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Consider a real symmetric matrix $A \in\mathbb{R}^{n\times n}$ and the following function

$$ f(A) = 2\cdot\text{tr}(A^\top A) - \Big(\text{tr}(A)\Big)^2 $$

To minimize $f(A)$ we can differentiate and set the derivative to zero. Which implies that $A^* = \frac{1}{2}\text{tr}(\text{A}^*)\mathbb{I}$. The optimal solution suggests that the matrix $A^*$ must be diagonal. However, what if diagonal matrices were not allowed ? Can we find a non-diagonal $A$ that minimizes $f(A)$ ?

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  • $\begingroup$ However, note that $\text{tr}(x \mathbb I) = n x$, so if $n \ne 2$ ... $\endgroup$ Nov 20, 2019 at 20:42

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No. Let $\langle A, B \rangle = \operatorname{tr} (A^TB)$. This is an inner product.

Let $\phi(A) = 2 \operatorname{tr}(A^TA) - (\operatorname{tr}(A))^2 = 2\langle A, A \rangle - ( \langle I , A \rangle )^2$.

Then $D\phi(A)(H) = 4 \langle A, H \rangle -2 \langle I, A \rangle\langle I, H \rangle = 2 \langle 2A - \langle I, A \rangle I , H \rangle$ and if we set the derivative to zero we get $2A = \langle I, A \rangle I$ or $A = ({1 \over 2} \operatorname{tr} A ) I$.

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  • $\begingroup$ That is what I have. However, there must be a non-diagonal $B$ such that $|f(B) - f(A^*) | \leq \epsilon$. Is it possible to find such a $B$ $\endgroup$
    – Kumar
    Nov 20, 2019 at 20:36
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    $\begingroup$ Of course. The function is continuous. $\|E_{1,n}\|_F = 1$. $\endgroup$
    – copper.hat
    Nov 20, 2019 at 20:37
  • $\begingroup$ So how do I find such a $B$ ? $\endgroup$
    – Kumar
    Nov 20, 2019 at 20:41
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    $\begingroup$ Let $E_{1,n}$ be the zero matrix except for the $1,n$ entry which is one. Compute $\phi(A+t E_{1,n})$. This will be a quadratic in $t$. Pick $t$ small enough to satisfy your above inequality. $\endgroup$
    – copper.hat
    Nov 20, 2019 at 20:48

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