Proving Eigenvalue Squared Is Eigenvalue of $A^2$

Proving Eigenvalue Squared Is Eigenvalue of $A^2$
$\begingroup$

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.

$\endgroup$
2

2 Answers

$\begingroup$

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. :)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

James H. Sterling
Author

James H. Sterling

James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.