Given a block diagonal matrix $A$ like so:
$$A=
\begin{bmatrix}
B & & \\
& C & \\
& & D \\
\end{bmatrix}
$$
Given $B$ is symmetric and has real, positive eigenvalues (obviously $B$ is positive-definite), $C$ is
$$C =
\begin{bmatrix}
11 & -4 \\
0 & 2
\end{bmatrix}
$$
and $D$ is
$$D =
\begin{bmatrix}
11 & -3 & 0 \\
2 & 3 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
$$
I'm asked to determine whether $C$, $D$ and $A$ are positive-definite. Well, $C$ and $D$ are asymmetric, so obviously we need to stretch the rules and use the definition of positive-definiteness that includes asymmetric matrices, which states that an asymmetric matrix is positive-definite iff its symmetric part (the sum of the matrix plus its transpose, divided by $2$ (optionally)) is positive-definite.
So, calculate the symmetric parts of $C$ and $D$, see if they're positive-definite, easy enough. However, not knowing the data of $B$ (besides its properties) doesn't allow me to calculate the symmetric part of $A$, so I need to work by properties only. What am I not seeing here?
Best Answer
Posting as answer by request.
You can apply the symmetric component rule to $A$, because its symmetric part is just the block diagonal made of the symmetric parts of $B$, $C$, and $D$. Then note that if the blocks of a block-diagonal matrix are p.d., then so is the matrix.
You already know that $B$ is p.d., and as user1551 stated, $B$ is itself its symmetric component. It just remains to show that $C$ and $D$ have p.d. symmetric components, but you've done that already.