# Eigenvalues of a "diagonal" block-matrix

Let $$n,n'\geq 1$$ and $$A_1 \in \mathbb{R}^{n \times n}, A_2 \in \mathbb{R}^{n' \times n'}$$ be two symmetric matrices. Let $$A = \begin{pmatrix} A_1 & 0\\\ 0 & A_2 \end{pmatrix} \in \mathbb{R}^{(n+n')^2}$$ (the $$0$$ blocks are of size $$n \times n'$$ and $$n' \times n$$).

Is it true that if we have $$A_1 \preccurlyeq B_1$$ and $$A_2 \preccurlyeq B_2$$ for some symmetric matrices $$B_1 \in \mathbb{R}^{n \times n}, B_2 \in \mathbb{R}^{n' \times n'}$$ , then $$A \preccurlyeq B = \begin{pmatrix} B_1 & 0\\\ 0 & B_2 \end{pmatrix}$$ ?

It would be sufficient to prove that the eigenvalues of $$A-B$$ are the eigenvalues of $$A_1-B_1$$ and $$A_2-B_2$$.

Yes. Your approach looks perfectly fine. You can relate the eigenvalues by looking at the characteristic polynomial of the block matrix: $$\det \begin{pmatrix} A_1 - B_1 - \lambda I & 0\\ 0 & A_2 - B_2 - \lambda I \end{pmatrix} = \det (A_1 - B_1 - \lambda I) \det (A_2 - B_2 - \lambda I).$$
A better alternative for showing $$A \preceq B$$ is to go back to the definition: $$A \preceq B$$ if and only if $$z^T A z \leq z^T B z$$ for all $$z \in \mathbb{R}^{n + n'}$$. Write $$z = \begin{pmatrix} x\\ y \end{pmatrix}$$. Then \begin{align*} z^T A z & = x^T A_1 x + y^T A_2 y\\ & \leq x^T B_1 x + y^T B_2 y\\ & = z^T B z. \end{align*}