# Eigenvalues and their Algebraic Multiplicities of a Matrix with a Variable

## Problem 206

Determine all eigenvalues and their algebraic multiplicities of the matrix
$A=\begin{bmatrix} 1 & a & 1 \\ a &1 &a \\ 1 & a & 1 \end{bmatrix},$ where $a$ is a real number.

## Proof.

To find eigenvalues we first compute the characteristic polynomial of the matrix $A$ as follows.
\begin{align*}
\det(A-tI)&=\begin{vmatrix}
1-t & a & 1 \\
a &1-t &a \\
1 & a & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
1-t & a\\
a& 1-t
\end{vmatrix}-a\begin{vmatrix}
a & a\\
1& 1-t
\end{vmatrix}+\begin{vmatrix}
a & 1-t\\
1& a
\end{vmatrix}
\end{align*}
We used the first row cofactor expansion in the second equality.

After we compute three $2 \times 2$ determinants and simply, we obtain
$\det(A-tI)=-t(t^2-3t+2-2a^2).$ The eigenvalues of $A$ are roots of this characteristic polynomial. Thus, eigenvalues are
$0, \quad \frac{3\pm \sqrt{1+8a^2}}{2}$ by the quadratic formula.

Now the only possible way to obtain a multiplicity $2$ eigenvalue is when
$\frac{3- \sqrt{1+8a^2}}{2}=0$ and it is straightforward to check that this happens if and only if $a=1$.

Therefore, when $a=1$ eigenvalues of $A$ are $0$ with algebraic multiplicity $2$ and $3$ with algebraic multiplicity $1$.

When $a \neq 1$, eigenvalues are
$0, \quad \frac{3\pm \sqrt{1+8a^2}}{2}$ and each of them has algebraic multiplicity $1$.

Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem)