linear algebra - Invertibility of block matrices, with the property of being symmetric, positive definite, and of full rank:

If A and B are real matrices, with A being symmetric, B having at least as many columns as rows, and

the matrix C defined as:
$$
\begin{bmatrix}
A & B^T \\
B & 0 \\
\end{bmatrix}
$$

how can I prove that:

1) C is invertible, if A is positive definite and B of full rank

and,

2) Is C always invertible, if A is invertible and B of full rank?

My attempt so far was to sketch the block matrices.

For part 1)
I let A be of size $nxn$. Since A is positive definite, it is invertible, and thus has full rank.

I let B be of size $(k-(n+1)+1)x(p)$ = $(k-n)x(p)$, so that $B^T$ is of size $(p)x(k-n)$. Then C is of size $kxk$. Then I try to argue that, this $kxk$ square matrix has full rank, with rank = k, which implies that C is invertible. B has full row rank, and $B^T$ has full column rank.

Am I sort of close to the answer? I'm basically trying to avoid the usage of determinants of block matrices, as I'm not all that comfortable with that method - but perhaps it's necessary for this question.

For part 2)
My work for part 1), if it's correct, would imply that, yes, C is always invertible if A is invertible and B of full rank. I get the feeling, though, that there is a counterexample.

Thanks in advance for your help,

Answer

Since $A$ is nonsingular, consider the following block factorization of $C$:

$$
C=\pmatrix{A&B^T\\B&0}
=
\pmatrix{I&0\\BA^{-1}&I}\pmatrix{A&0\\0&S}\pmatrix{I&A^{-1}B^T\\0&I},
$$
where $S:=-BA^{-1}B^T$. Since the triangular blocks are nonsingular, the matrix $C$ is nonsingular iff the Schur complement matrix $S$ is nonsingular.

Now if $A$ is SPD, it is easy to see that $S$ is SPD as well. First, the definiteness of $A$ implies that $A^{-1}$ is SPD. For a nonzero $x$, $B^Tx\neq 0$ since has $B$ has full row rank, and $$x^T(BA^{-1}B^T)x=(B^Tx)^TA^{-1}(B^Tx)>0.$$

Another way to see that $C$ is nonsingular if $A$ is SPD and $B$ has full row rank is as follows. Assume that $Cz=0$ for some nonzero $z=(x^T,y^T)^T$. Hence

$$\tag{1}
Ax+B^Ty=0, \quad Bx=0.
$$

None of the block components can be zero. If $x=0$ and $y\neq 0$ then $B^Ty=0$ which is impossible since $B$ has full row rank. If $x\neq 0$ and $y=0$ then $Ax=0$ which is impossible since $A$ is SPD. Hence both $x\neq 0$ and $y\neq 0$. Multiply the first equation in (1) with $x^T$ and the second with $y^T$ to get
$$
x^TAx+x^TB^Ty=0, \quad y^TBx=0.
$$
Since $x^TB^Ty=y^TBx=0$, we have $x^TAx=0$, which gives again a contradiction.

It is not sufficient that $A$ is nonsingular and $B$ of full rank for $S$ being nonsingular. Consider,
$$
A=\pmatrix{1 & 0 \\ 0 & -1},
\quad B=(1,1).
$$
It is easy to verify that $C$ is singular (actually $S=0$).