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$D$ is a positive definite matrix, $A$ and $B$ are both positive semidefinite matrices, $c$ is a postive integer. I want to know whether $trace\{(A+B+cI)^{-1}ABD\}=0$ implies that $AB=0$?

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up vote 4 down vote accepted

No. Let $$A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\!, \quad B = \begin{bmatrix} 1 \\ & 0 \end{bmatrix}\!, \quad D = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\!, \quad c = 1.$$ Then $$(A + B + cI)^{-1} A B D = \frac{1}{5}\begin{bmatrix} 2 & -1 \\ 4 & -2 \end{bmatrix}\!,$$ so $\mathop{\rm tr}((A + B + cI)^{-1} A B D) = 0$, but $$AB = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}.$$

Acknowledgment: Example fixed by Sebastien B's comments.

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I hope I did not make another dumb mistake, but when I compute $(A+B+cI)^{-1} ABD$ I get $(A+B+cI)^{-1}=\frac{1}{5}\begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$, the same $AB$ as you and thus $(A+B+cI)^{-1} ABD=\frac{1}{5}\begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix}$ whose trace is nonvanishing. –  Sebastien B Jul 13 '13 at 13:27
    
You're right; I had a typo in Mathematica code I used to calculate this. I'll try to look into it some more. Thanks. –  Vedran Šego Jul 13 '13 at 13:31
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But if you take $D=\begin{bmatrix} 1 & -1/2 \\ -1/2 & 1 \end{bmatrix}$, I think your example is fine: $(A+B+cI)^{-1} ABD=\frac{1}{5}\begin{bmatrix} 1 & -1/2 \\ 2 & -1 \end{bmatrix}$. –  Sebastien B Jul 13 '13 at 13:31
    
Thank you. I've multiplied your $D$ by $2$, to keep it visually nicer (less fractions). –  Vedran Šego Jul 13 '13 at 13:39
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Take the same example, for $e = 1$. This will give you $$D = \frac{1}{5} \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$$ and the rest will be more or less the same (factor $1/5$). This is due to the structure of $B$, so the second row of $D$ has no effect on your expression, and the rest is as above (just divided by $5$, which doesn't affect the fact that the trace is zero). –  Vedran Šego Jul 13 '13 at 15:15
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