Why two matrix sets defined by such LMI are equivalent?

Question

Suppose

• $\{(x_1,x_2) : x_1^2+x_2^2 = 1\}$ the unit circle.

Consider two sets defined by a quadratic constraint and LMI:

• $$\{Y\in R^{2\times 2}: \begin{bmatrix}x_1 & x_2 \end{bmatrix}\begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}\leq 1\}$$
• $$\{Y\in R^{2\times 2}: \begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix}\preceq I\}$$

How to show both constraints define the same set?

Suppose $$\begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix}=I,$$ then the equality of the first constraint holds.

Moreover, let $Y_{11}+Y_{22}=a, Y_{21}-Y_{12}=b$, we can rewrite the first constraint as:

$$(x_1^2-x_2^2)a+2x_1x_2b\leq 1$$ How to do the next step?

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My problem comes from the following:

http://arxiv.org/abs/1403.4914 (p.1328 top)

Answer 1

According to the description in your reference, a $Y\in\mathbb{R}^{2\times 2}$ belong to $SO(2)^o$ iff for $x_1^2+x_2^2=1$ (on the unit circle), we have $$\begin{bmatrix}x_1 & x_2 \end{bmatrix}\begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}\leq 1$$
which is the same as saying $$\begin{bmatrix}x_1 & x_2 \end{bmatrix}\left(\begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix} -\begin{bmatrix}1 & 0\\ 0& 1 \end{bmatrix} \right)\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}\leq 0.$$ Symbolically, in the language of negative-definiteness, it reads $$\begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix}\preceq I.$$

Answer 2

Thank T. Amdeberhan's nice answer. In the following, I come up with another answer that I just remember:

The first constraint means:

$$\max_{\|\hat{x}\|^2=1}\hat{x}^TA\hat{x}=\lambda_{\max}\leq 1$$

where $A = \begin{bmatrix}Y_{11}+Y_{22} & Y_{21}-Y_{12} \\ Y_{21}-Y_{12}&-Y_{11}-Y_{22} \end{bmatrix}$

Therefore it implies the second constraint.