# Diagonalize the 3 by 3 Matrix if it is Diagonalizable

## Problem 456

Determine whether the matrix
$A=\begin{bmatrix} 0 & 1 & 0 \\ -1 &0 &0 \\ 0 & 0 & 2 \end{bmatrix}$ is diagonalizable.

If it is diagonalizable, then find the invertible matrix $S$ and a diagonal matrix $D$ such that $S^{-1}AS=D$.

## How to diagonalize matrices.

For a general procedure of the diagonalization of a matrix, please read the post “How to Diagonalize a Matrix. Step by Step Explanation“.

## Solution.

We first determine the eigenvalues of the matrix $A$.
To do so, we compute the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
&p(t)=\det(A-tI)\\
&=\begin{vmatrix}
-t & 1 & 0 \\
-1 &-t &0 \\
0 & 0 & 2-t
\end{vmatrix}\6pt] &=(-1)^{3+3}(2-t)\begin{vmatrix} -t & 1\\ -1& -t \end{vmatrix} && \text{by the third row cofactor expansion}\\ &=(2-t)(t^2+1). \end{align*} Thus the eigenvalues of A are 2, \pm i. Since the 3\times 3 matrix A has three distinct eigenvalues, it is diagonalizable. To diagonalize A, we now find eigenvectors. For the eigenvalue 2, we compute \begin{align*} &A-2I=\begin{bmatrix} -2 & 1 & 0 \\ -1 &-2 &0 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{-R_2} \begin{bmatrix} -2 & 1 & 0 \\ 1 &2 &0 \\ 0 & 0 & 0 \end{bmatrix}\\[6pt] &\xrightarrow{R_1 \leftrightarrow R_2}\begin{bmatrix} 1 & 2 & 0 \\ -2 &1 &0 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_2+2R_1}\begin{bmatrix} 1 & 2 & 0 \\ 0 &5 &0 \\ 0 & 0 & 0 \end{bmatrix}\\[6pt] &\xrightarrow{\frac{1}{5}R_2}\begin{bmatrix} 1 & 2 & 0 \\ 0 &1 &0 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_1-2R_2} \begin{bmatrix} 1 & 0 & 0 \\ 0 &1 &0 \\ 0 & 0 & 0 \end{bmatrix}. \end{align*} Thus, the solutions \mathbf{x} of (A-2I)=\mathbf{0} satisfy x=y=0. Hence the eigenspace is \[E_2=\calN(A-2I)=\Span\left\{\, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \,\right\}.

For the eigenvalue $i$, we compute
\begin{align*}
A-iI=\begin{bmatrix}
-i & 1 & 0 \\
-1 &-i &0 \\
0 & 0 & 2-i
\end{bmatrix}
\xrightarrow{iR_1}
\begin{bmatrix}
1 & i & 0 \\
-1 &-i &0 \\
0 & 0 & 2-i
\end{bmatrix}\6pt] \xrightarrow{\substack{R_2+R_1\\ \frac{1}{2-i}R_3}} \begin{bmatrix} 1 & i & 0 \\ 0 &0 &0 \\ 0 & 0 & 1 \end{bmatrix} \xrightarrow{R_2 \leftrightarrow R_3} \begin{bmatrix} 1 & i & 0 \\ 0 & 0 & 1\\ 0 &0 &0 \end{bmatrix}. \end{align*} So the solutions \mathbf{x} of (A-iI)\mathbf{x}=\mathbf{0} satisfy \[x=-iy \text{ and } z=0. Thus, the eigenspace is
$E_i=\calN(A-iI)=\Span\left\{\, \begin{bmatrix} 1 \\ i \\ 0 \end{bmatrix} \,\right\}.$

Since $i$ and $-i$ are complex conjugate, their eigenspaces are also complex conjugate.
Hence the eigenspace for $-i$ is
$E_{-i}=\Span\left\{\, \begin{bmatrix} 1 \\ -i \\ 0 \end{bmatrix} \,\right\}.$
From these computations, we have obtained eigenvalues $2, i, -i$ and eigenvector corresponding to these are
$\mathbf{v}_{2}=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \mathbf{v}_i=\begin{bmatrix} 1 \\ i \\ 0 \end{bmatrix}, \mathbf{v}_{-i}=\begin{bmatrix} 1 \\ -i \\ 0 \end{bmatrix}.$

Let
$S=\begin{bmatrix} \mathbf{v}_2 & \mathbf{v}_i & \mathbf{v}_{-i} \\ \end{bmatrix}=\begin{bmatrix} 0 & 1 & 1 \\ 0 &i &-i \\ 1 & 0 & 0 \end{bmatrix}$ and
$D=\begin{bmatrix} 2 & 0 & 0 \\ 0 &i &0 \\ 0 & 0 & -i \end{bmatrix}.$ Then $S$ is invertible and we have $S^{-1}AS=D$ by the diagonalization process.

### 1 Response

1. 06/14/2017

[…] For a solution, check out the post “Diagonalize the 3 by 3 Matrix if it is Diagonalizable“. […]

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Determine all linear transformations of the $2$-dimensional $x$-$y$ plane $\R^2$ that take the line $y=x$ to the line $y=-x$.

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