# Commuting Matrices $AB=BA$ such that $A-B$ is Nilpotent Have the Same Eigenvalues

## Problem 587

Let $A$ and $B$ be square matrices such that they commute each other: $AB=BA$.
Assume that $A-B$ is a nilpotent matrix.

Then prove that the eigenvalues of $A$ and $B$ are the same.

## Proof.

Let $N:=A-B$. By assumption, the matrix $N$ is nilpotent.
This means that there exists a positive integer $n$ such that $N^n$ is the zero matrix $O$.

Let $\lambda$ be an eigenvalue of $B$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$. That is, we have $B\mathbf{v}=\lambda \mathbf{v}$ and $\mathbf{v}\neq \mathbf{0}$.
We prove that $\lambda$ is also an eigenvalue of $A$.

Note that since $A$ and $B$ commute each other, it follows that the matrices $N$ and $B-\lambda I$ commute each other as well.
Then we compute
\begin{align*}
(A-\lambda I)^n&=(N+B-\lambda I)^n\\
&=\sum_{i=0}^n \begin{pmatrix}
n \\
i
\end{pmatrix}
N^i(B-\lambda I)^{n-i},
\end{align*}
where the second equality follows by the binomial expansion. (Note that the binomial expansion is true for matrices commuting each other.)
Then we have
\begin{align*}
(A-\lambda I)^n \mathbf{v}&=\sum_{i=0}^{n-1} \begin{pmatrix}
n \\
i
\end{pmatrix}
N^i(B-\lambda I)^{n-i}\mathbf{v}+N^n\mathbf{v}=\mathbf{0}
\end{align*}
since $(B-\lambda I)\mathbf{v}=\mathbf{0}$ and $N^n=O$.

This implies that there exists an integer $k$, $0\leq k \leq n-1$ such that
$\mathbf{u}:=(A-\lambda I)^k\mathbf{v}\neq \mathbf{0} \text{ and } (A-\lambda I)^{k+1}\mathbf{v}=\mathbf{0}.$

It yields that $(A-\lambda I)\mathbf{u}=\mathbf{0}$ and $\mathbf{u}\neq \mathbf{0}$, or equivalently $A\mathbf{u}=\lambda \mathbf{u}$.
Hence $\lambda$ is an eigenvalue of $A$.

This proves that each eigenvalue of $B$ is an eigenvalue of $A$.
Note that if $A-B$ is nilpotent, then $B-A$ is also nilpotent.
Thus, switching the roles of $A$ and $B$, we also see that each eigenvalue of $A$ is an eigenvalue of $B$.
Therefore, the eigenvalues of $A$ and $B$ are the same.

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