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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 $A$.
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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 .
by Platinum (93,241 points)

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