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) 



Why two matrix sets defined by such LMI are equivalent?
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
$$\begin{bmatrix}Y\_{11}+Y\_{22} & Y\_{21}-Y\_{12} \\ Y\_{21}-Y\_{12}&-Y\_{11}-Y\_{22} \end{bmatrix}\preceq I.$$ 


