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.

The post Find Eigenvalues, Eigenvectors, and Diagonalize the 2 by 2 Matrix first appeared on Problems in Mathematics.

Consider the complex matrix
\[A=\begin{bmatrix}
\sqrt{2}\cos x & i \sin x & 0 \\
i \sin x &0 &-i \sin x \\
0 & -i \sin x & -\sqrt{2} \cos x
\end{bmatrix},\] where $x$ is a real number between $0$ and $2\pi$.

Determine for which values of $x$ the matrix $A$ is diagonalizable.
When $A$ is diagonalizable, find a diagonal matrix $D$ so that $P^{-1}AP=D$ for some nonsingular matrix $P$.

Solution.

Let us first find the eigenvalues of the matrix $A$.
To do so, we compute the characteristic polynomial $p(t)=\det(A-tI)$ of $A$ as follows.
Using Sarrus’s rule to compute the $3\times 3$ determinant, we have
\begin{align*}
&p(t)=\det(A-tI)\\[6pt] &=\begin{bmatrix}
\sqrt{2}\cos x -t & i \sin x & 0 \\
i \sin x & -t &-i \sin x \\
0 & -i \sin x & -\sqrt{2} \cos x-t
\end{bmatrix}\\[6pt] &=-t(\sqrt{2}\cos x-t)(-\sqrt{2}\cos x -t)
-\left(\, -(\sin^2 x) (-\sqrt{2}\cos x-t)-(\sin^2 x) (\sqrt{2}\cos x -t) \,\right)\\
&=-t^3+2(\cos^2 x-\sin ^2 x)t\\
&=-t^3+2\cos(2x) t.
\end{align*}

The eigenvalues of $A$ are the roots of
\[p(t)=-t^3+2\cos(2x) t=-t(t^2-2\cos(2x)).\] Hence the eigenvalues are
\[t=0, \quad\pm \sqrt{2\cos(2x)}.\]

Note that if $\sqrt{2\cos(2x)}=-\sqrt{2\cos(2x)}$ then we have $\cos(2x)=0$ and hence $x=\pi/4, 3\pi/4$.
It follows that if $x=\pi/4, 3\pi/4$, then the matrix $A$ has only one eigenvalue $0$ with algebraic multiplicity $3$.
Since $A$ is not the zero matrix, the rank of $A$ is greater than or equal to $1$.

Hence the nullity of $A$ is less than or equal to $2$ by the rank-nullity theorem.
It follows that the geometric multiplicity (=nullity) of the eigenvalue $0$ is strictly less than the algebraic multiplicity of $0$ and $A$ is not diagonalizable in this case.

Now suppose that $x\neq \pi/4, 3\pi/4$.
In this case, the matrix $A$ has three distinct eigenvalues $0, \pm \sqrt{2\cos(2x)}$.
This implies that $A$ is diagonalizable.

Let $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ be eigenvectors corresponding to eigenvalues $0, \pm \sqrt{2\cos(2x)}$, respectively.
Define the $3\times 3$ matrix $P$ by $P=\begin{bmatrix}
\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \\
\end{bmatrix}$.

It follows from the general procedure of the diagonalization that $P$ is a nonsingular matrix and
\[P^{-1}AP=D,\] where $D$ is a diagonal matrix
\[D=\begin{bmatrix}
0 & 0 & 0 \\
0 &\sqrt{2\cos(2x)} &0 \\
0 & 0 & -\sqrt{2\cos(2x)}
\end{bmatrix}.\]

Summary

In summary, when $x=\pi/4, 3\pi/4$ the matrix $A$ is not diagonalizable.

When $x \neq \pi/4, 3\pi/4$, the matrix $A$ is diagonalizable and we can take the diagonal matrix $D$ as
\[D=\begin{bmatrix}
0 & 0 & 0 \\
0 &\sqrt{2\cos(2x)} &0 \\
0 & 0 & -\sqrt{2\cos(2x)}
\end{bmatrix}.\] Click here if solved 19

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