# [solved] trace(A*A) is always positive?

1. Jul 23, 2010

### zeebek

I have a feeling that for any n x n non-trivial matrix A, trace(A*A) is always positive.
Is it true?

2. Jul 23, 2010

### zeebek

Nevermind, I found contre example.

3. Jul 26, 2010

### Landau

I'm eager to hear your counter-example, because the statement is true.

4. Jul 26, 2010

### HallsofIvy

Let $$A= \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$$

5. Jul 26, 2010

### Landau

@HallsofIvy: is this meant to be a counter-example? With this A, I get A*A=I (the identity matrix), which has trace equal to 2.

My proof: for an arbitrary matrix A, the product A*A is self-adjoint (because (A*A)*=A*A) and positive semidefinite (because $(A^*Av,v)=(A^*v,A^*v)=\|A^*v\|^2=\|Av\|^2$). Hence A*A has an orthonormal basis of eigenvectors, i.e. is diagonalizable with non-negative eigenvalues. The trace is then the sum of the eigenvalues, which is non-negative.

6. Jul 27, 2010

### HallsofIvy

Ah. I has assumed that "A*A" simply meant A times itself. Otherwise, as you say, there is no "counterexample".

7. Sep 1, 2010

### zeebek

Now I understand that I was wrong. A*A is indeed positive semidefinite
thanks to everybody