0
$\begingroup$

I have matrix $A \in \mathbb{R}^{N \times N}$, such that $A(i,j)=trace(B_iCB_j), \forall ij$.

Matrices $B_i$ and C are PSD and symmetric with positive entries. Can I prove that $A$ is PSD too?

In fact, I've tried different random matrices for $C$ and $B_i$ with different dimensions and $A$ was always PSD.

$\endgroup$
2
$\begingroup$

For any $x \in \mathbb{R}^n$ I will prove that $x^TAx \geq 0$. Let $x \in \mathbb{R}^n$, then $$\begin{align}x^TAx &= \sum_{i,j} x_i x_j \text{tr}(B_iCB_j) \\ &= \text{tr}(\sum_{i,j} x_i x_j B_iCB_j) \\ &= \text{tr}(\sum_{i} \sum_j (x_i B_i)C (x_j B_j)) \\ &= \text{tr}((\sum_{i} x_i B_i)C(\sum_j x_j B_j)) \\ \end{align}$$ Let $D = \sum_{i} x_i B_i$, then this expression equals $\text{tr}(DCD)$. Since $D$ is symmetric, $DCD=D^TCD$, which is positive semidefinite, so the trace is indeed nonnegative.

$\endgroup$
  • $\begingroup$ Note that I never used that the entries of the matrices are positive. $\endgroup$ – LinAlg Jan 18 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.