# Trace of PSD matrices [closed]

Let matrices $$A, B \in \mathbb{R}^{n \times n}$$ be positive semidefinite (PSD). Let $$B = V V^T$$, where $$V \in \mathbb{R}^{n \times n}$$. Why does the last equality hold in the following?

$$\mbox{tr} (A B) = \mbox{tr} \left(A V V^T\right) = \mbox{tr} \left(V^T A V\right)$$

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Feb 10 at 23:10
• Once more, this robot should be silent : the question is clear ! Feb 10 at 23:14
• @geetha290krm Please note that my comment was not about the facts you mention. Feb 10 at 23:19

That's due to the cyclic property of the trace operator: $$\operatorname{tr}(\mathbf{A}\mathbf{B}\mathbf{C}\mathbf{D}) = \operatorname{tr}(\mathbf{B}\mathbf{C}\mathbf{D}\mathbf{A}) = \operatorname{tr}(\mathbf{C}\mathbf{D}\mathbf{A}\mathbf{B}) = \operatorname{tr}(\mathbf{D}\mathbf{A}\mathbf{B}\mathbf{C}).$$
• Makes sense, I can't believe I didn't see $trace((AV)V^T) = trace(V^T(AV))$ Feb 11 at 0:52