# Common Eigenvector of Two Matrices $A, B$ is Eigenvector of $A+B$ and $AB$.

## Problem 382

Let $\lambda$ be an eigenvalue of $n\times n$ matrices $A$ and $B$ corresponding to the same eigenvector $\mathbf{x}$.

(a) Show that $2\lambda$ is an eigenvalue of $A+B$ corresponding to $\mathbf{x}$.

(b) Show that $\lambda^2$ is an eigenvalue of $AB$ corresponding to $\mathbf{x}$.

(The Ohio State University, Linear Algebra Final Exam Problem)

## Proof.

### (a) Show that $2\lambda$ is an eigenvalue of $A+B$ corresponding to $\mathbf{x}$.

Since $\lambda$ is an eigenvalue of $A$ and $B$, and $\mathbf{x}$ is a corresponding eigenvector, we have
$A\mathbf{x}=\lambda \mathbf{x} \text{ and } B\mathbf{x}=\lambda \mathbf{x} \tag{*}.$ Then we compute
\begin{align*}
(A+B)\mathbf{x}&=A\mathbf{x}+B\mathbf{x}\\
&=\lambda \mathbf{x}+ \lambda \mathbf{x} && \text {by (*)}\\
&=2\lambda \mathbf{x}.
\end{align*}

Since $\mathbf{x}$ is an eigenvector, it is a nonzero vector by definition.
Hence from the equality
$(A+B)\mathbf{x}=2\lambda \mathbf{x},$ we see that $2\lambda$ is an eigenvalue of the matrix $A+B$ and $\mathbf{x}$ is an associated eigenvector.

### (b) Show that $\lambda^2$ is an eigenvalue of $AB$ corresponding to $\mathbf{x}$.

We have
\begin{align*}
(AB)\mathbf{x}&=A(B\mathbf{x})\\
&=A(\lambda \mathbf{x}) && \text{by (*)}\\
&=\lambda (A\mathbf{x})\\
&=\lambda (\lambda \mathbf{x}) && \text{by (*)}\\
&=\lambda^2 \mathbf{x}.
\end{align*}

Since $\mathbf{x}$ is a nonzero vector as it is an eigenvector, it follows from the equality
$(AB)\mathbf{x}=\lambda^2 \mathbf{x}$ that $\lambda^2$ is an eigenvalue of the matrix $AB$ and $\mathbf{x}$ is a corresponding eigenvector.

##### Prove that the Length $\|A^n\mathbf{v}\|$ is As Small As We Like.
Consider the matrix $A=\begin{bmatrix} 3/2 & 2\\ -1& -3/2 \end{bmatrix} \in M_{2\times 2}(\R).$ (a) Find the eigenvalues and corresponding eigenvectors...