Commit 9593797d by Jim Hefferon

### computational exercises, including one of similarity

parent 88d60eaa
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 ... ... @@ -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. ... ...
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 ... ... @@ -103,11 +103,11 @@ So we have this matrix representation of the map. The matrix changing bases from $B$ to $D$ is $\rep{\identity}{B,D}$. We find these by eye \begin{equation*} \rep{1}{D}=\colvec{1 \\ 0 \\ 0} \rep{\identity(1)}{D}=\colvec{1 \\ 0 \\ 0} \quad \rep{x}{D}=\colvec{-1 \\ 1 \\ 0} \rep{\identity(x)}{D}=\colvec{-1 \\ 1 \\ 0} \quad \rep{x^2}{D}=\colvec{0 \\ -1 \\ 1} \rep{\identity(x^2)}{D}=\colvec{0 \\ -1 \\ 1} \end{equation*} to get this. \begin{equation*} ... ... @@ -144,7 +144,7 @@ To check that, and to underline what the arrow diagram says V_{\wrt{D}} @>t>\hat{T}> V_{\wrt{D}} \end{CD} \end{equation*} we calculate $T$ directly. we calculate $\hat{T}$ directly. The effect of the map on the basis elements is $d/dx(1)=0$, $d/dx(1+x)=1$, and $d/dx(1+x+x^2)=1+2x$. Representing of those with respect to $D$ ... ... @@ -155,7 +155,7 @@ Representing of those with respect to $D$ \quad \rep{1+2x}{D}=\colvec{-1 \\ 2 \\ 0} \end{equation*} gives the same matrix $\hat{T}=\rep{d/dx}{D,D}$ as we found above. gives the same matrix $\hat{T}=\rep{d/dx}{D,D}$ as above. \end{frame} \begin{frame} The definition doesn't require that we consider the underlying maps. ... ... @@ -455,115 +455,6 @@ Not every vector is simply rescaled. \begin{frame} Matrices that are similar have the same eigenvalues, but needn't have the same eigenvectors. \ex These two are similar \begin{equation*} T= \begin{mat} 4 &0 &0 \\ 0 &8 &0 \\ 0 &0 &12 \end{mat} \qquad S= \begin{mat}[r] 6 &-1 &-1 \\ 2 &11 &-1 \\ -6 &-5 &7 \end{mat} \end{equation*} since $S=PTP^{-1}$ for this $P$. \begin{equation*} P= \begin{mat}[r] 1 &-1 &0 \\ 0 &1 &-1 \\ 2 &1 &1 \end{mat} \qquad P^{-1}= \begin{mat}[r] 1/2 &1/4 &1/4 \\ -1/2 &1/4 &1/4 \\ -1/2 &-3/4 &1/4 \end{mat} \end{equation*} \end{frame} \begin{frame} \noindent Suppose that $\map{t}{\C^3}{\C^3}$ is represented by $T$ with respect to the standard basis. Then this is the action of $t$. \begin{equation*} \colvec{x \\ y \\ z}\mapsunder{t}\colvec{4x \\ 8y \\ 12z} \end{equation*} \pause By eye we see that three eigenvalues of~$t$ are $\lambda_1=4$, $\lambda_2=8$, and~$\lambda_3=12$. For instance this holds. \begin{equation*} T\cdot\colvec{1 \\ 0 \\ 0} =\begin{mat} 4 &0 &0 \\ 0 &8 &0 \\ 0 &0 &12 \end{mat}\colvec{1 \\ 0 \\ 0} =4\cdot\colvec{1 \\ 0 \\ 0} \end{equation*} \end{frame} \begin{frame} Contrast that with $S=PTP^{-1}$, which represents the same function, but with respect to a different basis. \begin{equation*} \begin{CD} V_{\wrt{\stdbasis_3}} @>t>T> V_{\wrt{\stdbasis_3}} \\ @V{\scriptstyle\identity} VV @V{\scriptstyle\identity} VV \\ V_{\wrt{B}} @>t>S> V_{\wrt{B}} \end{CD} \end{equation*} We can easily find the basis~$B$. Since $P^{-1}=\rep{\identity}{B,\stdbasis_3}$, its first column is $\rep{\identity(\vec{\beta}_1)}{\stdbasis_3}=\rep{\vec{\beta}_1}{\stdbasis_3}$. With respect to the standard basis any vector is represented by itself so the first basis element $\vec{\beta}_1$ is the first column of $P^{-1}$. The same goes for the other two columns. \begin{equation*} B=\sequence{\colvec[r]{1/2 \\ -1/2 \\ -1/2}, \colvec[r]{1/4 \\ 1/4 \\ -3/4}, \colvec[r]{1/4 \\ 1/4 \\ 1/4}} \end{equation*} \end{frame} \begin{frame} % We know that the transformation~$t$ has eigenvalues of $4$, $8$, and~$12$. % For instance $t(\vec{e}_1)=4\vec{e}_1$. Now, since each represents the transformation~$t$, the matrices~$T$ and $S$ reflect the same action $\vec{e}_1\mapsto4\vec{e}_1$. \begin{align*} &\rep{t}{\stdbasis_3,\stdbasis_3}\cdot\rep{\vec{e}_1}{\stdbasis_3} =T\cdot\rep{\vec{e}_1}{\stdbasis_3} =4\cdot\rep{\vec{e}_1}{\stdbasis_3} \\ &\rep{t}{B,B}\cdot\rep{\vec{e}_1}{B} =S\cdot\rep{\vec{e}_1}{B} =4\cdot\rep{\vec{e}_1}{B} \end{align*} But, while in the two equations the $4$'s are the same, the vectors representations are not. \begin{align*} T\cdot\rep{\vec{e}_1}{\stdbasis_3} =T\colvec{1 \\ 0 \\ 0} &=4\cdot\colvec{1 \\ 0 \\ 0} \\ S\cdot\rep{\vec{e}_1}{B} =S\cdot\colvec{1 \\ 0 \\ 2} &=4\cdot\colvec{1 \\ 0 \\ 2} \end{align*} So the two matrices have the same eigenvalues but different eigenvectors. \end{frame} \begin{frame}{Computing eigenvalues and eigenvectors} \ex ... ... @@ -791,6 +682,121 @@ These are for $\lambda_2=2$. \end{frame} \begin{frame} Matrices that are similar have the same eigenvalues, but needn't have the same eigenvectors. \ex These two are similar \begin{equation*} T= \begin{mat} 4 &0 &0 \\ 0 &8 &0 \\ 0 &0 &12 \end{mat} \qquad S= \begin{mat}[r] 6 &-1 &-1 \\ 2 &11 &-1 \\ -6 &-5 &7 \end{mat} \end{equation*} since $S=PTP^{-1}$ for this $P$. \begin{equation*} P= \begin{mat}[r] 1 &-1 &0 \\ 0 &1 &-1 \\ 2 &1 &1 \end{mat} \qquad P^{-1}= \begin{mat}[r] 1/2 &1/4 &1/4 \\ -1/2 &1/4 &1/4 \\ -1/2 &-3/4 &1/4 \end{mat} \end{equation*} For the first matrix \begin{equation*} \colvec{1 \\ 0 \\ 0} \end{equation*} is an eigenvector associated with the eigenvalue~$4$ but that does not hold for the second matrix. \end{frame} % \begin{frame} % \noindent Suppose that $\map{t}{\C^3}{\C^3}$ is % represented by $T$ with respect to the standard basis. % Then this is the action of $t$. % \begin{equation*} % \colvec{x \\ y \\ z}\mapsunder{t}\colvec{4x \\ 8y \\ 12z} % \end{equation*} % \pause % By eye we see that three % eigenvalues of~$t$ are $\lambda_1=4$, $\lambda_2=8$, and~$\lambda_3=12$. % For instance this holds. % \begin{equation*} % T\cdot\colvec{1 \\ 0 \\ 0} % =\begin{mat} % 4 &0 &0 \\ % 0 &8 &0 \\ % 0 &0 &12 % \end{mat}\colvec{1 \\ 0 \\ 0} % =4\cdot\colvec{1 \\ 0 \\ 0} % \end{equation*} % \end{frame} % \begin{frame} % Contrast that with $S=PTP^{-1}$, which represents the same function, but % with respect to a different basis. % \begin{equation*} % \begin{CD} % V_{\wrt{\stdbasis_3}} @>t>T> V_{\wrt{\stdbasis_3}} \\ % @V{\scriptstyle\identity} VV @V{\scriptstyle\identity} VV \\ % V_{\wrt{B}} @>t>S> V_{\wrt{B}} % \end{CD} % \end{equation*} % We can easily find the basis~$B$. % Since $P^{-1}=\rep{\identity}{B,\stdbasis_3}$, its first column is % $\rep{\identity(\vec{\beta}_1)}{\stdbasis_3}=\rep{\vec{\beta}_1}{\stdbasis_3}$. % With respect to the standard basis any vector is represented by itself % so the first basis element $\vec{\beta}_1$ is the first column of $P^{-1}$. % The same goes for the other two columns. % \begin{equation*} % B=\sequence{\colvec[r]{1/2 \\ -1/2 \\ -1/2}, % \colvec[r]{1/4 \\ 1/4 \\ -3/4}, % \colvec[r]{1/4 \\ 1/4 \\ 1/4}} % \end{equation*} % \end{frame} % \begin{frame} % % We know that the transformation~$t$ has eigenvalues of $4$, $8$, and~$12$. % % For instance $t(\vec{e}_1)=4\vec{e}_1$. % Now, since each represents the transformation~$t$, the matrices~$T$ and $S$ % reflect the same action $\vec{e}_1\mapsto4\vec{e}_1$. % \begin{align*} % &\rep{t}{\stdbasis_3,\stdbasis_3}\cdot\rep{\vec{e}_1}{\stdbasis_3} % =T\cdot\rep{\vec{e}_1}{\stdbasis_3} % =4\cdot\rep{\vec{e}_1}{\stdbasis_3} \\ % &\rep{t}{B,B}\cdot\rep{\vec{e}_1}{B} % =S\cdot\rep{\vec{e}_1}{B} % =4\cdot\rep{\vec{e}_1}{B} % \end{align*} % But, while in the two equations the $4$'s are the same, the vectors % representations are not. % \begin{align*} % T\cdot\rep{\vec{e}_1}{\stdbasis_3} % =T\colvec{1 \\ 0 \\ 0} % &=4\cdot\colvec{1 \\ 0 \\ 0} \\ % S\cdot\rep{\vec{e}_1}{B} % =S\cdot\colvec{1 \\ 0 \\ 2} % &=4\cdot\colvec{1 \\ 0 \\ 2} % \end{align*} % So the two matrices have the same eigenvalues but different eigenvectors. % \end{frame} \begin{frame}{Characteristic polynomial} ... ...
