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Let $A$ be orthogonal matrix such that $A=$\begin{bmatrix} A_{11} & A_{12}\\A_{21} & A_{22}\end{bmatrix}

Show that $A_{11}$ is invertible $\iff $ $A_{22}$ is invertible.

Note :$A_{11}$ is square matrix.

Attempt:

Suppose that $A_{11}$ is non-invertible then there exists $v_0\neq 0$ such that $A_{11}v=0$

Since $A$ is othogonal $\implies A_{11}A_{21}+A_{12}A_{22}=0;A_{11}A_{12}+A_{21}A_{22}=0;A_{11}^2+A_{12}^2=I$

For any $v\neq 0$ we have $(A_{11}^2+A_{12}^2)v=v\neq 0\implies A_{12}v_0\neq 0$

Similarly $A_{21}v_0\neq 0$

But I could not proceed further.Will you please give some hints.

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2 Answers2

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If $A_{11}$ is singular, $A_{11}v=0$ for some $v\neq 0$. Then $$ Bv:=\begin{bmatrix} A_{11}\\A_{21} \end{bmatrix} v =: \begin{bmatrix} 0\\w \end{bmatrix}, $$ where $w\neq 0$ since $B$ has orthonormal columns. The orthogonality of $A$ gives $$ A_{22}^TA_{21}=-A_{12}^TA_{11} $$ and hence multiplying with $v$ from right gives $$ A_{22}^Tw=0. $$ Since $w\neq 0$, $A_{22}$ is singular.

Algebraic Pavel
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We have $$A_{11}A_{11}^T=I-A_{12}A_{12}^T$$ from the first entry of $AA^T=I$, and $$A_{22}^TA_{22}=I-A_{12}^TA_{12}$$ from the last entry of $A^TA=I$.

If $A_{11}$ is invertible, then so is $A_{11}A_{11}^T$, so that $I-A_{12}A_{12}^T$ is invertible. Since $ab$ and $ba$ have the same nonzero eigenvalues$^1$, $I-A_{12}^TA_{12}$ is also invertible, which implies $A_{22}^TA_{22}$ is invertible, which implies that $A_{22}$ is invertible$^2$. The other direction is similar.

$^1$ If you know how to show this is the case for square matrices (related to a general fact about rings), you can get the rectangular case by adding rows/columns of $0$s to make them square.

$^2$ Used implicitly is that one-sided inverses are two-sided inverses for finite dimensional matrices, which implies that if $ab$ is invertible, then so are $a$ and $b$, applied here with $A_{22}^T=a$ and $A_{22}=b$.

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