Eigenvalues of conjugate transpose matrix

• Sep 6th 2009, 03:12 AM
sssouljah
Eigenvalues of conjugate transpose matrix
Hello
how do i prove that all the eigenvalues of the product A*A (where A* is the conjugate transpose matrix of A) are real non negative numbers. note A is not necessarily a square matrix ?

Letting t be an eueigenval of A*A, with eigenvector v. Then we have A*Av = tv, and taking the inner product with v on both sides implies that v*A*Av = tv*v.

But v*A*Av = |Av|2,

How do I finish it off from here?

Cheers

• Sep 6th 2009, 04:11 AM
HallsofIvy
Quote:

Originally Posted by sssouljah
Hello
how do i prove that all the eigenvalues of the product A*A (where A* is the conjugate transpose matrix of A) are real non negative numbers. note A is not necessarily a square matrix ?

Letting t be an eueigenval of A*A, with eigenvector v. Then we have A*Av = tv, and taking the inner product with v on both sides implies that v*A*Av = tv*v.

But v*A*Av = |Av|2,

How do I finish it off from here?

Cheers

The crucial point is the basic property of the "conjugate transpose": $(Au)\cdot v= u\cdot (A*v)$, where " $\cdot$ is an inner product over the complex numbers: $u\cdot v= \overline{v\cdot u}$.
Suppose $\lambda$ is an eigenvalue of A*A and v is an eigenvector, corresponding to $\lambda$ with length 1. Then
$\lambda= \lambda(v\cdot v)= (\lambda v)\cdot v= (A*Av)\cdot v$ $= (Av)\cdot(A**v)= (Av)\cdot(Av)= v\cdot (A*Av)= v\cdot(\lambda v)$ $= \overline{\lambda}(v\cdot v)= \overline{\lambda}$.

That is, $\lambda= \overline{\lambda}$ and so $\lambda$ is real.