Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 4 down vote accepted

(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 $\lambda$.

share|improve this answer
    
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
1  
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.