TITLE: Proving Eigenvalue squared is Eigenvalue of $A^2$
QUESTION [13 upvotes]: The question is: Prove that if $\lambda$ is an eigenvalue of a matrix A with corresponding eigenvector x, then $\lambda^2$ is an eigenvalue of $A^2$ with corresponding eigenvector x.
I assume I need to start with the equation $Ax=\lambda x$ and end up with $A^2 x=\lambda^2 x$ but between those I am kind of lost. I have manipulated the equations several different ways and just can't seem to end up where I need to be. Help would be greatly appreciated as I believe this will be on a test tomorrow.

REPLY [27 votes]: We know $Ax = \lambda x$.  Then $A \lambda x = \lambda(Ax) = \lambda^2x = A^2x$.  Putting this into a more readable mathematical sentence, we get:
$$A^2x = A(Ax) = A\lambda x = \lambda(Ax) = \lambda^2x$$
You were done and didn't realize it.  :)

REPLY [5 votes]: You are on the right way: let $x$ an eigenvector of $A$ associated to the eigenvalue $\lambda$ so
$$Ax=\lambda x$$
and then apply $A$ we find
$$A(Ax)=A^2 x=A(\lambda x)=\lambda A x=\lambda\lambda x=\lambda^2 x$$
and conclude.