I want to prove the following:
If the matrix
$$
M=\begin{pmatrix}
0&\vec{q}^T \\
\vec{q}&N
\end{pmatrix}
$$
is PSD, then $\vec{q}=\vec{0}$.
The only three properties of a positive semidefinite matrix that I know is the non-negative eigenvalues, $\vec{x}^TM\vec{x}\geq0$, and $\exists U$ such that $M=U^TU$, but I have no idea to apply them into this question. Any hints?
Best Answer
This text uses that N is a PSD matrix (see comments)
Let $\vec y$ be the vector $\vec x$ without the first element.
By direct summation: $$\vec{x}^TM\vec{x} = 2 x_1 \sum_{i=2}^N x_i q_i + \vec y^T N \vec y $$
Now for $\vec q = \vec 0$ this is PSD. Conversely, if $\vec q \ne \vec 0$, then you can always find a $\vec x$ for which it is not PSD. If $q_2 \ne 0$ (otherwise choose any other nonzero element of $\vec q$), then choose $x_i = 0$ for $i=3 \cdots N$ and $x_1 \ne0$, $x_2 \ne 0$. This gives $\vec{x}^TM\vec{x} = 2 x_1 x_2 q_2 + (x_2, 0, \cdots ,0)^TN(x_2, 0, \cdots ,0) $. Since N is PSD, we have $(x_2, 0, \cdots ,0)^TN(x_2, 0, \cdots ,0) = R \ge 0$.
If $R=0$, you select $x_1 = -x_2 q_2$ which makes the expression negative. If $R>0$, you select $x_1 = -x_2 q_2 R /(q_2^2 x_2^2)$ which also makes the expression negative.
So you can always construct cases for $\vec q \ne \vec 0$ which violate the PSD condition.
So you need $\vec q = \vec 0$ to obtain PSD.