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Let $A,B$ be $n\times n$ matrix, denote by $A\circ B=(a_{ij}b_{ij})$.

Let $C$ be an invertible real matrix. $D=diag(x_1,\cdots,x_n)$, $F=CDC^{-1}$. Show that $\sum(F\circ F)\geq x_1^2+\cdots+x_n^2$, where $\sum$ means the sum of all entries of $F\circ F$.

If $C$ is the identity matrix, it is OK. What about other matrices?

xldd
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    Is $F \circ F$ just matrix product? If so, use the cyclic property of the trace $$\mathrm{tr}(F^2) = \mathrm{tr}(CDC^{-1} CDC^{-1}) = \mathrm{tr}(CD^2C^{-1}) = \mathrm{tr}(D^2C^{-1}C) = \mathrm{tr}(D^2) = \sum x_i^2.$$ – William M. Aug 12 '21 at 01:02
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    @WillM. It seems to be element-wise multiplication, not matrix product. – angryavian Aug 12 '21 at 01:03
  • @WillM. The operation $\circ$ is defined in the question. – azif00 Aug 12 '21 at 01:05
  • Oh, yes, I clearly only read the title. I suppose it will have to do with the cycle property but it is not so straightforward then. – William M. Aug 12 '21 at 01:05
  • @WillM. Since the function of interest has shifted to $\operatorname{tr}(F^TF):;$ if $C$ is orthogonal, then $F$ is symmetric, which takes us back to the equality in your original comment. – greg Aug 12 '21 at 16:51
  • @greg OP (original poster) also asked it here: https://math.stackexchange.com/questions/4222739/a-lower-bound-for-the-frobenius-norm – William M. Aug 12 '21 at 17:00

1 Answers1

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$ \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} $Here is a small $(n=3)\,$ counter-example $$\eqalign{ &C = \m{ -1 & 6 & -6 \\ 4 & 3 & 1 \\ -3 & 2 & 4 \\} \quad x = \m{1\\2\\6} \\ &F = C\cdot\Diag{X}\cdot C^{-1} = \frac 1{113}\m{ 425 & -174 & -336 \\ -14 & 186 & 102 \\ -151 & 182 & 406 \\} \\\\ &\trace{F\circ F} = \frac{425^2+186^2+406^2}{113^2} \approx 29.76404 \\ &\|x\|^2 = 41 \\ }$$

greg
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