# 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 $\det(AB-BA)=0$.

Add to solve later

Sponsored Links

Contents

## Steps.

- Write down eigenequations of $A$ and $B$ with the eigenvector $\mathbf{x}$.
- Show that AB-BA is singular.
- 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.

Add to solve later

Sponsored Links