# Problem on Matrix. help

1. May 17, 2009

### BiBByLin

Suppose A is a non-singular matrix. AT is the transpose matrix of A. Therefore, the eigenvalue of A can be expressed as:
S-1AS=Λ
(S-1AS)TT
(AS)TS-T=STATS-T
So, AT and A share the same eigenvalue and eigenvector.
Here, x is the base eigenvector of A. Hence, span{x} is the eigenvector of A.
ATAx=ATΛx.
We consider the new eigenvector Λx as the eigenvector of A or AT.
Therefore:
ATAx=ATΛx=ΛΛx.
So, ATA is symmetric positive definite.

Is my proof right?

It is very urgent for my program, pls help me. Thanks a lot!

2. May 18, 2009

### HallsofIvy

I can see a dozen different things wrong with it. First you talk about "the" eigenvalue when a matrix may have many eigenvalues. Second,you say "S-1AS= =Λ" with out saying what S or Λ are!

You say "Here, x is the base eigenvector of A. Hence, span{x} is the eigenvector of A."
What do you mean the "the base eigenvector"? span{x} is a subspace spanned by x, not a vector at all so it cannot be "the eigenvector of A"- and A is unlikely to have only one eigenvector so it makes no sense to talk about "the eigenvector of A".

Some of these (the use of "the" in particular) are probably just English language problems but I suspect some basic misunderstanding as well.

Last edited by a moderator: May 19, 2009
3. May 18, 2009

### maze

Consider the SVD.

4. May 18, 2009

### BiBByLin

Thanks HallsofIvy.
I just want to know whether ATA is symmetric positive definite if A is a N*N matrix, and A is a non-singular matrix.

5. May 19, 2009

### maze

Do you know about the singular value decomposition?

6. May 19, 2009

### BiBByLin

Not, I dont know.

I will study it. Can you give me some hint?

7. May 19, 2009

### maze

Hmm. If you don't know the SVD, it may be easier to prove it in a more direct manner, so keep that in mind. The SVD is a decomposition that any matrix has, with the form M=UEV*, where U and V are unitary and E is diagonal with positive or zero entries on the diagonal in decreasing order. The SVD is unique up to phase factors in U and V, and can be made unique if a convention is chosen.

Already we see the SVD is very similar to the eigenvalue decomposition, but it is not quite the same - U and V are always unitary, and not necessairily inverses of each other, whereas in the eigenvalue decomposition (PDP-1), P and P-1 are inverses that might not be unitary. The entries of E are nonnegative real numbers, whereas D could have negative or complex entries on the diagonal. Also, every matrix has a SVD, whereas not all matrices have an eigenvalue decomposition.

The geometric intuition behind the SVD is that, a matrix maps the unit sphere to a hyperellipse. The vectors in U are the principal axes of the hyperellipse, and the vectors in V are the vectors on the unit sphere that get mapped to the principal axes. The values on the diagonal in E are the scaling factors for each ellipse axis.

To apply the SVD to your problem, try:
A*A = (UEV*)*(UEV*) = VEU*UEV* = VE2V*

Since E2 is diagonal and V-1 = V*, this is the eigenvalue decomposition of A*A. I will leave it to you to verify that, VE2V* is symmetric positive definite.

8. May 19, 2009

### jbunniii

It's hard to tell exactly what you are assuming and what you are deducing. It's also not clear if you are assuming that you are working with a vector space over the REAL field. If not, you need to use the conjugate (Hermitian) transpose below.

This is certainly NOT true in general. It is possible if and only if A is diagonalizable, which is true if and only there exists a full linearly independent set of eigenvectors. If you're working over the real field, there's not even any guarantee that your matrix has ANY eigenvalues or eigenvectors, even if it's nonsingular. (Example: rotation matrix.)

Second, your manipulation doesn't show that A and $$\Lambda$$ have the same eigenvectors. You seem to be assuming that S is symmetric, but why should it be?

Fortunately, your question can be answered without reference to eigenvalues or eigenvectors. A symmetric matrix (in a real vector space - otherwise replace with "Hermitian") is positive definite if and only if

$$x^T M x > 0$$ for all nonzero vectors x

Why don't you use this test directly for $$M = A^T A$$? The answer should pop right out at you.

Last edited: May 19, 2009
9. May 24, 2009

### BiBByLin

great! Thank you very much! Maze, and JBunniii