The task:

To solve a system of equations with a nonsingular block matrix:
$$begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}f\gend{bmatrix}$$
where $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$ we use the following block Gauss-Seidel method:
$$Ax_{n+1}+By_n=f,\B^{operatorname{T}}x_{n+1}-y_{n+1}=g$$
Prove that if $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2<1$, then the iteration converges to the solution of the system of equations for any initial $x_0,y_0$.

Solution attempt:

In the general case: An iterative method $Mmathcal{X}_{n+1}=mathcal{B}+Zmathcal{X}_n$, solving a system of equations $mathcal{A}mathcal{X}=mathcal{B}$, where $mathcal{A}=M-Z$, converges for any $mathcal{X_0}$ if $left|left|mathfrak{B}right|right|<1$, for $mathfrak{B}:=mathrm{I}-M^{-1}mathcal{A}$.

Here we have $mathcal{A}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix},mathcal{X}=begin{bmatrix}x\yend{bmatrix},mathcal{B}=begin{bmatrix}f\gend{bmatrix}$. _{Sorry for using fonts to distinguish tokens, but as you can see we have crazy conflicts here!} Firstly let's try to find $M$:

$$begin{cases}Ax_{n+1}+By_n=f\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}iffbegin{cases}Ax_{n+1}=f-By_n\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}$$
In matrix form:
$$begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x_{n+1}\y_{n+1}end{bmatrix}=begin{bmatrix}f\gend{bmatrix}+begin{bmatrix}0&-B\0&0end{bmatrix}begin{bmatrix}x_n\y_nend{bmatrix}$$

Since clearly $begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}-begin{bmatrix}0&-B\0&0end{bmatrix}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=A$,

we can conclude that $begin{cases}M=begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}\Z=begin{bmatrix}0&-B\0&0end{bmatrix}end{cases}$.

Now let's try to find $M^{-1}$. We have $M^{-1}M=operatorname{I}$, so:
$$begin{bmatrix}M^{-1}_{1,1}&M^{-1}_{1,2}\M^{-1}_{2,1}&M^{-1}_{2,2}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$

Filling in the blanks we easily get:
$$begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$
So $M^{-1}=begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}$.

Now let's compute $mathfrak{B}$:

$$mathfrak{B}=mathrm{I}-M^{-1}mathcal{A}=mathrm{I}-begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}-begin{bmatrix}mathrm{I}&A^{-1}B\0&mathrm{I}+B^{operatorname{T}}A^{-1}Bend{bmatrix}=begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}$$

Now don't really know what to do next. It would seem I have to somehow show that $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2geqleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2$, but I don't know how to do this.

I'm not sure if this is correct, but I think I remember that for any matrix $mathfrak{A}$, either all $p$-norms are $<1$, $=1$ or $>1$. If this holds, then we could perhpas do something like this: $$left|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2<1iffleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_infty<1iffmaxleft(left|left|0right|right|_infty+left|left|A^{-1}Bright|right|_{infty},left|left|0right|right|_infty+left|left|B^{operatorname{T}}A^{-1}Bright|right|_{infty}right)<1iff\maxleft(left|left|A^{-1}Bright|right|,left|left|B^{operatorname{T}}A^{-1}Bright|right|right)<1$$

So if this reasoning is correct (I wouldn't bet a penny on that), the task would boil down to proving that $left|left|A^{-1}Bright|right|leqleft|left|B^{operatorname{T}}A^{-1}Bright|right|$, but I don't know how to prove it nor if it even holds.

And I haven't yet used the fact that $A$ is positive definite!

How to proceed?