself:~$PZP^{-1}=PZ=Z$. The identity matrix has the same property:~$PIP^{-1}=PP^{-1}=I$. % \end{example} \begin{example} Consider the derivative transformation $\map{d/dx}{\polyspace_2}{\polyspace_2}$, and two bases for that space. \begin{equation*} B=\sequence{x^2,x,1} \qquad D=\sequence{1,1+x,1+x^2} \end{equation*} and two bases for that space $B=\sequence{x^2,x,1}$ and $D=\sequence{1,1+x,1+x^2}$ We will compute the four sides of the arrow square. \begin{equation*} \begin{CD} ... ... @@ -108,7 +55,7 @@ We will compute the four sides of the arrow square. {\polyspace_2\,}_{\wrt{D}} @>d/dx>\hat{T}> {\polyspace_2\,}_{\wrt{D}} \end{CD} \end{equation*} First the top. The top is first. The effect of the transformation on the starting basis~$B$ \begin{equation*} x^2\mapsunder{d/dx} 2x ... ... @@ -161,7 +108,7 @@ gives the matrix~$\hat{T}$. Third, computing the matrix for the right-hand side involves finding the effect of the identity map on the elements of~$B$. Of course, the identity map does not transform them at all so to find $\rep{id}{B,D}$ we represent $B$'s elements with respect to find the matrix we represent $B$'s elements with respect to~$D$. \begin{equation*} \rep{x^2}{D}=\colvec{-1 \\ 0 \\ 1} ... ... @@ -180,9 +127,9 @@ So the matrix for going down the right side is the concatenation of those. \end{mat} \end{equation*} With that, we can compute in two ways the matrix for going up on With that, we have two options to compute the matrix for going up on left side. Fpr direct computation, represent elements of~$D$ with The direct computation represents elements of~$D$ with respect to~$B$ \begin{equation*} \rep{1}{B}=\colvec{0 \\ 0 \\ 1} ... ... @@ -191,7 +138,7 @@ respect to~$B$ \quad \rep{1+x^2}{B}=\colvec{1 \\ 0 \\ 1} \end{equation*} and concatenate to make the matrix. and concatenates to make the matrix. \begin{equation*} \begin{mat} 0 &0 &1 \\ ... ... @@ -199,8 +146,8 @@ and concatenate to make the matrix. 1 &1 &1 \end{mat} \end{equation*} The other way to compute the matrix for going up on the left is to find it as the inverse of the matrix~$P$ for The other option to compute the matrix for going up on the left is to take the inverse of the matrix~$P$ for going down on the right. \begin{equation*} \begin{pmat}{ccc|ccc} ... ... @@ -222,10 +169,60 @@ going down on the right. 0 &0 &1 &1 &1 &1 \end{pmat} \end{equation*} The definition expresses the relationship in the second way, as $\hat{T}=PTP^{-1}$. \end{example} \begin{definition} \label{df:Similar} %<*df:Similar> The matrices $$T$$ and $\hat{T}$ are \definend{similar}\index{matrix!similarity}% \index{equivalence relation!matrix similarity}\index{similar matrices} if there is a nonsingular $$P$$ such that $\hat{T}=PTP^{-1}$. % \end{definition} \noindent Since nonsingular matrices are square, $T$ and $\hat{T}$ must be square and of the same size. \nearbyexercise{exer:SimIsEquivRel} checks that similarity is an equivalence relation. \begin{example} The definition does not require that we consider a map. Calculation with these two \begin{equation*} P= \begin{mat}[r] 2 &1 \\ 1 &1 \end{mat} \qquad T= \begin{mat}[r] 2 &-3 \\ 1 &-1 \end{mat} \end{equation*} gives that $T$ is similar to this matrix. \begin{equation*} \hat{T}= \begin{mat}[r] 12 &-19 \\ 7 &-11 \end{mat} \end{equation*} \end{example} \begin{example} \label{ex:OnlyZeroSimToZero} %<*ex:OnlyZeroSimToZero> The only matrix similar to the zero matrix is itself:~$PZP^{-1}=PZ=Z$. The identity matrix has the same property:~$PIP^{-1}=PP^{-1}=I$. % \end{example} Matrix similarity is a special case of matrix equivalence so if two matrices are similar then they are matrix equivalent. What about the converse:~if they are square, ... ... @@ -329,7 +326,7 @@ if and only if they have the same rank). \end{mat} \end{multline*} \end{answer} \recommended \item \item \nearbyexample{ex:OnlyZeroSimToZero} shows that the only matrix similar to a zero matrix is itself and that the only matrix similar to the identity ... ... @@ -375,25 +372,127 @@ if and only if they have the same rank). \end{equation*} \end{exparts} \end{answer} \recommended \item Show that these matrices are not similar. \recommended \item Consider this transformation of~$\C^3$ \begin{equation*} \begin{mat}[r] 1 &0 &4 \\ 1 &1 &3 \\ 2 &1 &7 \end{mat} \qquad \begin{mat}[r] 1 &0 &1 \\ 0 &1 &1 \\ 3 &1 &2 \end{mat} t(\colvec{x \\ y \\ z})=\colvec{x-z \\ z \\ 2y} \end{equation*} and these bases. \begin{equation*} B=\sequence{\colvec{1 \\ 2 \\ 3}, \colvec{0 \\ 1 \\ 0}, \colvec{0 \\ 0 \\ 1}} \qquad D=\sequence{\colvec{1 \\ 0 \\ 0}, \colvec{1 \\ 1 \\ 0}, \colvec{1 \\ 0 \\ 1}} \end{equation*} We will compute the parts of the arrow diagram to represent the transformation using two similar matrices. \begin{exparts} \partsitem Draw the arrow diagram, specialized for this case. \partsitem Compute $T=\rep{t}{B,B}$. \partsitem Compute $\hat{T}=\rep{t}{D,D}$. \partsitem Compute the matrices for other the two sides of the arrow square. \end{exparts} \begin{answer} Gauss's Method shows that the first matrix represents maps of rank two while the second matrix represents maps of rank three. \begin{exparts} \partsitem \begin{equation*} \begin{CD} \C^3_{\wrt{B}} @>t>T> \C^3_{\wrt{B}} \\ @V{\scriptstyle\identity} VV @V{\scriptstyle\identity} VV \\ \C^3_{\wrt{D}} @>t>\hat{T}> \C^3_{\wrt{D}} \end{CD} \end{equation*} \partsitem For each element of the starting basis~$B$ find the effect of the transformation \begin{equation*} \colvec{1 \\ 2 \\ 3}\mapsunder{t}\colvec{-2 \\ 3 \\ 4} \qquad \colvec{0 \\ 1 \\ 0}\mapsunder{t}\colvec{0 \\ 0 \\ 2} \qquad \colvec{0 \\ 0 \\ 1}\mapsunder{t}\colvec{-1 \\ 1 \\ 0} \end{equation*} and represented those outputs with respect to the ending basis~$B$ \begin{equation*} \rep{\colvec{-2 \\ 3 \\ 4}}{B}=\colvec{-2 \\ 7 \\ 10} \qquad \rep{\colvec{0 \\ 0 \\ 2}}{B}=\colvec{0 \\ 0 \\ 2} \qquad \rep{\colvec{-1 \\ 1 \\ 0}}{B}=\colvec{-1 \\ 3 \\ 3} \end{equation*} to get the matrix. \begin{equation*} T=\rep{t}{B,B}= \begin{mat} -2 &0 &-1 \\ 7 &0 &3 \\ 10 &2 &3 \end{mat}