# Prove the following about an eigenvector, $w$

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 $$w\mu = wB$$?

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