3
$\begingroup$

Let $A,C\in\mathbb{R}^{m\times n}$, $n\ge m$, $B\in\mathbb{R}^{n\times m}$, and $P$ be a real positive definite $m\times m$ matrix. Denote by $\mathcal{S}^n$ the space of $n\times n$ real symmetric matrices. Suppose that $AB$ is non-singular. Let us define

$$F := (AB)^{-1} CB (AB)^{-1}P(AB)^{-\top} + (AB)^{-1}P(AB)^{-\top}(CB)^\top (AB)^{-\top}.$$

Next, consider the scalar function $$ c(X)=\mathrm{tr}(B^{\top}XBF) $$ where $X\in\mathcal{S}^n$ is such that $B^{\top}XB\ne 0$ and $\mathrm{tr}(\cdot)$ denotes the trace operator.

Is it possible to find a suitable $X$ such that $c(X)$ turns into a quadratic expression, i.e., something of the form $$ c(X)=\mathrm{tr}(L^\top Y L), $$ where $L$ and $Y$ are functions of $A,B,C,P$ and $Y$ is positive definite?

Note. I made several attempts with different $X$'s, but without any success yet. Hence, I wonder whether there is a rationale to follow either to get a positive answer to my question or to prove that this is actually not doable.

Improve this question
$\endgroup$
2
  • $\begingroup$ What about $X=0,L=0,Y=I$? $\endgroup$
    – Federico Poloni
    Commented Dec 20, 2017 at 19:39
  • $\begingroup$ @FedericoPoloni: Right. I should have added that $B^\top XB\ne 0$ to rule out trivial cases. Just edited now. $\endgroup$
    – Ludwig
    Commented Dec 20, 2017 at 20:05

1 Answer 1

Reset to default
2
$\begingroup$

Nope. That would mean $c(X)$ is always nonnegative, but that can't be that way (unless it's always $0$) because it switches sign if you replace $C$ with $-C$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Right. Lastly, I wonder whether we could argue that if $c(X)=0$ then $CB$ must have some special form (besides the trivial one $CB=0$). $\endgroup$
    – Ludwig
    Commented Dec 21, 2017 at 6:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .