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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 \(A\).
in Mathematics by Platinum (108k points) | 411 views

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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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