Constructing a determinantal inequality The following is from page 3410 of the paper Quadratically constrained attitude control via semidefinite programming.
Consider a polynomial:     
$$\mu_1(p_1^Tx)^2+ \cdots + \mu_n(p_n^Tx)^2\leq a$$
where


*

*$x\in R^n$, $p_i\in R^n$    

*$\mu_i,a\in R_+$  


Can that polynomial assume the following?   
$$A = \begin{bmatrix}I_n & l(x) \\l(x)^T & a \end{bmatrix}\geq 0$$
where $$l(x) = \begin{bmatrix}\sqrt{\mu_1}p_1^Tx  \\\vdots \\ \sqrt{\mu_n}p_n^Tx   \end{bmatrix}$$.     

I try to derive from the following:      


*

*determinant:   From https://en.wikipedia.org/wiki/Determinant
$\text{det}(A) = \text{det}(I_na)-\text{det}(l(x)l(x)^T) = a$ since $l(x)l(x)^T$ is rank $1$ so there must exist eigenvalue $0$; therefore, $\text{det}(l(x)l(x)^T)=0$.     Not correct.


How to show both are equivalent?   
 A: If $\mu_i > 0$,
$$\mu_1 (\mathrm{p}_1^T \mathrm{x})^2 + \mu_2 (\mathrm{p}_2^T \mathrm{x})^2 + \cdots + \mu_n (\mathrm{p}_n^T \mathrm{x})^2 = \| (\operatorname{diag}(\mu))^{\frac{1}{2}} P^T \mathrm{x}\|_2^2$$
where the $i$-th column of $P \in \mathbb{R}^{n \times n}$ is $\mathrm{p}_i$. The polynomial inequality
$$\mu_1 (\mathrm{p}_1^T \mathrm{x})^2 + \mu_2 (\mathrm{p}_2^T \mathrm{x})^2 + \cdots + \mu_n (\mathrm{p}_n^T \mathrm{x})^2 \leq a$$
can then be written as the following determinantal inequality
$$\det \begin{bmatrix} I_n & (\operatorname{diag}(\mu))^{\frac{1}{2}} P^T \mathrm{x}\\ \mathrm{x}^T P (\operatorname{diag}(\mu))^{\frac{1}{2}} & a\end{bmatrix} \geq 0$$
as the determinant of the block matrix above is
$$\begin{array}{rl}\det \begin{bmatrix} I_n & (\operatorname{diag}(\mu))^{\frac{1}{2}} P^T \mathrm{x}\\ \mathrm{x}^T P (\operatorname{diag}(\mu))^{\frac{1}{2}} & a\end{bmatrix} &= a - \mathrm{x}^T P (\operatorname{diag}(\mu))^{\frac{1}{2}} (\operatorname{diag}(\mu))^{\frac{1}{2}} P^T \mathrm{x}\\\\ &= a - \| (\operatorname{diag}(\mu))^{\frac{1}{2}} P^T \mathrm{x} \|_2^2\end{array}$$
