I need a check on the following exercise, especially on the first point
Let $A$ be a real $m$-by-$n$ matrix with $m>n$ with full rank, and define
$$M = \begin{bmatrix}I & A \\ A^T & 0 \end{bmatrix}$$
i) Show that the $x$ component of the solution of $$M \begin{bmatrix} -r \\ x\end{bmatrix} = \begin{bmatrix} b \\ 0 \end{bmatrix}$$ minimize $||Ax-b||_2$.
ii) Find an explicit formula for $M^{-1}$ in terms of $A$ and $A^T$
i) Solving the block system I get the equations $$Ax-b = r$$ and $$A^T r = 0$$
Therefore $||A x- b|| = ||r||$. Now, I have to use the fact that the matrix $A$ has full rank, i.e. it is $\operatorname{rank}(A)=n$, but I don't know how to move. I would say that this implies that $r=0$ is a solution, therefore the norm is minimized, but I don't know if it's right.
ii) This one is easy: reasoning with blocks, I found $$M^{-1}= \frac{1}{\det{A^TA}}\begin{bmatrix}0 & -A \\ -A^T & I \end{bmatrix}$$
Part (i): suppose $Ax_0-b=r$ and $A^Tr=0$. For any vector $x$, let $v=x-x_0$. Then \begin{aligned} \|Ax-b\|^2 &= \|A(v+x_0)-b)\|^2\\ &= \|Av+r\|^2 \\ &= (Av+r)^T(Av+r)\\ &= v^TA^TAv + v^TA^Tr + r^TAv + r^Tr\\ &= v^TA^TAv + r^Tr\\ &= \|Av\|^2 + \|r\|^2\\ &\geq \|r\|^2 = \|Ax_0-b\|^2.\\ \end{aligned} Therefore $x_0$ minimizes $\|Ax-b\|$.
Part (ii): your answer is wrong. $M^{-1}=\pmatrix{I-A(A^TA)^{-1}A^T&A(A^TA)^{-1}\\ (A^TA)^{-1}A^T&-(A^TA)^{-1}}$.