MATH 240

Fri. November 1st, 2019

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Change of Basis Review

Let $B=\{b_1,...,b_n\}$ and $C=\{c_1,...c_n\}$ be bases of a vector space $V$. Then there is a unique $n\times n$ matrix $P_{C\leftarrow B}$ such that $[x]_C=P_{C\leftarrow B}[x]_B$.

The columns of $P_{C\leftarrow B}$ are the $C$-coordinate vectors of the vectors in the basis $B$, i.e. $P_{C\leftarrow B}=\begin{bmatrix} [b_1]_C,...,[b_n]_C \end{bmatrix}$.

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Ex. Let $b_1=\begin{pmatrix} -9\\1 \end{pmatrix},\ b_2=\begin{pmatrix} -5\\1 \end{pmatrix},\ c_1=\begin{pmatrix} 1\\-4 \end{pmatrix},\ c_2=\begin{pmatrix} 3\\-5 \end{pmatrix}$.

$B=\{b_1,b_2\},\ C=\{c_1,c_2\}$ form bases of $\R^2$. Find the change of basis matrix $P_{C\leftarrow B}$.

$\begin{aligned} \begin{bmatrix} c_1&c_2 \end{bmatrix}[b_1]_C&=b_1\\ \begin{bmatrix} c_1&c_2 \end{bmatrix}[b_2]_C&=b_2\\ \end{aligned}$ The operations to reduce $\begin{bmatrix} c_1&c_2 \end{bmatrix}$ to the identity will be the same for both equations. Therefore to simplify computations we can place $\begin{bmatrix} b_1&b_2 \end{bmatrix}$ to the right of it, reduce it to the identity, and then whatever ends up on the right will be $P_{C\leftarrow B}$ (similar to the technique we used before to find the inverse of a matrix).

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Eigenvalues and Eigenvectors

Given an $n\times n$ matrix $A$, what points will remained fixed under the transformation $x\rightarrow Ax$?

This question is significant because the result of applying the transformation to that point any number of times will be easily computable.

Definition. Let $A$ be an $n\times n$ matrix. A scalar $\lambda$ is called an eigenvalue of $A$ if there is a non-trivial (non-zero) solution $x$ of $Ax=\lambda x$. Such a vector $x$ is called an eigenvector of $\lambda$.

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Ex.
If $x$ is an eigenvector of $A$ with an eigenvalue of $2$, then $Ax=2x$.
$A^2(x)=A(Ax)=A(2x)=2A(x)=4x$.
$A^3(x)=A(A(Ax))=A(4x)=4A(x)=8x.$
$A^n(x)=2^nx$.
This point moves away from the origin in a straight line under repeated multiplication with $A$.

If $x$ is an eigenvector of $A$ with an eigenvalue of $\frac{1}{2}$, then $Ax=\frac{1}{2}x$.
$A^2(x)=A(Ax)=A(\frac{1}{2}x)=\frac{1}{2}A(x)=\frac{1}{4}x$.
$A^3(x)=A(A(Ax))=A(\frac{1}{4}x)=\frac{1}{4}A(x)=\frac{1}{8}x.$
$A^n(x)=2^{-n}x$.
This point moves towards the origin in a straight line under repeated multiplication with $A$.

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Ex.
Let $A=\begin{pmatrix} 1&6\\5&2 \end{pmatrix}$, $u=\begin{pmatrix} 6\\-5 \end{pmatrix}$, and $v=\begin{pmatrix} 3\\-2 \end{pmatrix}$. Are either $u$ or $v$ eigenvectors of $A$?

$\begin{aligned} Au&=\begin{pmatrix} -24\\20 \end{pmatrix}\\ &=-4\begin{pmatrix} 6\\-5 \end{pmatrix}\\ &=-4u\\ &=\lambda u\\ \\ Av&=\begin{pmatrix} -9\\11 \end{pmatrix}\\ &\neq \lambda v \end{aligned}$
$u$ is an eigenvector of $A$ with an eigenvalue $-4$ because $Au=-4u$.
$v$ is not an eigenvector of $A$ because there is no scalar $\lambda$ such that $Av=\lambda v$.

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Ex.
Show that $7$ is an eigenvalue of $\begin{pmatrix} 1&6\\5&2 \end{pmatrix}$.

$\begin{aligned} Ax&=7x\\ Ax-7x&=0\\ (A-7I)x&=0\\ \begin{pmatrix} -6&6\\5&-5 \end{pmatrix}x&=0 \end{aligned}$