Block matrix

Show that a Hermitian block matrix $$\begin{pmatrix} \boldsymbol A & \boldsymbol 0_{nm} \\ \boldsymbol 0_{mn} & \boldsymbol B \end{pmatrix}$$ is positive definite if and only if the matrices $$\boldsymbol A$$ and $$\boldsymbol B$$ are both positive definite.
• 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 $$\boldsymbol B$$.