 ... ... @@ -284,17 +284,19 @@ Thus here is the contrast. \begin{frame}{The determinant is unique} Recall the process by which we are developing the determinant. We gave four conditions that any determinant function must satisfy. From that definition it is not evident that a function satisfying those conditions exists. If such a function exists, from the definition it also is not immediately evident that the function is unique; perhaps there are $f_1$ and $f_2$ that give different outputs for some inputs. We now settle the second issue. Recall our definition, that a function is a determinant if it satisfies four conditions. This approach does not make evident that such function is unique. (An analogy: imagine defining a function $\map{f}{\N}{\N}$ to be an `even-maker' under the condition that its output is an even constant. There is such a function, but also there is more than one.) We now handle that issue; later we will handle the issue of showing that such a function exists at all. \pause \lm[lm:DetFcnIsUnique] \ExecuteMetaData[../det1.tex]{lm:DetFcnIsUnique} ... ... @@ -303,10 +305,10 @@ We now settle the second issue. \ExecuteMetaData[../det1.tex]{pf:DetFcnIsUnique} \qed \medskip So if there is a function mapping $\matspace_{\nbyn{n}}$ to $\Re$ that satisfies the four conditions of the definition then there is only one such function. % \medskip % So if there is a function mapping $\matspace_{\nbyn{n}}$ to $\Re$ that % satisfies the four conditions of the definition then there is only one such % function. \end{frame} \begin{frame}{More process discussion} We are left with the possibility that such a function does not exist. ... ... @@ -331,14 +333,14 @@ such a thing, \pause The rest of this section gives an alternative way to compute the value of a determinant, a formula. Because it does not involve Gauss's Method, this formula the determinant, a formula. This formula does not involve Gauss's Method and makes plain that the determinant is a function, that it returns well-defined outputs. As mentioned earlier, using this formula As mentioned earlier, computing a determinant with this formula is less practical than using the algorithm of Gauss's Method since it is slow. But it is very valuable for theory. But it nonetheless is invaluable for the theory. \end{frame} ... ... @@ -445,6 +447,11 @@ determinants also break along a plus sign one row at a time. \begin{frame} \ExecuteMetaData[../det1.tex]{pf:DetsMultilinear3} \qed \medskip \noindent (\textit{Remark}. Some authors use multilinearity to define the determinant in place of our four conditions that lead to Gauss's Method.) \end{frame} ... ... @@ -692,12 +699,6 @@ There are $3\cdot 2\cdot 1=6$ of these. \noindent After bringing out each entry from the original matrix, we are left with matrices that are all $0$'s except for a single~$1$ in each row and column. So, the only one thing remains to be done in our process of justifying the definition of determinant by finding a way to express determinants without using Gauss's Method:~give a formula for the determinant of such matrices (not involving Gauss's Method). \end{frame} ... ... @@ -825,8 +826,14 @@ Renaming the matrix entries gives the familiar $\nbyn{2}$ formula. \begin{frame} The next subsection is optional, so we give the statements of its results here. The only thing remaining in our process of finding a formula for the determinant (not involving Gauss's Method) is to give a formula for the determinant of such matrices. We do that in the next subsection. \pause That subsection is optional so we state its results here. \th[th:DetsExist] \ExecuteMetaData[../det1.tex]{th:DetsExist} ... ... @@ -1085,19 +1092,16 @@ So $\sgn(\phi)=+1$. %.......... \begin{frame}{Determinants exist} Recall the process by which we are validating the determinant definition. That definition is given as four conditions and it is not clear that for each input matrix there is one and only one associated output, that the determinant gives a well-defined value. Performing Gauss's Method on the input matrix shows that for each input there is at least one possible output. But Gauss's Method can be done in more than one way so to show there is exactly one we want a formula that gives an obviously well-defined value. \begin{frame}{Process finished} We are in the process of showing that a function exists that satisfies the four conditions in the definition of determinant. We must show that for each input square matrix there is a well-defined output value~\Dash Gauss's Method can be done in more than one way so it isn't obvious that by keeping track of signs and multiplying down the diagonal we always get the same output. Consequently we have turned to getting an alternate formula that obviously gives only one output. \pause \ExecuteMetaData[../det1.tex]{DefiningDFunction} ... ...
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