MHB Singular Values and Eigenvalues

[linearishard]

Hi, one more question!

How do I prove that A has eigenvalues equal to its singular values iff it is symmetric positive definite? I think I have the positive definite down but I can't figure out the symmetric part. Thanks!

 

[Greg]

It will help if you post the work you already have.

 

[linearishard]

All I have is that if a singular value is the eigenvalue of ATA, then A must be positive semi definite or the signs will be different on at least one eigenvalue. I don't know where to start with symmetry or if my assumption is correct.

 

[I like Serena]

Are you familiar with the Spectral Theorem?
That is that every real symmetric matrix is diagonalizable?

So if the matrix is symmetric and has real numbers as its elements, it is diagonalizable, which means that it has a full set of real eigenvalues and corresponding eigenvectors that span the vector space.

 

[Euge]

Hello again, linearishard,

Usually, positive definite matrices are assumed to be symmetric (or Hermitian for complex matrices). Does your teacher's definition of positive definite exclude the symmetry assumption? In any case, the problem statement is not true. For since singular values of a matrix can be zero, having eigenvalues of $A$ equal to the singular values of $A$ does not necessarily result in every eigenvalue being positive (which is what you need to claim positive definiteness).

The forward conditional is true, however. if $A$ is symmetric positive definite, the eigenvalues of $A$ are positive. The eigenvalues of $A$ are square roots of the eigenvalues of $A^2 = A^TA$, so the singular values of $A$ are the eigenvalues of $A$.