For a square matrix $M$ call any square matrix M' of the form $$\left(\begin{array}{cc} M & A\\ B & C \end{array}\right)$$ an extension of $M$. Does it follow that if $M$ is not invertible that all extensions $M'$ are not invertible? I believe the answer is no. If not, is there an extension that is invertible? Can we prove that there always is?

Robert Lewis 06/12/2018.

My answer is similar in spirit to Lord Shark the Unknown's, but provides a few more details.

First of all, as Lord Shark affirms, for any matrix square matrix $M$, the extension matrix $E_M$,

$E_M = \begin{bmatrix} M & I \\ I & 0 \end{bmatrix}, \tag 1$

is invertible. The easiest way to see this is to show that

$\ker E_M = \{ 0 \}; \tag 2$

now if

$\text{size}(M) = n, \tag 3$

that is, $M$ is an $n \times n$ matrix over some field $\Bbb F$, then

$\text{size}(E_M) = 2n; \tag 4$

thus $E_M$ may be considered as operating on the $2n$-dimensional vector space $\Bbb F^{2n}$, any vector $v \in \Bbb F^{2n}$ of which may be written in "stacked form"

$v = \begin{pmatrix} x \\ y \end{pmatrix}, \tag 5$

where $x, y \in \Bbb F^n$; then if

$E_M v = 0, \tag 6$

we have

$\begin{pmatrix} Mx + y \\ x \end{pmatrix} = \begin{bmatrix} M & I \\ I & 0 \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0, \tag 7$

from which we conclude

$x = 0, \tag 8$

and

$y = -Mx = -M(0) = 0; \tag 9$

thus

$v = 0, \tag{10}$

which shows that (2) binds and thus that $E_M$ is invertible. We may in fact find the inverse $E_M^{-1}$ by setting

$E_M v = w, \tag{11}$

where

$w = \begin{pmatrix} s \\ t \end{pmatrix}; \tag{12}$

then as above we have

$Mx + y = s, \; x = t, \tag{13}$

whence

$y = s - Mt; \tag{14}$

(13) and (14) together show that

$E_M^{-1} = \begin{bmatrix} 0 & I \\ I & -M \end{bmatrix}, \tag{15}$

which we may easily check:

$E_M^{-1} w = \begin{bmatrix} 0 & I \\ I & -M \end{bmatrix} \begin{pmatrix} s \\ t \end{pmatrix} = \begin{pmatrix} t \\ s - Mt \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}, \tag{16}$

using (13) and (14).

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