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I'm trying to show that Tr$\left(\sqrt{(\mathbf{PXP}^\top)}\right) \le \text{c Tr}\left(\mathbf{P}\sqrt{\mathbf{X}}\mathbf{P}^\top \right)$

where Tr is the trace operator, $\mathbf{X}$ is symmetric positive semi-definite matrix, $\mathbf{P}$ is a non-symmetric low rank matrix, and $c$ is a positive constant. Is this statement always true and if so how can it be shown?

user113440
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  • If you want this question to be accessible to the widest possible readership then you might like to define the square root for a matrix. Is it true that

    $$\sqrt{I_2} \subset \left{ X = \left[\begin{array}{cc} a & b \ \frac{1-a^2}{b} & -a \end{array}\right] : a,b \in \mathbb{C}, \ b \neq 0 \right}$$

    All have the property that $X^2 = I_2$.

     – Fly by Night Dec 04 '13 at 17:52
  • What do you mean by "low rank" exactly? – Greg Martin Dec 04 '13 at 18:17
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    @FlybyNight In functional analysis, at least, the square root of a positive (i.e., symmetric positive semi-definite) matrix or operator is always taken to be its unique positive (i.e., symmetric positive semi-definite) square root. So, $\sqrt{I_2} = I_2$ without any ambiguity at all. I can't speak for other branches of mathematics, though. – Branimir Ćaćić Dec 04 '13 at 18:27
  • is $c$ given? Note that for any choice of those matrices, one can choose an arbitrarily large $c$ and make the inequality true. – dineshdileep Dec 05 '13 at 04:46
  • if $c=1$, then this is not true... – dineshdileep Dec 05 '13 at 04:49
  • To answer Greg's question, I guess I would say a low rank matrix is a rank-$r$ matrix $\mathbf{P}$ of size $m \times n$, where $r << \min(m,n)$. – user113440 Dec 05 '13 at 19:02
  • To answer dineshdileep's question, I guess I should re-state my problem to one where I want to find out if the RHS happens to be an upper bound of the LHS of the equation in the original posting above, where c is some positive constant. I'm looking to solve an engineering problem defined by the LHS, but I think I can solve the RHS. So I want to show that the true solution is bounded by what I obtain..... – user113440 Dec 05 '13 at 19:06

1 Answers1

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This is not true. Let $X=I_n$ and consider the family of real matrices of the form $P=\pmatrix{0&\epsilon\\ 0&0}\oplus0_{(n-2)\times(n-2)}$ with $\epsilon>0$. The inequality in question then becomes $\epsilon\le c\epsilon^2$ for all $\epsilon>0$, which is obviously wrong.

user1551
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