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I want to see what the unit nuclear norm ball looks like.

So I think of matrices whose singular values add up to $1$. For simplicity, let's talk about symmetric, $2\times 2$ matrices (so that I can limit myself to $3$ dimensions). These matrices can be thought of as points in a $3$-dimensional space, and the coordinate values tell us about the entries in the matrix. But what shape will the matrices that have nuclear norm $1$ form?

I have seen figures showing it to be like a solid cylinder, but I can't see why.

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I have considered a symmetric matrix $$ A=\left( \begin{array}{cc} a & b\\ b &c\end{array} \right)$$ where $a, b, c$ are real numbers. I calculated the singular values $s_{1,2}$ of $A$ as positive square roots of the eigenvalues of $A^2$. If I have not make a mistake, then $$ s_{1,2}=\frac{1}{\sqrt{2}}\sqrt{a^2+2b^2+c^2\pm|a+c|\sqrt{(a-c)^2+4b^2}}. $$ Now we want that $s_1+s_2=1$. After squaring I received $$ a^2+2b^2+c^2+\sqrt{(a^2+2b^2+c^2)^2-(a+c)^2((a-c)^2+4b^2)}=1$$ and consequently $$ a^2+2b^2+c^2+2|b^2-ac|=1. $$ Thus, if $ac\geq b^2$, then $$(a+c)^2=1,$$ and if $ac<b^2$, then $$ (a-c)^2+4b^2=1.$$

I hope that I have not make a mistake in my calculations.

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  • $\begingroup$ I plotted the functions and the second one does look like an elliptic cylinder. thanks for the answer. $\endgroup$ – nemo Jan 11 '15 at 11:53
  • $\begingroup$ Hi, I post a related question "math.stackexchange.com/questions/1914889/…", which is deeper but related to this answer. $\endgroup$ – sleeve chen Sep 5 '16 at 2:19
  • $\begingroup$ Also, the first equality has a typo: $(a-b)^2$ should be $(a-c)^2$ $\endgroup$ – sleeve chen Sep 5 '16 at 2:20
  • $\begingroup$ @sleeve chen Thank you! $\endgroup$ – Janko Bracic Oct 24 '17 at 7:16

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