How to make a block matrix positive semi-definite?

Question

I have a matrix $A=\begin{bmatrix} \textbf{0}_{N\times N} & S\\ S^T & \textbf{0}_{M\times M} \end{bmatrix},$ where $S\in R^{N\times M}$. What $S$ would make $A$ a positive semi-definite matrix?

Answer 1

Take any $z = (z_1,z_2)^T$. We have

\begin{align*} & \left(z_1^T z_2^T \right) A \begin{pmatrix} z_1 \\ z_2 \end{pmatrix} \\ & = z_1^T S z_2 + z_2^T S^T z_1 \\ & = k(z) + k(z)^T \\ & = 2k(z) \end{align*}

where $k(z) = z_1^T S z_2$.

Now suppose $k \geq 0$. If you take the vector $z^\prime = (-z_1 , z_2)^T$, $k(z^\prime) \leq 0$.

So positive semi-definiteness can only be achieved if $k(z) = 0$ for all $z$ which is the case if $S = 0$.

Answer 2

Since $A$ is real symmetric, $A$ is diagonalizable and similar to some diagonal matrix $D$.

If $A$ is positive semi-definite, so does $D$ and hence all entries in $D$ are non-negative numbers.

Since $A$ is traceless, so does $D$ and hence all entries of $D$ are zero.

This means $D$ is the zero matrix and so does $A$.