My question is about a logical equivalence in Robert Freund's lecture notes "An Introduction to Semidefinite Programming (SDP)" from MIT's OCW.
The claim is that for $Q$ an $n\times n$ real symmetric PSD matrix with factorization $M^{T}M$, $q$ an $n$ dimensional column vector, and $c$ a real number, $$\mathcal{S} = \begin{pmatrix} I & Mx \\ x^{T}M^{T} & -c - q^{T}x \end{pmatrix}$$ is PSD if and only if \begin{equation} x^{T}Qx + q^{T}x + c \leq 0. \end{equation}
I've worked out $u^{T}\mathcal{S}u$ (for $u \in \mathbb{R}^{n+1}$) and would like to know how the statement: \begin{align} u^{T}\mathcal{S}u &= \sum_{i=1}^n u_i^2+2u_{n+1}\sum_{i=1}^nu_i\sum_{j=1}^nM_{ij}x_j -(u_{n+1})^2(c+q^{T}x)\\ & \geq 0 \text{ for all $u$} \end{align} is equivalent to the condition on the quadratic function of $x$. I've done the computation a couple of times, but it is possible I went wrong somewhere and this is the wrong expansion of the quadratic form.
It looks like doing this by computing eigenvalues or even computing all principle minors (and showing their being non-negative is equivalent to the quadratic condition) is possible with work. I wouldn't mind knowing the "right" or easiest way to see this equivalence, but this question is about understanding why $u^{T}\mathcal{S}u \geq 0 \forall u$ is equivalent to $x^{T}Qx+q^{T}x + c \leq 0$.