Positive Definite Matrix (Block Matrix)

Question

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?

Answer 1

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 $(a + b)^2$?

Answer 2

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?

Answer 3

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 $0$