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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 (108k points)

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