% This is lecture no. 12 of MH1201/: Linear Algebra II.

\chapter{Eigenvectors \& Eigenvalues}%
\label{cha:eigenvectors_&_eigenvalues}

The study of eigenvectors and eigenvalues is used to help solve the diagonalization problem.
\bigbreak
\begin{exmp}
	Find an expression of the reflection $\mathsf{T}$ about the line $y=2x$.
	\begin{sol}
		We can use theorem \ref{uniq}.
		Note that the point $(1,2)$ lies ont he straight line $y=2x$ and the point $(-2,1)$ lies on the line perpendicular to the first one.
		Then it is clear that
		\[
		\mathsf{T}
		\begin{bmatrix}
			1\\
			2\\
		\end{bmatrix}
		=
		\begin{bmatrix}
			1\\
			2\\
		\end{bmatrix}
		,
		\qquad
		\mathsf{T}
		\begin{bmatrix}
			-2\\
			1\\
		\end{bmatrix}
		=
		-
		\begin{bmatrix}
			-2\\
			1\\
		\end{bmatrix}
		=
		\begin{bmatrix}
			2\\
			-1\\
		\end{bmatrix}
		.\]
		Therefore, if we take
		\[
		\beta^{\prime}=\left\{
			\begin{bmatrix}
				1\\
				2\\
			\end{bmatrix}
			,
			\begin{bmatrix}
				-2\\
				1\\
			\end{bmatrix}
		\right\}
		\]
		as an ordered basis for $\mathbb{R}^2$, then we have
		\[
		\mathsf{T}
		\begin{bmatrix}
			1\\
			2\\
		\end{bmatrix}
		=
		\begin{bmatrix}
			1\\
			2\\
		\end{bmatrix}
		= 1
		\begin{bmatrix}
			1\\
			2\\
		\end{bmatrix}
		+ 0
		\begin{bmatrix}
			-2\\
			1\\
		\end{bmatrix}
		\]
		and
		\[
		\mathsf{T}
		\begin{bmatrix}
			-2\\
			1\\
		\end{bmatrix}
		=
		\begin{bmatrix}
			2\\
			-1\\
		\end{bmatrix}
		= 0
		\begin{bmatrix}
			1\\
			2\\
		\end{bmatrix}
		-1
		\begin{bmatrix}
			-2\\
			1\\
		\end{bmatrix}
		.\]
		That is,
		\[
		[\mathsf{T}]_{\beta^{\prime}}=
		\begin{bmatrix}
			1 & 0\\
			0 & -1\\
		\end{bmatrix}
		.\]
		\begin{note}
			This matrix is \textbf{diagonal}.
		\end{note}
		Furthermore, if $\beta$ is the standard basis for $\mathbb{R}^2$, then we can determine the matrix that changes $\beta^{\prime}$-coordinates into $\beta$-coordinates, $Q$, and its inverse $Q^{-1}$:
		\[
		Q = [I]^{\beta}_{\beta^{\prime}} =
		\begin{bmatrix}
			1 & -2\\
			2 & 1\\
		\end{bmatrix}
		,
		\qquad
		Q^{-1}=\frac{1}{5}
		\begin{bmatrix}
			1 & 2\\
			-2 & 1\\
		\end{bmatrix}
		.\]
		Then since $Q^{-1}[\mathsf{T}]_{\beta}Q=[\mathsf{T}]_{\beta^{\prime}}$, we can solve the equation to get
		\[
		[\mathsf{T}]_{\beta} = Q[\mathsf{T}]_{\beta^{\prime}}Q^{-1}=\frac{1}{5}
		\begin{bmatrix}
			-3 & 4\\
			4 & 3\\
		\end{bmatrix}
		.\]
	\end{sol}
\end{exmp}
