invertible block matrix equivalence

Question

Let $$A=\begin{pmatrix} R&v \\u^T & 0 \end{pmatrix}$$ where $R$ is an invertible upper triangular matrix and $u,v \in \mathbb R^n$

Prove that $A$ is invertible if and only if $u^TR^{-1}v\neq0$

Would appreciate any help

Answer 1

Let x = $0 \in \mathbb{R}^n$

Let $$B=\begin{pmatrix} R^{-1}&x \\x^T & 1 \end{pmatrix}$$

$B$ is invertible => $A$ is invertible iff $C = AB$ is invertible

$$C=\begin{pmatrix} E&v \\u^TR^{-1} & 0 \end{pmatrix}$$

So, you can prove it only for $R = E$

Answer 2

If $u^TR^{-1}v=0$ then $$\left(\begin{array}{cc} R & v \\ u^T & 0 \end{array}\right)\left(\begin{array}{c} R^{-1}v \\ -1\end{array}\right) = 0$$ which means that there is a vector $x\neq 0$ that satisfies $Ax=0$, $A$ is not invertible.

If $c= u^TR^{-1}v\neq 0$, we find $$\left(\begin{array}{cc} R & v \\ u^T & 0 \end{array}\right)\left(\begin{array}{cc} cR^{-1}-R^{-1}vu^TR^{-1} & R^{-1}v \\ u^TR^{-1} & -1 \end{array}\right) = cI$$

Answer 3

According to the determinant formula for block matrices $$ \det A=\det\begin{bmatrix}R & v\\u^T & 0\end{bmatrix}=\det R\cdot\det(0-u^TR^{-1}v)=-\det R\cdot u^TR^{-1}v. $$ Since $\det R\ne 0$ we have $\det A\ne 0$ iff $u^TR^{-1}v\ne 0$.