Find All Matrices Satisfying a Given Relation

Tokyo University Linear Algebra Exam Problems and Solutions

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)

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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.


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