Block matrix
Task number: 4444
Solution
Directly by definition, for \(\boldsymbol z=(x_1,\dots,x_n,y_1,\dots,y_m)^T\in \mathbb C^{n+m}\), then \( \boldsymbol z^H \begin{pmatrix} \boldsymbol A & \boldsymbol 0_{nm} \\ \boldsymbol 0_{mn} & \boldsymbol B \end{pmatrix} \boldsymbol z = \boldsymbol x^H\boldsymbol A\boldsymbol x + \boldsymbol y^H\boldsymbol B\boldsymbol y \)
If \(\boldsymbol z\) is nontrivial, then one summand on the right is positive and the other is nonnegative.
Conversely, if we extend \(\boldsymbol x\) to \(\boldsymbol z\) with zeros, we get \( \boldsymbol x^H\boldsymbol A\boldsymbol x = \boldsymbol z^H \begin{pmatrix} \boldsymbol A & \boldsymbol 0_{nm} \\ \boldsymbol 0_{mn} & \boldsymbol B \end{pmatrix} \boldsymbol z \gt 0\).
For \(\boldsymbol B\) we can proceed similarly.
Alternative solutions
One can use also the characterization using decompositions, since the decompositions \(\boldsymbol A\) and \(\boldsymbol B\) yield the decomposition of the block matrix and vice versa.
Also, we can use the eigenvalue characterization because the characteristic polynomial of the block matrix is the product of \(p_\boldsymbol A(t)p_\boldsymbol B(t)\), and thus the eigenvalues of the block matrix are the union of the eigenvalues of \(\boldsymbol A\) and \(\boldsymbol B\).
