Inverse of a triangular block matrix (sufficient and necessary conditions for the existence) Consider the following theorem:
Let $\mathrm{\mathbf{A}}\in\mathbb{C}^{n\times n}$ be an upper triangular block matrix with $2\times2$ blocks ($p=2$), i. e.,
$\mathrm{\mathbf{A}}=\left(\begin{array}{cc}
\mathrm{\mathbf{A}_{11}} & \mathrm{\mathbf{A}_{12}}\\
0 & \mathrm{\mathbf{A}_{22}}
\end{array}\right)$. 
Then $\mathrm{\mathbf{A}}$ is invertible if and only if $\mathbf{A}_{11}$ and $\mathbf{A}_{22}$ are invertible matrices.
(a) Prove the theorem.
(b) Generalize the theorem to $\mathrm{\mathbf{A}}\in\mathbb{C}^{n\times n}$ upper triangular block matrix with $p\times p$ blocks, i. e.,
$\mathrm{\mathbf{A}}=\left(\begin{array}{cccc}
\mathrm{\mathbf{A}_{11}} & \mathrm{\mathbf{A}_{12}} & \cdots & \mathrm{\mathbf{A}}_{1n}\\
 & \mathrm{\mathbf{A}_{22}} & \cdots & \mathrm{\mathbf{A}}_{2n}\\
 &  & \ddots & \vdots\\
 &  &  & \mathrm{\mathbf{A}}_{nn}
\end{array}\right)$, 
and prove. (Note that it's not necessary to calculate the inverse matrix)
MY ATTEMPT
(a) In this case $\mathrm{\mathbf{A}^{-1}}\in\mathbb{C}^{n\times n}$ is
$\mathrm{\mathbf{A}}^{-1}=\left(\begin{array}{cc}
\mathrm{\mathbf{A}_{11}^{-1}} & \mathrm{-\mathrm{\mathbf{A}_{11}^{-1}}\mathbf{A}_{12}}\mathrm{\mathbf{A}_{22}^{-1}}\\
0 & \mathrm{\mathbf{A}_{22}^{-1}}
\end{array}\right)$.
Therefore, $\mathrm{\mathbf{A}}$ is invertible if and only if $\mathbf{A}_{11}$ is invertible and $\mathbf{A}_{22}$ is invertible. $\square$
(b) I think I have to use induction on the number of blocks ($p$), but I don't know how to do
Can you help me? Thanks.
 A: a/ -- You have only shown "if $\mathbf{A}_{11}$ and $\mathbf{A}_{22}$ are invertible then $\mathbf{A}$ is invertible". It still remains to show the other implication.
b/ A matrix $\mathbf{A}$ is invertible if and only if the equation $$ \mathbf{A}x = 0$$ has a unique (zero) solution.
$\mathbf{A}_{kk}$ is invertible for all $k = 1 ... n$ We are solving the equation $$\mathbf{A}x = 0.$$

As $\mathbf{A}_{nn}$ is invertible then $x[(n-1)p + 1:np] = 0$ is the unique solution of the equation $$\mathbf{A}_{nn} x[(n-1)p +1: np] = 0$$ i.e. the last $p$ elements of $x$ are zero. Thus, the rightmost column
$[\mathbf{A}_{1n};...;\mathbf{A}_{nn}]$ has no effect (it is multiplied by 0).  It means that $x[1:(n-1)p]$ is determined by the equation 
$$ \left(\begin{array}{ccc}
\mathbf{A}_{11} & \ldots & \mathbf{A}_{1,n-1} \\
 & \ddots & \vdots \\
 \mathbf{0} & & \mathbf{A}_{n-1,n-1}
\end{array}\right) x[1:(n-1)p] = \mathbf{0}. $$
Now repeat the same argument inductively for matrices $\mathbf{A}_{n-1,n-1}, ..., \mathbf{A}_{11}$.
$\mathbf{A}$ is invertible
Suppose that $\mathbf{A}_{11}$ is not invertible and there exists $z$ such that $\mathbf{A}_{11} z = 0$. Then a vector $y = [z;0;0;...;0]$ is a solution to $\mathbf{A} y = 0$ while $y \neq 0$. This is a contradiction to assumption that $\mathbf{A}$ is invertible and thus $\mathbf{A}_{11}$ is invertible too.  That means you can express $$x[1:p] = - \mathbf{A}_{11}^{-1}[\mathbf{A}_{12},...,\mathbf{A}_{1n}]x[p+1:np],$$ and the rest of $x$ is solution to $$\left(\begin{array}{ccc}
\mathbf{A}_{22} & \ldots & \mathbf{A}_{2,n} \\
 & \ddots & \vdots \\
 \mathbf{0} & & \mathbf{A}_{n,n}
\end{array}\right) x[p+1:np] = \mathbf{0}.$$ You can now proceed recursively.
