Good! Minced meat for NCA! (Nod)

Here's a topological proof I found. It's much less direct but maybe the idea can be used elsewhere! Let's equip \(\displaystyle M=\mathbb{R}_{n\times n}\) with the Euclidean metric. Note that \(\displaystyle G=\mbox{GL}_n(\mathbb{R})\) is dense in \(\displaystyle M\). Let \(\displaystyle F=M-G\). Note that \(\displaystyle F\) is not dense in \(\displaystyle M\).

If \(\displaystyle T\) is invertible, then the theorem is trivial with \(\displaystyle T_1=T_2=T/2\).

Therefore suppose \(\displaystyle T \in F\). Let \(\displaystyle U=G+T=\{f+T : f\in G\}\). It's clear that \(\displaystyle U\) is dense in \(\displaystyle M\) since \(\displaystyle G\) is dense in \(\displaystyle M\). If \(\displaystyle U\subset F\), then \(\displaystyle M = \overline U \subset \overline F\) and therefore \(\displaystyle F\) is dense in \(\displaystyle M\) which is false. Therefore \(\displaystyle U \cap G \not=\emptyset \). \(\displaystyle \ \ \ \ \ \square\)