I have a matrix of the following form, where $a,b,c>0$
\begin{align*}
A = \left[
\begin{array}{cccccc}
aM_{12}^2 & aM_{12}M_{13} & 0 & 0 & 0 & 0 & 0 \\
aM_{13}M_{12} & aM_{13}^2 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & bM_{21}^2 & bM_{21}M_{23} & bM_{21}M_{24} & 0 & 0 \\
0 & 0 & bM_{23}M_{21} & bM_{23}^2 & bM_{23}M_{24} & 0 & 0 \\
0 & 0 & bM_{24}M_{21} & bM_{24}M_{23} & bM_{24}^2 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & cM_{31}^2 & cM_{31}M_{32} \\
0 & 0 & 0 & 0 & 0 & cM_{32}M_{31} & cM_{32}^2
\end{array}
\right]
\end{align*}
Is this matrix positive semi-definite? Since it is block-diagonal, I am thinking that it should be easy to check something to do with determinants, but I'm not sure how to write it out. Also, can the above matrix be written in a more concise fashion?
Update:
Is it sufficient to check that the all leading principal minors (from Wiki: "the kth leading principal minor of a matrix A is the determinant of its upper-left k by k sub-matrix") of the sub-matrices are non-negative?
Best Answer
Look at the first block, which is of the form $C = [\alpha^2 \ \alpha \beta; \alpha \beta \ \beta^2]$. Let $z = [x; y]$ be a column $2$-vector. Then: $z' C z = [\alpha x; \beta y]' [1 \ 1; 1 \ 1] [\alpha x; \beta y] \geq 0$, because the matrix $[1 \ 1; 1 \ 1]$ is sp(semi)d. The same happens with the $3\times3$ block. Therefore, your matrix is sp(semi)d.