Consider the block upper triangular matrix $$A = \left[ \begin{matrix} A_{11} & A_{12} \\ 0 & A_{22} \end{matrix} \right], $$ where $A\in\mathbb R^{n\times n}$ and $A_{11}\in\mathbb R^{k\times k}$ with $1 \le k \le n$. Suppose $A_{12} \ne 0$ and that we want to block diagonlize $A$ via the similartiy transformation $$T = \left[ \begin{matrix} I & X \\ 0 & I \end{matrix} \right], $$ where $X\in\mathbb R^{k\times (n-k)}$, i.e., $$ T^{-1}AT = \left[ \begin{matrix} A_{11} &0 \\ 0 & A_{22} \end{matrix} \right]. $$ Find a matrix equation $X$ must satisfy for this to be possible.

**Important**: my question is not how to solve the problem. I have question regarding finding $T^{-1}$.

The solution presented said that $$T^{-1} = \left[ \begin{matrix} I & -X \\ 0 & I \end{matrix} \right]\label{eq:star} \tag{$\star$}.$$ Quesutions:

- Is $\eqref{eq:star}$ true?
- Why is it true?

Explanations are great! Thanks!