# Eigen Values Proof

a.) Let A and B be $n$ x $n$ matrices. Prove that the matrix products $AB$ and $BA$ have the same eigenvalues.

b.) Prove that every eigenvalue of a matrix A is also an eigenvalue of its transpose $A^T$. Also, prove that if v is an eigenvector of A with eigenvalue $\lambda$ and w is an eigenvector of $A^T$ with a different eigenvalue $\mu \ne \lambda$, then v and w are orthogonal vectors with respect to the dot product.

For a, I know that if their eigenvalues are the same then their eigenvectors must relate too.

For the first part of b, is it similar to proving that the $det(A) =det (A^T)$? And, I do not know how to do the second part.

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(a) Let $\lambda$ be an eigenvalue of $AB$. Then there is a vector $v$ such that $$ABv=\lambda v.$$ Do you notice anything - in terms of eigenvectors of $BA$ - if you left-multiply by $B$? You should be able to show from here that every eigenvalue of $AB$ is also an eigenvalue of $BA$. Then do the same in reverse.

(b) You are right about the first part being related to equality of the determinant of a square matrix and the determinant of its transpose. A necessary and sufficient condition for $\lambda$ to be an eigenvalue of $A$ is that $\det(A-\lambda I)=0$, where $I$ is the $n\times n$ identity. Can you see what this would tell you about $\det(A^T-\lambda I)$?

For the second part, look at the equation $$Av\cdot w=w\cdot Av$$ and use the representation $x\cdot y=x^Ty$. On the left side, you should be able to bring in the eigenvalue $\mu$ and on the right side you can bring in the eigenvalue $\lambda$.

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I think you're missing a transpose in that second displayed equation. –  Gerry Myerson Nov 8 '12 at 1:56
Thank you very much! –  diimension Nov 8 '12 at 8:12
@Gerry - ? $x\cdot y=y\cdot x$ for $x,y\in\mathbb{R}^n$ which I understand is the context. –  user12477 Nov 8 '12 at 11:53
Sorry, I read too fast, thought you were claiming $Av\cdot w=v\cdot Aw$. –  Gerry Myerson Nov 8 '12 at 11:58
On the left, $Av\cdot w=(Av)^Tw=v^T(A^Tw)=v^T(\mu w)=\mu (v\cdot w)$. On the right, $w\cdot Av=w\cdot(\lambda v)=\lambda (w\cdot v)=\lambda(v\cdot w)$. So $(\lambda-\mu)v\cdot w=0$ and $\lambda\neq\mu$ gives the result. –  user12477 Nov 12 '12 at 14:12