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If $w$ is an eigenvector of the $n \times n$ matrix $A$ with a corresponding eigenvalue $\lambda$, prove that $w$ is also an eigenvector of the $n \times n$ matrix $B = A^2$. Find the eigenvalue $\mu$ corresponding to this eigenvector of $B$.

Linear algebra really trips me up, but we know that $w \lambda = wA$ but how does that help prove anything about the eigenvectors and eigenvalues of $B$? Is there a trick I'm not using to manipulate $w \lambda = wA$ to get $w\mu = wB$?

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If $Aw = \lambda w$, then $Bw = A^2 w = A(Aw) = A(\lambda w) = \lambda A w = \lambda^2 w$

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