Is trace(A*A) always positive?

[zeebek]

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

 

[zeebek]

Nevermind, I found contre example.

 

[Landau]

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

 

[HallsofIvy]

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

 

[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.

 

[HallsofIvy]

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

 

[zeebek]

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