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.

Solution 1:

(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 $\lambda$.