Let $$M := \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$$ where both $A$ and $B$ are $N \times N$ symmetric positive definite (SPD) matrices. Show that $M$ is also SPD.
$M$ is clearly symmetric. Now to show that the eigenvalues of $M$ are positive, is it enough to observe that the eigenvalues of $A$ and $B$ are positive, and so are those of $M$ (as they are the same: How to find the eigenvalues of a block-diagonal matrix?)?
Let $$ v=\begin{pmatrix} v_1 \\ v_2 \\ \end{pmatrix} $$where $v_1$ and $v_2$ are $n\times 1 $ vectors so $v$ is $2n\times 1$ vector. Therefore we have:$$v^TMv=\begin{pmatrix} v_1^T & v_2^T \\ \end{pmatrix}M\begin{pmatrix} v_1 \\ v_2 \\ \end{pmatrix}=\begin{pmatrix} v_1^T & v_2^T \\ \end{pmatrix}\begin{pmatrix} A & 0 \\ 0 & B \\ \end{pmatrix}\begin{pmatrix} v_1 \\ v_2 \\ \end{pmatrix}=v_1^TAv_1+v_2^TBv_2$$ since both $v_1^TAv_1$ and $v_2^TBv_2$ are positive so is $v^TMv$ which implies positive definition of $M$.
