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Does fixing diagonal entries and minimizing nuclear norm under weighted sum of entries conditions produce a rank $1$ matrix? I think the answer for this is no.

At least could it be true in $2\times2$ case?

Singular values of $2\times 2$ matrix $$M=\pmatrix{m_{00}&m_{01}\\m_{10}&m_{11}}\in\mathbb R^{2\times2}$$ are $s_x=Q+R$ and $s_y=Q-R$ where $Q=\sqrt{E^2+H^2}; R=\sqrt{F^2+G^2}$ and

$$E=\frac{m_{00}+m_{11}}{2}; F=\frac{m_{00}-m_{11}}{2}; G=\frac{m_{10}+m_{01}}{2}; H=\frac{m_{10}-m_{01}}{2}$$

holds.

Suppose we know $m_{00},m_{11}$ and also know weighted sum of entries of $M$ at some positive real weights.

Then if we seek to complete the matrix by nuclear norm minimization on these conditions is it guaranteed that the matrix will be rank $1$?

If not what minimal conditions would one need to force rank $1$ under known diagonal, known weighted sum condition with nuclear norm minimization?

What about general $n\times n$ case?

Note without the sum condition the problem is trivial.

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  • $\begingroup$ In the 2x2 case if the weights for the diagonal elements do not sum to zero you can find a symmetric matrix fullfilling the sum condition. According to your formula any symmetric matrix with the prescribed diagonal entries has minimal nuclear norm. $\endgroup$ – user35593 Nov 7 '18 at 6:48
  • $\begingroup$ @user35593 why symmetric matrices minimize nuclear norm here? $\endgroup$ – Anon. Nov 11 '18 at 19:40
  • $\begingroup$ According to your formula the nuclear norm is 2Q. E depends only on diagonal entries and is therefore fixed. $H^2$ is minimal if M symmetric. However this is only true if we assume Q>=R. In the other case the offdiagonal part of M needs to be skew-symmetric. $\endgroup$ – user35593 Nov 11 '18 at 21:27
  • $\begingroup$ @user35593 Not sure your condition is sufficient. $\endgroup$ – Anon. Nov 14 '18 at 6:31

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