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

  1. Jul 23, 2010 #1
    I have a feeling that for any n x n non-trivial matrix A, trace(A*A) is always positive.
    Is it true?
     
  2. jcsd
  3. Jul 23, 2010 #2
    Nevermind, I found contre example.
     
  4. Jul 26, 2010 #3

    Landau

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    I'm eager to hear your counter-example, because the statement is true.
     
  5. Jul 26, 2010 #4

    HallsofIvy

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    Let [tex]A= \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}[/tex]
     
  6. Jul 26, 2010 #5

    Landau

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    @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.
     
  7. Jul 27, 2010 #6

    HallsofIvy

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    Ah. I has assumed that "A*A" simply meant A times itself. Otherwise, as you say, there is no "counterexample".
     
  8. Sep 1, 2010 #7
    Now I understand that I was wrong. A*A is indeed positive semidefinite
    thanks to everybody
     
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