# Eigenvalues of A transpose A

1. Dec 7, 2011

### 3.141592654

1. The problem statement, all variables and given/known data

Let A be an m x n matrix with rank(A) = m < n. As far as the eigenvalues of $A^{T}A$ is concerned we can say that...

2. Relevant equations

3. The attempt at a solution

If eigenvalues exist, then

$A^{T}A$x = λx where x ≠ 0.

The only thing I think I can show is that 0 is an eigenvalue:

If 0 is an eigenvalue for $A^{T}A$ then

$A^{T}A$x = (0)x where x ≠ 0.

N(A) ≠ {0}, so Ax = 0 where x ≠ 0.

Therefore $A^{T}(Ax) = 0$ where x ≠ 0. So λ = 0 is an eigenvalue for $A^{T}A$.

Is there anything else that can be said about the eigenvalues for this matrix?

2. Dec 7, 2011

### I like Serena

Did you hear about the spectral theorem for symmetric matrices?

3. Dec 7, 2011

### 3.141592654

No, that wasn't covered in my course so I suppose that's not what the professor is looking for. Is it relatively ea

4. Dec 7, 2011

### Dick

If A^T*A*x=lambda*x what happens if you multiply both sides on the left by x^T? No, you don't need the spectral theorem.

5. Dec 8, 2011

### I like Serena

Spectral theorem says that a symmetric matrix is diagonalizable.
In particular, a real nxn symmetric matrix has n real eigenvalues.