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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 $A_2-B_2$.

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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*}

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