Let $V=(\mathbb R^N,(\cdot,\cdot))$, where $(\cdot,\cdot)$ denotes the standard Euclidean inner product. Let $A$ be a $N \times N$ positive definite matrix. Let $B$ be a $M \times N$ matrix, with $M \le N-1$ with $\text{rank }B=M$.
Show that the following linear system has one and only one solution $(u,\lambda)\in \mathbb R^N \times \mathbb R^M$ $$ \begin{cases} Au +B^T \lambda &=c,\\ Bu&=d. \end{cases} $$
Thanks in advance!
I found this useful formula in the book by Bernstein: Matrix Mathematics, Princeton Univeristy Press, 2005. I applied it to the studied problem and the outcome is the following ($\mathbb O$ denotes the zero matrix)
$$ \begin{pmatrix} A&B^T\\ B&\mathbb O \end{pmatrix}^{-1}\\ \quad= \begin{pmatrix} A^{-1}+A^{-1} B^T(\mathbb O- BA^{-1}B^T)^{-1}BA^{-1} & -A^{-1}B^T(\mathbb O-BA^{-1}B^T)^{-1} \\ -(\mathbb O-BA^{-1}B^T)^{-1}BA^{-1} & (\mathbb O-BA^{-1}B^T)^{-1} \end{pmatrix} $$
but I need to show that $B A^{-1}B^T$ is invertible and I do not know how to proceed.
Well $B^T$ has rank $M$ so you have a left inverse $C$ such that $CB=I_M$.
Similarly $B$ has a right inverse $C'$ such that $BC'=I_M$.
So $C'AC$ is a suitable inverse for $BA^{-1}B^T$.