Revisit "How can I visualize the nuclear norm ball"

Question

_{Revisiting How can I visualize the nuclear norm ball?}

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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 $s_{1,2}$:

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

Answer 1

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).

Answer 2

$A$ 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 $$y=\pm\sqrt{x(1-x)},\quad z=1-x;$$