Is trace(A*A) always positive?

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Discussion Overview

The discussion centers around the question of whether the trace of the product of a matrix A with itself, denoted as trace(A*A), is always positive for any non-trivial n x n matrix A. The scope includes theoretical considerations and mathematical reasoning.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses a belief that trace(A*A) is always positive for any non-trivial matrix A.
  • Another participant claims to have found a counter-example, prompting interest in its details.
  • A specific matrix A is provided as a potential counter-example, which is a rotation matrix.
  • A subsequent participant calculates A*A for the provided matrix and finds that it equals the identity matrix, leading to a trace of 2, which is positive.
  • A proof is presented that argues A*A is self-adjoint and positive semidefinite, suggesting that its trace must be non-negative.
  • One participant acknowledges a misunderstanding regarding the notation A*A, realizing it does not refer to A multiplied by itself in a different context.
  • A final participant concedes that A*A is indeed positive semidefinite, indicating a shift in their understanding.

Areas of Agreement / Disagreement

The discussion reflects disagreement initially regarding the positivity of trace(A*A), but later contributions suggest a consensus that A*A is positive semidefinite, thus supporting the claim that trace(A*A) is non-negative.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the notation A*A and the definitions of non-trivial matrices. The resolution of the counter-example remains dependent on the interpretation of the matrix product.

zeebek
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I have a feeling that for any n x n non-trivial matrix A, trace(A*A) is always positive.
Is it true?
 
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Nevermind, I found contre example.
 
I'm eager to hear your counter-example, because the statement is true.
 
Let [tex]A= \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}[/tex]
 
@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 [itex](A^*Av,v)=(A^*v,A^*v)=\|A^*v\|^2=\|Av\|^2[/itex]). 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.
 
Ah. I has assumed that "A*A" simply meant A times itself. Otherwise, as you say, there is no "counterexample".
 
Now I understand that I was wrong. A*A is indeed positive semidefinite
thanks to everybody
 

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