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

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..

• You could try to factor $S$ in $VV^{*}$, or you could also try straightforwardly the definition $x^{T}Sx \ge 0$, $\forall x$. These are also possible points of attack. – megas Dec 25 '14 at 2:07

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.

• I'm having some difficulty carrying the block multiplication, @JimmyK4542. Can you elaborate just a bit on how to do it? I'm pretty sure that the way I'm doing it is way too wordy and probably incorrect... – User001 Dec 25 '14 at 3:24
• ...for example, is your [x,y] one nx1 vector or two vectors that add up to size nx1? (Assuming S is of size nxn) – User001 Dec 25 '14 at 3:29
• If $R$ and $Q$ are $n \times n$ matrices, then $S$ is a $2n \times 2n$ matrix. So, we need to show that $v^*Sv \ge 0$ for all $2n \times 1$ vectors $v$. All I did was break up $v$ into two $n \times 1$ vectors $x$ and $y$. – JimmyK4542 Dec 25 '14 at 3:30
• yes, that's exactly what I started with. but then i started the multiplication of $[x^* y^*]$ with S, but I'm going through every column of S. is there a shorter way to do this? haha...might just be a rule that I am missing...(I know I'm supposed to end up with a 1x2n matrix to multiply against the column vector [x,y]...) – User001 Dec 25 '14 at 3:34
• So I'm breaking out R into its individual entries, R_11, R_21 ...etc... – User001 Dec 25 '14 at 3:35

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