Positive Definite Matrix (Block Matrix) Let $B$ be an $m\times n$ matrix. Is $$ A=\begin{pmatrix} I & B \\ B^T & I+B^TB \end{pmatrix} $$ positive definite?
Attempt:
Let $\mathbf{z}=\begin{pmatrix} \mathbf{x} \\ \mathbf{y} \end{pmatrix}$. To show that $A$ is positive definite, $\mathbf{z}^TA\mathbf{z}>0$. Expanding $\mathbf{z}^TA\mathbf{z}>0$ gives $\mathbf{x}^T\mathbf{x}+\mathbf{y}^T\mathbf{y}+(B\mathbf{y})^T(B\mathbf{y})+2\mathbf{x}^TB\mathbf{y}$. The first three terms are positive, but what can be concluded about the $2\mathbf{x}^TB\mathbf{y}$ term?
 A: One hint could be to try to rewrite with sum of self-outer products:
$$A = \begin{bmatrix}I&B\\B^T&I+B^TB\end{bmatrix}= \cdots\\=\begin{bmatrix}I\\B^T\end{bmatrix} \begin{bmatrix}I&B\end{bmatrix} + \begin{bmatrix}0\\I\end{bmatrix} \begin{bmatrix}0&I\end{bmatrix}$$
What can we say about it now?
A: Hint: can you rewrite $x^T x + 2x^T (By) + (By)^T (By)$ in other way? If $a, b \in \mathbb{R}$, then how to expand formula $(a + b)^2$?
A: Sylvester's Law of Inertia. 
We have your symmetric $A.$ We are going to construct $P^T A P = D$ where $\det P \neq 0$ and $D$ is diagonal. Then the count of positive eigenvalues for $A$ is the same as the count of positive eigenvalues of $D$
$$
\left(
\begin{array}{cc}
I&0 \\
-B^T& I \\
\end{array}
\right)
\left(
\begin{array}{cc}
I&B \\
B^T&I+B^T B \\
\end{array}
\right)
\left(
\begin{array}{cc}
I&-B \\
0&I \\
\end{array}
\right) =
\left(
\begin{array}{cc}
I&0 \\
0&I \\
\end{array}
\right)
$$
Sylvester says that your $A$ is positive definite. 
Or, given my invertible
$$
P=
\left(
\begin{array}{cc}
I&-B \\
0&I \\
\end{array}
\right)
$$
let us switch to $Q = P^{-1}$
$$
Q=
\left(
\begin{array}{cc}
I&B \\
0&I \\
\end{array}
\right)
$$
and write $Q^T D Q = A.$ Since $D$ turned out to be the identity matrix 
$$
\left(
\begin{array}{cc}
I&0 \\
B^T&I \\
\end{array}
\right)
\left(
\begin{array}{cc}
I&B \\
0&I \\
\end{array}
\right) =
\left(
\begin{array}{cc}
I&B \\
B^T&I+B^T B \\
\end{array}
\right)
$$
As $Q$ is nonsingular this again says $Q^T Q = A$ is positive definite. We can write using dot products, you wrote a column vector
 $$
z=
\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)
$$
after which
 $$
Qz=
\left(
\begin{array}{c}
x + B y \\
y \\
\end{array}
\right)
$$
and
$$ (Qz)^T (Qz) = |x+By|^2 + |y|^2  $$
If $y\neq0$ this is positive. If $y=0,$ we are left with $|x|^2,$ and this is positive unless $x$ is also $0$
