Equivalence involving diagonalization of block matrix

Question

Given $T^\top L T = diag(A, B, C)$ being a block diagonalization with

• $L$ symmetric
• $A,B$ positive definite (p.d.)
• $T = \left( \begin{array}{ccc}I & 0 & T_2 \\ T_1 & I & T_3 \\ 0 & 0 & I\end{array}\right)$, where $T_2, T_3$ are not quadratic and we don't know whether $T_1$ is regular.

I read in a paper that $L$ is p.d. iff $C$ is p.d. I know that if $L$ is p.d. and $T$ is regular, then also $T^\top L T$ is p.d. and thus every sub matrix of $T^\top L T$, therefore also $C$.

• Why does the converse hold?
• Why is $T$ regular?

Answer 1

$T$ is a block triangular matrix: $$ T = \left( \begin{array}{cc|c}I & 0 & T_2 \\ T_1 & I & T_3 \\ \hline0 & 0 & I\end{array}\right). $$ It is invertible if and only if its two diagonal blocks, $\pmatrix{I & 0\\ T_1 & I}$ and $I$, are invertible. Yet this is trivial. Since $T$ is invertible, $L=(T^{-1})^\top\operatorname{diag}(A,B,C)\,T^{-1}$. If $C$ is p.d., $\operatorname{diag}(A,B,C)$ is also p.d. and by Sylvester's law of inertia, $L$ is p.d..