Matrix invertiblity and its Inverse I'd like to prove that the Matrix $L:={ M }^{ T }M$ is invertible and determine its inverse (in dependence of $A$ and $B$).
$M:=\begin{pmatrix}A & B \\ 0_{q\times p}& I_q\end{pmatrix}$ and $A\in K^{p\times p},B\in K^{p\times q}$. 
Further is given that the matrix $A$ has rank $p$.
I tried to form $L=\begin{pmatrix}{ A }^T A& A^T B\\ B^T A &{ B^T B+I }_{ q }\end{pmatrix}$ into the Unity Blockmatrix using elementary row operations.
Since A has a full rank it must be invertible and because A is a $p\times p$ matrix, its transpose must be invertible too, with this knowledge its fairly easy to get ${ a }_{ 21 }=0$ and  I'm stuck in getting ${ a }_{ 12 }=0$ since I know nothing about the invertibility of $B$.
Are there any other ways to solve this problem?
 A: Computing the inverse of $ L $ given the inverse of $ M $ is easy:  $ L^{-1} = ( M^T M)^{-1} = M^{-1} M^{-T} $.
So, the question really is how to find the inverse of $ M $.
But the inverse of this partitioned triangular matrix is given by
$$ \left( \begin{array}{c | c}
A & B \\ \hline
0 & I 
\end{array} \right)^{-1} =
\left( \begin{array}{c | c}
A^{-1} & -A^{-1} B  \\ \hline
0 & I 
\end{array} \right) $$
(just multiply out.)
So, you can now put the pieces together.
A: We can compute the inverse of $M$ with "row-operations" as follows:
$$
\left[\begin{array}{cc|cc}
A&B&I&0\\0&I&0&I
\end{array}\right] \to \\
\left[\begin{array}{cc|cc}
I&A^{-1}B&A^{-1}&0\\0&I&0&I
\end{array}\right] \to\\
\left[\begin{array}{cc|cc}
I&0&A^{-1}&-A^{-1}B\\0&I&0&I
\end{array}\right]
$$
So that 
$$M^{-1} = \pmatrix{A^{-1} & -A^{-1}B\\0&I}$$
From there, we can compute
$$
(M^TM)^{-1} = M^{-1}(M^T)^{-1} = M^{-1}(M^{-1})^T =\\
\pmatrix{A^{-1} & -A^{-1}B\\0&I} \pmatrix{A^{-1} & -A^{-1}B\\0&I}^T =\\
\pmatrix{A^{-1} & -A^{-1}B\\0&I} \pmatrix{(A^{-1})^T & 0\\-B^T(A^{-1})^T&I}^T = \\
\pmatrix{A^{-1}(A^{-1})^T + A^{-1}BB^T(A^{-1})^T & -A^{-1}B\\
-B^T(A^{-1})^T & I} = \\
\pmatrix{(A^TA)^{-1} + (A^{-1}B)(A^{-1}B)^T & -A^{-1}B\\
-(A^{-1}B)^T & I}
$$
