# Common Eigenvector of Two Matrices and Determinant of Commutator

## Problem 13

Let $A$ and $B$ be $n\times n$ matrices.
Suppose that these matrices have a common eigenvector $\mathbf{x}$.

Show that $\det(AB-BA)=0$.

Contents

## Steps.

1. Write down eigenequations of $A$ and $B$ with the eigenvector $\mathbf{x}$.
2. Show that AB-BA is singular.
3. A matrix is singular if and only if the determinant of the matrix is zero.

## Proof.

Let $\alpha$ and $\beta$ be eigenvalues of $A$ and $B$ such that the vector $\mathbf{x}$ is a corresponding eigenvector.
Namely we have $A \mathbf{x}=\alpha \mathbf{x}$ and $B\mathbf{x}=\beta \mathbf{x}$.

Then we have
\begin{align*}
(AB-BA)\mathbf{x}&=AB\mathbf{x}-BA\mathbf{x}=A(\beta \mathbf{x}) -B( \alpha \mathbf{x}) \\
& = \beta A \mathbf{x}- \alpha B\mathbf{x} =\beta \alpha -\alpha \beta=0.
\end{align*}

By the definition of eigenvector,  $\mathbf{x}$ is a non-zero vector. Thus the matrix $AB-BA$ is singular.
Equivalently the determinant of $AB-BA$ is zero.

## Comment.

This is a simple necessary condition that $A$ and $B$ have a common eigenvector.

Here are few derived questions.

• Is this a sufficient condition?
• If so prove it.
• If not give a counterexample,
• and find a necessary and sufficient condition.

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