# Revisit “How can I visualize the nuclear norm ball”

Revisit "https://math.stackexchange.com/questions/1099320/how-can-i-visualize-the-nuclear-norm-ball"

Two eigenvalues are reproduced as following:

$$s_{1,2}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2\pm|x+z|\sqrt{(x-z)^2+4y^2}}.$$

According to the following (from a paper) If a symmetric matrix: $$A=\left( \begin{array}{cc} x & y\\ y & z\end{array} \right)$$ is rank $$1$$, then $$y=\sqrt{xz}$$, which comes from the fact that $$vv^T$$ is rank $$1$$ and any rank $$1$$ matrix can be represented in this form.

$$\left[\begin{array}{cc} v_1\\ v_2\end{array}\right]\left[\begin{array}{cc} v_1 & v_2\end{array}\right]=\left( \begin{array}{cc} v_1^2 & v_1v_2\\ v_1v_2 & v_2^2\end{array} \right)$$

My question: how to explain the red circle in figure (b) is the $$2\times 2$$ symmetric unit-Euclidean-norm rank $$1$$ matrix?
This is a circle in $$3$$-D, how to get the equation of this circle through the rank $$1$$ matrix provided above?

I believe just replace $$y=\pm\sqrt{xz}$$ in $$s_{1,2}$$ and can get the answer.

So I choose the larger one of $$s_{1,2}$$:

$$s_{\max}=\frac{1}{\sqrt{2}}\sqrt{x^2+2y^2+z^2 + |x+z|\sqrt{(x-z)^2+4y^2}}= \sqrt{2(x+z)^2}=1 (\text{unit norm})$$

• Not sure how you got that formula, but I get $\frac12 (x + z \pm \sqrt{x^2 - 2xz + z^2 + 4y^2})$ as the eigenvalues of $A$. – user856 Sep 5 '16 at 2:40
• @Rahul you mean $s_{1,2}$? that is from the link I provide. – sleeve chen Sep 5 '16 at 2:43
• Maybe it's equivalent, but it seems more complicated than necessary. Also only positive semidefinite rank 1 matrices can be represented as $vv^T$; to represent all rank 1 matrices you need $\pm vv^T$, equivalently $y=\pm\sqrt{xz}$. – user856 Sep 5 '16 at 2:47

The answer to the linked question shows that the curved side of the "cylinder" satisfies the equation $$(x-z)^2+4y^2=1$$ and its planar caps satisfy $$(x+z)^2=1.$$ The red curves are the intersection of the two surfaces, so they satisfy both equations. Therefore, they must also satisfy their sum, $$x^2+4y^2+z^2=2.$$ Thus the curves lie on the intersection of the ellipsoid $x^2+4y^2+z^2=2$ and the planes $x+z=\pm1$, so they are ellipses (but not necessarily circles).
$A$ must be rank-$1$ and symmetric. Thus $$A=vv^T$$ This implies that $$y^2=xz$$ In this problem, we also have $\|v\|=1$, so $$x+z=1$$ All of these result in $$y=\pm\sqrt{x(1-x)},\quad z=1-x;$$