a) Counterexample: Consider the matrices

\[A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\]

It follows that \(A B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0\). The eigenvector of \(B\) with eigenvalue 1 is \([0,1]^T\), but this is not an eigenvector of \(A\), as \(A[0,1]^T=[0,0]^T\). Hence, the statement is false.

b) Proof: If \(A B=0\), then for any eigenvector \(v\) of \(B\) with eigenvalue \(\lambda\), we have

\[A B v=\lambda B v=\lambda v\]

Since \(A B=0\), it follows that

\[A v=0\]

Hence, every eigenvector of \(B\) is also an eigenvector of \(A\), with eigenvalue 0 .