Collection of Maths Problems

Block matrix 2

Task number: 4445

• 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.