# Find All Matrices Satisfying a Given Relation

## Problem 43

Let $a$ and $b$ be two distinct positive real numbers. Define matrices
$A:=\begin{bmatrix} 0 & a\\ a & 0 \end{bmatrix}, \,\, B:=\begin{bmatrix} 0 & b\\ b& 0 \end{bmatrix}.$

Find all the pairs $(\lambda, X)$, where $\lambda$ is a real number and $X$ is a non-zero real matrix satisfying the relation
$AX+XB=\lambda X. \tag{*}$

(The University of Tokyo Linear Algebra Exam)

Contents

## Hint.

1. Let $X=\begin{bmatrix} x_1 & x_2\\ x_3& x_4 \end{bmatrix}$ and compute (*).
2. Compare entries of matrices.
3. Rewrite it as a matrix equation.
4. The problem is now translated to a problem of eigenvalue/eigenvectors.

## Solution.

Let $X=\begin{bmatrix} x_1 & x_2\\ x_3& x_4 \end{bmatrix}$.
Then the relation (*) becomes
$a\begin{bmatrix} x_3 & x_4\\ x_1& x_2 \end{bmatrix} +b \begin{bmatrix} x_2 & x_1\\ x_4& x_3 \end{bmatrix} = \lambda \begin{bmatrix} x_1 & x_2\\ x_3& x_4 \end{bmatrix}.$ Comparing the entries of matrices, we obtain four equations
\begin{align*}
ax_3+b x_2 &=\lambda x_1\\
ax_4+ b x_1 &=\lambda x_2\\
ax_1+b x_4 &=\lambda x_3 \\
a x_2+b x_3 &=\lambda x_4.
\end{align*}
We rewrite these equations into the matrix equation
$\begin{bmatrix} -\lambda & b & a & 0 \\ b &-\lambda & 0 & a \\ a & 0 & -\lambda & b \\ 0 & a & b & -\lambda \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}.$ Thus the problem is to find all the pairs $(\lambda, X)$ satisfying this matrix equation.
Let
$C= \begin{bmatrix} 0 & b & a & 0 \\ b & 0 & 0 & a \\ a & 0 & 0 & b \\ 0 & a & b & 0 \end{bmatrix}.$

Then the problem is equivalent to find all eigenvalues and eigenvectors of the matrix $C$.
So we first compute the characteristic polynomial of $C$ to find eigenvalues.
We have
\begin{align*}
\det(C-\lambda I)&= \begin{vmatrix}
-\lambda & b & a & 0 \\
b &-\lambda & 0 & a \\
a & 0 & -\lambda & b \\
0 & a & b & -\lambda
\end{vmatrix}\\
&=\lambda^4-2(a^2+b^2)\lambda^2+(a^2-b^2)^2.
\end{align*}
(Use cofactor expansion and simplify to obtain this.)
We solve
$\lambda^4-2(a^2+b^2)\lambda^2+(a^2-b^2)^2=0$ for $\lambda$.
By the quadratic formula we have
\begin{align*}
\lambda^2&=a^2+b^2\pm \sqrt{(a^2+b^2)^2-(a^2-b^2)^2}\\
&=a^2+b^2\pm2ab =(a\pm b)^2.
\end{align*}
Hence we obtained four eigenvalues $\lambda=\pm (a\pm b)$.
Note that since we have four distinct eigenvalues, each eigenspace is one dimensional.

Now, let us find eigenvectors. First consider the eigenvalue $\lambda=a+b$.
In this case,
\begin{align*}
C-\lambda I =\begin{bmatrix}
-a-b & b & a & 0 \\
b & -a-b & 0 & a \\
a & 0 & -a-b & b \\
0 & a & b & -a-b
\end{bmatrix}.
\end{align*}
The eigenvector is the solutions of $(C-\lambda I)\mathbf{x}=\mathbf{0}$. The typical method to find eigenvector is to reduce the matrix $C-\lambda I$ into row echelon form.
But as we noted earlier each eigenspace is one dimensional, so if we find one nonzero vector solution, then it is a basis of the eigenspace.

For this specific matrix, we see that $\mathbf{x}=\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$ satisfies $(C-\lambda I)\mathbf{x}=\mathbf{0}$. Therefore all the eigenvectors corresponding to eigenvalue $\lambda=a+b$ is
$\mathbf{x}=\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}t$ for any nonzero scalar $t$.

Similarly, we find that
$\begin{bmatrix} -1 \\ 1 \\ 1 \\ -1 \end{bmatrix}t, \quad \begin{bmatrix} 1 \\ -1 \\ 1 \\ -1 \end{bmatrix}t, \quad \begin{bmatrix} 1 \\ 1 \\ -1 \\ -1 \end{bmatrix}t$ for any nonzero $t$ are all the eigenvector corresponding to eigenvalue $\lambda=-a-b, a-b, -a+b$, respectively.

Therefore the solution to the problem is that the pair $(\lambda, X)$ is one of the following paris.
\begin{align*}
\left(a+b, t\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}\right), \,\,
\left(-a-b, t \begin{bmatrix}
-1 & 1\\
1& -1
\end{bmatrix}\right),\\
\left(a-b, t\begin{bmatrix}
1 & -1\\
1& -1
\end{bmatrix}\right), \,\,
\left(-a+b, t\begin{bmatrix}
1 & 1\\
-1& -1
\end{bmatrix}\right)
\end{align*}
for nonzero $t$ (because $X$ must be a nonzero matrix.)

## Comment.

When we find an eigenvector, you could have used elementary row operations to reduce a matrix.
The reduction for this matrix is not impossible, but it takes time and a sheet of paper.

So I think it is better to use the fact that the eigenspace is 1 dimensional (for the current problem)
and guess one nonzero eigenvector like I did above.

Let $A$ be a $4\times 4$ real symmetric matrix. Suppose that $\mathbf{v}_1=\begin{bmatrix} -1 \\ 2 \\ 0 \\ -1 \end{bmatrix}$...