# How to show that $A$ is invertible?

Let A be an orthogonal matrix partitioned as $\displaystyle A=\begin{pmatrix}A_{11} & A_{12}\\A_{21} & A_{22}\end{pmatrix}$ where $A_{11}$ is a square matrix. Prove that $A_{11}$ is invertible if and only if $A_{22}$ is invertible.

I was looking this problem as $AA^{-1}$=I as I don't how to proceed further

please help me ... but it's my humble request please don't downvote my question

## 1 Answer

Clearly it suffices to show $\det A_{11}=0 \Longrightarrow \det A_{22}=0$, because if we can do this, we can also show $\det A_{22}=0 \Longrightarrow \det A_{11}=0$ by the symmetry of the problem and then we are done.

Investigating the condition $A^TA=I$ yields:

$$A_{12}^TA_{11}=-A_{22}^TA_{21}$$ $$A_{11}^TA_{11}+A_{21}^TA_{21}=I$$ $$A_{22}^TA_{22}+A_{12}^TA_{12}=I$$

Note that the dimensions of the $I$'s might be different, but we do not have to bother.

Assume $\det A_{11}=0$ and let $v \neq 0$ be contained in the kernel. The second equation yields $A_{21}^TA_{21}v=v \neq 0$, in particular $A_{21}v \neq 0$. The first equation yields $0=A_{22}^TA_{21}v$, i.e. $0 \neq A_{21}v \in \operatorname{ker} A_{22}^T$. This shows the desired $\det A_{22} = \det A_{22}^T=0$.