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 $S^{-1}AS=D$.

 
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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.

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