Given $A\in \mathcal M_{k\times l}(\Bbb F)$, prove that matrix $M=\begin{pmatrix} \ I_k & A \\ A^T & -I_l \end{pmatrix}$ is invertible

Question

Given $A\in \mathcal M_{k\times l}(\Bbb F)$, prove that the following matrix $$M=\begin{pmatrix} \ I_k & A \\ A^T & -I_l \end{pmatrix}$$ is invertible.

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I have tried to find its determinant and its rank, but it was not helpful in figuring it out. Could someone enlighten me on how I can find out whether the matrix is invertible?

Answer 1

For $A,B,C,D$ matrices of appropriate sizes, you have the formulae

$$\det(M)= \det(A-BD^{-1}C)\det(D)$$ where

$$M = \begin{pmatrix} A & B\\ C & D \end{pmatrix}.$$

Which gives in your particular case that

$$\det(M) = \det(I_k+A^TA)\det(-I_l).$$

We're left to prove that $\det(I_k+A^TA) \neq 0$, i.e. that $-1$ is not an eigenvalue of $A^TA$. And indeed if that was the case, we would be able to find $X \neq 0$ such that

$$\Vert AX \Vert^2 = X^T A^T A X = -X^TX = -\Vert X \Vert^2$$ and that can't be as the left side of the equality is non negative while the right side is strictly negative.

Note: following the answer of "levap", I made here the assumption that $\mathbb F$ is $\mathbb R$ or a formally real field.

Answer 2

The result is false as stated. For example, we can take $A = (i) \in M_{1 \times 1}(\mathbb{C})$ and then

$$ M = \begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix} $$

with

$$ \det(M) = -1 - i^2 = 0. $$

However, note that we have

$$ M^2 = \begin{pmatrix} I & A \\ A^T & -I \end{pmatrix}\begin{pmatrix} I & A \\ A^T & -I \end{pmatrix} = \begin{pmatrix} I + AA^T & 0 \\ 0 & I + A^T A \end{pmatrix} $$

and so

$$ \det(M^2) = \det(M)^2 = \det(I - AA^T) \det (I - A^T A). $$

If $\mathbb{F} = \mathbb{R}$, the matrices $AA^T$ and $A^TA$ are positive semi-definite and so have only non-negative eigenvalues. Hence, $1$ is not an eigenvalue of $AA^T,A^TA$ and so $I - AA^T,I - A^TA$ are invertible and so is $M$.