# Find Eigenvalues, Eigenvectors, and Diagonalize the 2 by 2 Matrix

## Problem 630

Consider the matrix $A=\begin{bmatrix} a & -b\\ b& a \end{bmatrix}$, where $a$ and $b$ are real numbers and $b\neq 0$.

(a) Find all eigenvalues of $A$.

(b) For each eigenvalue of $A$, determine the eigenspace $E_{\lambda}$.

(c) Diagonalize the matrix $A$ by finding a nonsingular matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

## Solution.

### (a) Find all eigenvalues of $A$.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI) = \begin{vmatrix}
a-t & -b\\
b& a-t
\end{vmatrix}\6pt] &=(a-t)^2+b^2. \end{align*} The eigenvalues of A are roots of p(t). So we solve p(t)=0. We have \begin{align*} & \quad (a-t)^2+b^2=0\\ \Leftrightarrow & \quad (a-t)^2=-b^2\\ \Leftrightarrow &\quad a-t =\pm i b\\ \Leftrightarrow &\quad t= a \pm ib. \end{align*} Here i=\sqrt{-1}. Thus, the eigenvalues of A are a\pm ib. ### (b) For each eigenvalue of A, determine the eigenspace E_{\lambda}. We first determine the eigenspace E_{\lambda} for \lambda = a+ib. Recall that by definition E_{\lambda}=\calN(A-\lambda I), the nullspace of A-\lambda I. We compute \begin{align*} A-(a+ib)I=\begin{bmatrix} -ib & -b\\ b& -ib \end{bmatrix} \xrightarrow{\frac{i}{b}R_1} \begin{bmatrix} 1 & -i\\ b& -ib \end{bmatrix} \xrightarrow{R_2-bR_1} \begin{bmatrix} 1 & -i\\ 0& 0 \end{bmatrix}. \end{align*} Note that in the above row reduction, we needed the assumption b\neq 0. It follows that the general solution of the system is x_1=i x_2. Hence, we have \[E_{a+ib} =\Span \left(\, \begin{bmatrix} i \\ 1 \end{bmatrix} \,\right).

Note that the other eigenvalue $a-ib$ is the complex conjugate of $a+ib$.
It follows that the eigenspace $E_{a-ib}$ is obtained by conjugating the eigenspace $E_{a+ib}$.
Thus,
$E_{a-ib} =\Span \left(\, \begin{bmatrix} -i \\ 1 \end{bmatrix} \,\right).$

### (c) Diagonalize the matrix $A$

From part (b), we see that
$\begin{bmatrix} i \\ 1 \end{bmatrix} \text{ and } \begin{bmatrix} -i \\ 1 \end{bmatrix}$ form an eigenbasis for $\C^2$.

So, we set
$S=\begin{bmatrix} i & -i\\ 1& 1 \end{bmatrix} \text{ and } D=\begin{bmatrix} a+ib & 0\\ 0& a-ib \end{bmatrix},$ and we obtain $S^{-1}AS=D$ by the diagonalization procedure.

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##### Diagonalize a 2 by 2 Symmetric Matrix

Diagonalize the $2\times 2$ matrix $A=\begin{bmatrix} 2 & -1\\ -1& 2 \end{bmatrix}$ by finding a nonsingular matrix $S$ and a...

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