Question

Let \(A\) and \(B\) be \(n \times n\) matrices with the property that \(A B=0 .\) For each of the following give a proof or counterexample.

a) Every eigenvector of \(B\) is also an eigenvector of \(A\).
b) At least one eigenvector of \(B\) is also an eigenvector of \(A\).

Answer 1

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 .