Given a block matrix $A = \begin{pmatrix}A_{11} \quad A_{12} \\ A_{21}\quad A_{22}\end{pmatrix}$ that is real orthogonal, where $A_{11}$ and $A_{22}$ are square matrices, I am asked to show that $A_{11}$ is invertible if and only if $A_{22}$ is.
What I've done so far:
From orthogonality, one concludes that the inverse of $A$ is $A^T$, which is writable in block form as $\begin{pmatrix}A_{11}^T \quad A_{21}^T \\ A_{12}^T\quad A_{22}^T\end{pmatrix}$. Ordinary multiplication of $A$ with $A^T$ (i.e. $AA^T + I$) helps us conclude that $$ A_{11}A^T_{11} + A_{12}A_{12}^T = I = A_{21}A_{21}^T + A_{22} A_{22}^T \\ A_{11} A_{21}^T + A_{12} A_{22}^T = 0 = A_{21} A_{11}^T + A_{22} A_{12}^T $$
Now, there is no expression involving $A_{11}$ and $A_{22}$ coming together, hence I am unsure what to do next.
Let's assume $A_{11}$ is invertible. Then the second line can be changed to $$A^T_{21}=-A^{-1}_{11}A_{12}A^T_{22}$$ Plugging this into the first gives $$I=-A_{21}A^{-1}_{11}A_{12}A^T_{22}+A_{22}A^T_{22}=(-A^{-1}_{11}A_{21}A_{12}+A_{22})A^T_{22}$$ The other directions works the same way.