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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?

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    $\begingroup$ M is invertible, since it has full rank, then is $M^T$ invertible too.Then you can say since $GL_n (F) $ is a group. $MM^T $ is invertible. $\endgroup$
    – user519338
    Apr 2, 2018 at 14:11

2 Answers 2

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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 $$ (M^TM)^{-1} = M^{-1}(M^T)^{-1} = M^{-1}(M^{-1})^T =\\ \pmatrix{A^{-1} & -A^{-1}B\\0&I} \pmatrix{A^{-1} & -A^{-1}B\\0&I}^T =\\ \pmatrix{A^{-1} & -A^{-1}B\\0&I} \pmatrix{(A^{-1})^T & 0\\-B^T(A^{-1})^T&I}^T = \\ \pmatrix{A^{-1}(A^{-1})^T + A^{-1}BB^T(A^{-1})^T & -A^{-1}B\\ -B^T(A^{-1})^T & I} = \\ \pmatrix{(A^TA)^{-1} + (A^{-1}B)(A^{-1}B)^T & -A^{-1}B\\ -(A^{-1}B)^T & I} $$

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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.

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