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Is the following statement true: If $A\succeq 0$ and $B \preceq 0$ are two $n\times n$ real-valued matrices, then $AB \preceq 0$.

If not, is it true that $\forall x\ge 0$ (i.e., all vectors in the positive orthant), $x^TABx \le 0$?

If not, is the second statement true if all entries of $B$ are non-positive.

I'd really appreciate help with this.

Thanks!

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In MATLAB notation (sorry I don't know how to type matrices here), if A = [1 -2; -2 4] and B = [-1 -2; -2 -4] then $A$ is nonnegative definite, $B$ is nonpositive definite and has nonpositive entries, and $[1,~0] AB [1,~0]^T=3$. –  alex o. May 17 '12 at 5:48

1 Answer 1

up vote 6 down vote accepted

No, $AB$ won't even be Hermitian in general. The correct formulation is that $A^{1/2}BA^{1/2} \preceq 0$.

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However, it is true that all eigenvalues of $AB$ are negative. This is because $AB=A^{1/2}(A^{1/2}BA^{1/2})A^{-1/2}$. –  Federico Poloni May 17 '12 at 8:01

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