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Let $X\in\mathbb{C}^{n\times n}$, $X\ne \frac{I}{n}$, be a positive definite unit trace matrix and let $\bar{X}(\gamma):=X+\gamma\left(\frac{I}{n}-X\right)$ with $\gamma\in\mathbb{R}$ and $I$ being the $n\times n$ identity matrix. Define $$ P_X := \bar{X}\left(\min_{\,\gamma\in\mathbb{R}} \left\{\, |\gamma|\,:\, \bar{X}(\gamma) \text{ is singular}\, \right\}\right) $$

My question: Is $P_X$ the best singular unit trace matrix approximation of $X$ w.r.t. the Frobenius norm?

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  • $\begingroup$ What does $\min \{\overline{X} : \overline{X} \text{ is singular }\}$ mean? The minimum value of a set of complex matrices ? $\endgroup$ – Gribouillis Aug 25 '17 at 9:08
  • $\begingroup$ @Gribouillis: $\bar{X}$ is a function of a real parameter $\gamma$ and the minimum is taken over $\gamma$'s. $\endgroup$ – Ludwig Aug 25 '17 at 9:10
  • $\begingroup$ So you mean $P_X = \bar{X}(\min\{\gamma : \bar{X}(\gamma) \text{ is singular}\})$. Why would such a $\gamma$ exist ? $\endgroup$ – Gribouillis Aug 25 '17 at 9:12
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    $\begingroup$ @Gribouillis: Yes, I made a mistake with the notation. Hope that now is clearer. To see that such a $\gamma$ exists, consider the eigenvalue decomposition of $X=T D T^\top$, it holds $\bar{X}=T(D+\gamma(I/n-D))T^\top$. Next, you just need to pick a $\gamma$ which sets to zero one diagonal entry of $D+\gamma(I/n-D)$. $\endgroup$ – Ludwig Aug 25 '17 at 9:24
  • $\begingroup$ You could as well suppose that $X=D$ because $T$ is unitary, which implies that the map $A \to T A T^\top$ preserves the Frobenius norm. $\endgroup$ – Gribouillis Aug 25 '17 at 9:54
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The answer is negative. Here is a counterexample for all $n\ge3$: let $$ \begin{align*} X &= \operatorname{diag}\left(\frac{n^2+1}{n(n+1)},\ \frac 1{n(n+1)},\ \ldots,\ \frac1{n(n+1)}\right)\\ &= \frac In + \operatorname{diag}\left(\frac{n-1}{n+1},\ \frac{-1}{n+1},\ \ldots,\ \frac{-1}{n+1}\right)\\ &= \frac In + D. \end{align*} $$ The closest singular matrix to $X$ of the form $X-\gamma D$ is given by $P_X=X+\frac1nD$. Hence $\|P_X-X\|_F=\|\frac1nD\|_F>\frac1nd_{11}=\frac{n-1}{n(n+1)}$. However, let $B$ be the matrix whose only nonzero entries are the $(2,3)$-th and $(3,2)$-th, each of value $\frac1{n(n+1)}$. Then $X+B$ is also a singular matrix of the same trace as $X$, but $\|B\|_F=\frac{\sqrt{2}}{n(n+1)}<\frac{n-1}{n(n+1)}$.

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