# matrix product with trace zero

$D$ 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 $AB=0$?

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 $$AB = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}.$$
• 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
• 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
• 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