## Block matrix 2

The goal of the task is to prove the following statement:

A Hermitian block matrix $$\begin{pmatrix} \boldsymbol A & \boldsymbol B^H\\ \boldsymbol B & \boldsymbol C \end{pmatrix}$$ is positive definite, if and only if the matrix $$\boldsymbol A$$ and the matrix $$\boldsymbol C - \boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H$$ are both positive definite.

• #### Step 1

Show first that if the $$\begin{pmatrix} \boldsymbol A & \boldsymbol B^H\\ \boldsymbol B & \boldsymbol C \end{pmatrix}$$ positive definite, then $$\boldsymbol A$$ is positive definite.
• #### Solution

Suppose that $$\boldsymbol A$$ is of order $$n$$, $$\boldsymbol C$$ is of order $$m$$, or $$\boldsymbol B\in \mathbb C^{m\times n}$$. Given a non-trivial $$\boldsymbol x=(x_1,\ldots,x_n)^T\in \mathbb C^n$$, we add $$m$$ zeros to $$\boldsymbol x'=(x_1,\dots,x_n,0,\dots,0)^T\in \mathbb C^{n+m}$$. Then $$\boldsymbol x^H\boldsymbol A\boldsymbol x = (\overline{x_1},\ldots,\overline{x_n},0,\ldots,0) \begin{pmatrix} \boldsymbol A & \boldsymbol B^H \\ \boldsymbol B & \boldsymbol C \end{pmatrix} (x_1,\ldots,x_n,0,\ldots,0)^T = \boldsymbol x'^H \begin{pmatrix} \boldsymbol A & \boldsymbol B^H \\ \boldsymbol B & \boldsymbol C \end{pmatrix} \boldsymbol x' > 0$$.
• #### Step 2

What elementary transformations can be used to eliminate the matrix $$\boldsymbol B$$? Can you express these by a matrix product?
• #### Solution

Since we have shown that $$\boldsymbol A$$ is positive definite, it is regular and thus it has its inverse matrix $$\boldsymbol A^{-1}$$.

If we take the first $$n$$ rows, or the matrix $$(\boldsymbol A|\boldsymbol B^H)$$, and multiply it from the left $$\boldsymbol B\boldsymbol A^{-1}$$, we get the matrix $$(\boldsymbol B|\boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H)$$.

Elimination of the matrix $$\boldsymbol B$$ corresponds to subtracting $$\boldsymbol B\boldsymbol A^{-1}$$ times the first $$n$$ rows from the remaining $$m$$ rows.

The whole arrangement can be represented by a product with the regular matrix $$\begin{pmatrix} \mathbf I_n & \boldsymbol 0_{nm}\\ -\boldsymbol B\boldsymbol A^{-1} & \mathbf I_m \end{pmatrix}$$, ie.

$\begin{pmatrix} \mathbf I_n & \boldsymbol 0_{nm} \\ -\boldsymbol B\boldsymbol A^{-1} & \mathbf I_m \end{pmatrix} \begin{pmatrix} \boldsymbol A & \boldsymbol B^H \\ \boldsymbol B & \boldsymbol C \end{pmatrix} = \begin{pmatrix} \boldsymbol A & \boldsymbol B^H \\ \boldsymbol 0_{mn} & \boldsymbol C-\boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H \end{pmatrix}$
• #### Step 3

Then use the knowledge from the previous two tasks.
• #### Solution

In addition, we multiply the equality derived in the previous step from the right by the Hermitian transpose of the matrix $$\begin{pmatrix} \mathbf I_n & \boldsymbol 0_{nm} \\ -\boldsymbol B\boldsymbol A^{-1} & \mathbf I_m \end{pmatrix}$$.

Specifically, we get: $\begin{pmatrix} \mathbf I_n & \boldsymbol 0_{nm} \\ -\boldsymbol B\boldsymbol A^{-1} & \mathbf I_m \end{pmatrix} \begin{pmatrix} \boldsymbol A & \boldsymbol B^H \\ \boldsymbol B & \boldsymbol C \end{pmatrix} \begin{pmatrix} \mathbf I_n & \boldsymbol 0_{nm} \\ -\boldsymbol B\boldsymbol A^{-1} & \mathbf I_m \end{pmatrix}^H=$ $= \begin{pmatrix} \boldsymbol A & \boldsymbol B^H\\ \boldsymbol 0_{mn} & \boldsymbol C-\boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H \end{pmatrix} \begin{pmatrix} \mathbf I_n & -\boldsymbol A^{-1}\boldsymbol B^H \\ \boldsymbol 0_{mn} & \mathbf I_m \end{pmatrix} = \begin{pmatrix} \boldsymbol A & \boldsymbol 0_{nm} \\ \boldsymbol 0_{mn} & \boldsymbol C-\boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H \end{pmatrix}$

We found that the matrix $$\begin{pmatrix} \boldsymbol A & \boldsymbol B^H\\ \boldsymbol B & \boldsymbol C \end{pmatrix}$$ a $$\begin{pmatrix} \boldsymbol A & \boldsymbol 0_{nm} \\ \boldsymbol 0_{mn} & \boldsymbol C-\boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H \end{pmatrix}$$ are either both positive definite or none, with the latter matrix being positive definite if and only if both $$\boldsymbol A$$ and $$\boldsymbol C-\boldsymbol B\boldsymbol A^{-1}\boldsymbol B^H$$ are positive definite.