# Diagonalize a 2 by 2 Symmetric Matrix

## Problem 629

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

## Solution.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\begin{align*}
p(t)&=\det(A-tI)=\begin{vmatrix}
2-t & -1\\
-1& 2-1
\end{vmatrix}\6pt] &=(2-t)^2-1 =t^2-4t+3\\ &=(t-1)(t-3). \end{align*} It follows that the eigenvalues of A are \lambda=1, 3 with algebraic multiplicities are both 1. Hence, the geometric multiplicities are 1 and thus any nonzero vector in eahc eigenspace forms a eigenbasis. Now let us find a eigenbasis for each eigenspace E_{\lambda}=\calN(A-\lambda I). For the eigenvalue 1, we have \[A-I=\begin{bmatrix} 1 & -1\\ -1& 1 \end{bmatrix}\xrightarrow{R_2+R_1} \begin{bmatrix} 1 & -1\\ 0& 0 \end{bmatrix} This yields that the eigenvectors corresponding to the eigenvalue $1$ are $x_2\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ with $x_2\neq 0$. Hence
$\mathbf{v}_1=\begin{bmatrix} 1 \\ 1 \end{bmatrix} \in E_1$ is an eigenbasis for $E_1$.

Similarly, as we have
$A-3I=\begin{bmatrix} -1 & -1\\ -1& -1 \end{bmatrix} \xrightarrow{-R_1}\begin{bmatrix} 1 & 1\\ -1& -1 \end{bmatrix} \xrightarrow{R_2+R_1} \begin{bmatrix} 1 & 1\\ 0& 0 \end{bmatrix},$ we see that
$\mathbf{v}_2=\begin{bmatrix} -1 \\ 1 \end{bmatrix} \in E_3$ is an eigenbasis for $E_3$.

Let
$S=\begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 \end{bmatrix}= \begin{bmatrix} 1 & -1\\ 1& 1 \end{bmatrix} \text{ and } D=\begin{bmatrix} 1 & 0\\ 0& 3 \end{bmatrix}.$

Then the diagonalization procedure yields that $S$ is nonsingular and $S^{-1}AS= D$.

### 1 Response

1. 01/03/2018

[…] 1 end{bmatrix} text{ and } begin{bmatrix} -1 \ 1 end{bmatrix},] respectively. (See the post Diagonalize a 2 by 2 Symmetric Matrix for […]

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