(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.

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