What Are the Eigenvalues of A Transpose A?

  • #1
3.141592654
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Homework Statement



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

Homework Equations





The Attempt at a Solution



If eigenvalues exist, then

[itex]A^{T}A[/itex]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 [itex]A^{T}A[/itex] then

[itex]A^{T}A[/itex]x = (0)x where x ≠ 0.

N(A) ≠ {0}, so Ax = 0 where x ≠ 0.

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

Is there anything else that can be said about the eigenvalues for this matrix?
 
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  • #2
Did you hear about the spectral theorem for symmetric matrices?
 
  • #3
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
3.141592654 said:
No, that wasn't covered in my course so I suppose that's not what the professor is looking for. Is it relatively ea

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
3.141592654 said:
No, that wasn't covered in my course so I suppose that's not what the professor is looking for. Is it relatively ea

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

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