# trace of product of positive definite matrix

For positive definite matrices $A$ and $C$, positive semidefinite matrices $B$ and $D$, I want to know whether $tr\{ABCD\}=0$ implies that $tr\{BD\}=0$.

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If I add a constraint on the setting. For positive definite matrices $A$ and $C$, positive semidefinite matrices $B$ and $D$. This time $A=(B+D+eI)^{-1}$, $C=(B+D+fI)^{-1}$ where $I$ is the identity matrix, $e$ and $f$ are both positive numbers. It is obvious that under this setting $tr\{BD\}=0$ implies $tr\{ABCD\}=0$. However I want to know whether $tr\{ABCD\}=0$ implies that $tr\{BD\}=0$. – nuse_li Jun 21 '13 at 12:42

No. Let $B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $D=\begin{bmatrix}0&0\\0&1\end{bmatrix}$, and $A=C=\begin{bmatrix}2&1\\1&1\end{bmatrix}$. Then $\mathrm{tr}(BD)=0$, but $\mathrm{tr}(ABCD)=1$.

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Thanks! but whether $tr\{ABCD\}=0$ implies $tr\{BD\}=0$? – nuse_li Jun 21 '13 at 8:29
Jim, you showed him the wrong implication. – Vedran Šego Jun 21 '13 at 12:16
@VedranŠego He didn't. The OP edited the question after Jim answered. – user1551 Jun 21 '13 at 13:02
Ah, I apologize then. – Vedran Šego Jun 21 '13 at 13:03

No. Let

$$A := {\rm I}_2 := \begin{bmatrix} 1 \\ & 1 \end{bmatrix}, \quad B := \begin{bmatrix} 1 \\ & 0 \end{bmatrix}, \quad C := \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}, \quad D := \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}.$$

Then: $$ABCD = 0, \quad BD = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix},$$ so $$\mathop{\rm tr}(ABCD) = 0, \quad \mathop{\rm tr}(BD) = 1.$$

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