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

Diagonalization Problems and Solutions in Linear Algebra

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

 
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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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