Square block matrix, with Hermitian, non-negative definite blocks, prove that the matrix is also non-negative definite,

Question

Consider the square block matrix

$$S= \begin{bmatrix} R & RQ^* \\ QR & QRQ^* \\ \end{bmatrix} $$

where $R$ is a Hermitian, non-negative definite square matrix and $Q$ and $Q^*$ are square Hermitian conjugate matrices. Prove that $S$ is also non-negative definite.

Edit: My first thoughts would be to try and do a similarity transformation to see whether I can diagonalize S, or put S in some form where I can then observe its eigenvalues, which are invariant under a similarity transform. Would I be on the right track...? I need to show the eigenvalues are non-negative..

Answer 1

You are on the right track with the similarity transform idea. The blocks on the bottom row have $Q$ on the left and the blocks on the right column have $Q^*$ on the right. Hence, we can write $S$ as:

$$S = \begin{bmatrix}R & RQ^* \\ QR & QRQ^*\end{bmatrix} = \begin{bmatrix}I & 0 \\ 0 & Q\end{bmatrix}\begin{bmatrix}R & R \\ R & R\end{bmatrix}\begin{bmatrix}I & 0 \\ 0 & Q^*\end{bmatrix}$$

Thus, $S = \begin{bmatrix}R & RQ^* \\ QR & QRQ^*\end{bmatrix}$ is similar to $\begin{bmatrix}R & R \\ R & R\end{bmatrix}$.

Now, can you show that $\begin{bmatrix}x^* & y^*\end{bmatrix}\begin{bmatrix}R & R \\ R & R\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} \ge 0$ for any $x,y \in \mathbb{C}^n$?

Try multiplying that product out and factoring it.

EDIT: Showing that:

$$\begin{bmatrix}x^* & y^*\end{bmatrix}\begin{bmatrix}R & RQ^* \\ QR & QRQ^*\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = (x^*+y^*Q)R(x+Q^*y) \ge 0$$

for all $x,y \in \mathbb{C}^n$ is pretty easy as well, so I guess the similarity transform wasn't really needed.

Answer 2

$\mathbf{R}$ is PSD by assumption. Hence, there exists $\mathbf{V}$ such that $$ \mathbf{R} = \mathbf{V}\mathbf{V}^{*}. $$ To show that $\mathbf{S}$ is also PSD, it suffices to show that there exists a matrix $\mathbf{B}$ such that $\mathbf{S} = \mathbf{B}\mathbf{B}^{*}$.

For $$ \mathbf{B} = \begin{bmatrix} \mathbf{V}\\ \mathbf{Q}\mathbf{V} \end{bmatrix}, $$ we have $$ \mathbf{B} \mathbf{B}^{*} = \begin{bmatrix} \mathbf{V}\\ \mathbf{Q}\mathbf{V} \end{bmatrix} \begin{bmatrix} \mathbf{V}^{*}& \mathbf{V}^{*}\mathbf{Q}^{*} \end{bmatrix} = \begin{bmatrix} \mathbf{V}\mathbf{V}^{*}& \mathbf{V}\mathbf{V}^{*}\mathbf{Q}^{*}\\ \mathbf{Q}\mathbf{V}\mathbf{V}^{*} & \mathbf{Q}\mathbf{V}\mathbf{V}^{*}\mathbf{Q}^{*} \end{bmatrix} = \begin{bmatrix} \mathbf{R}& \mathbf{R}\mathbf{Q}^{*}\\ \mathbf{Q}\mathbf{R}& \mathbf{Q}\mathbf{R}\mathbf{Q}^{*} \end{bmatrix} = \mathbf{S}. $$