# Theoretical question about rank and invertibility of a block matrix,

Let A and B be real matrices, A is symmetric, and B has at least as many columns as rows.

$$C= \begin{bmatrix} A & B^t \\ B & 0 \\ \end{bmatrix}$$

a) Prove that C is invertible if A is positive-definite and B of full rank.

b) What if A is positive-definite and B is not of full rank?

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

EDIT: part (a) is it simply just using the positive-definiteness of A (positive, non-zero determinant = full rank) to claim that A has full rank, and since B has full rank, too, then the whole matrix C has full rank and is thus invertible. Is this argument ok? Thanks,

## 1 Answer

For $\mathbf{C}$ to be invertible, its rows (or columns) must be linearly independent. Consider the lower rows of $\mathbf{C}$ which are $[\mathbf{B}\:\: \mathbf{0}]$. For these rows to be linearly independent, rows of $\mathbf{B}$ must be linearly independent and since $\mathbf{B}$ is a fat matrix (more columns than rows), this means that $\mathbf{B}$ must be full-rank. Therefore, a necessary condition for $\mathbf{C}$ to be invertible is that $\mathbf{B}$ must be full-rank. :D

This answers part (b). Nothing comes to my mind about $\mathbf{A}$ right now. So (a) and (c) need to be answered yet.

• Thanks so much. So that answers both parts a) and b). What about c)? C) sounds like a trick question with a counterexample... @user298935 – user303506 Jan 8 '16 at 3:21
• Especially because B is a fat matrix...hmmm... – user303506 Jan 8 '16 at 3:22
• No, my answer is for part (b) only, I'll edit my answer now. – user298935 Jan 8 '16 at 3:23