1. ## [SOLVED] Linear transformations

Show that every linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is the sum of two invertible linear transformations $T_1, T_2$.

2. Originally Posted by Bruno J.
Show that every linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is the sum of two invertible linear transformations $T_1, T_2$.
that is actually true for any field $F$ with $\text{card}(F) \geq n+2.$ let $A$ be the matrix of $T$ in the standard basis. let $I$ be the $n \times n$ identity matrix and $f(x)=\det(A-xI) \in \mathbb{R}[x],$ which is a polynomial

of degree $n$ and so it has (at most) $n$ roots in $\mathbb{R}.$ choose $0 \neq \lambda \in \mathbb{R}$ such that $f(\lambda) \neq 0.$ then $B=A-\lambda I$ is invertible and $A=\lambda I + B.$

3. Good! Minced meat for NCA!

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

If $T$ is invertible, then the theorem is trivial with $T_1=T_2=T/2$.

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

4. Originally Posted by Bruno J.
Good! Minced meat for NCA!

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

If $T$ is invertible, then the theorem is trivial with $T_1=T_2=T/2$.

Therefore suppose $T \in F$. Let $U=G+T=\{f+T : f\in G\}$. It's clear that $U$ is dense in $M$ since $G$ is dense in $M$. If $U\subset F$, then $M = \overline U \subset \overline F \subset M$ and therefore $F$ is dense in $M$ which is false. Therefore $U \cap G \not=\emptyset$. $\ \ \ \ \ \square$
So, I'm not super knowledgeable about this kind of stuff, but isn't $M_{n\times n}\approx\mathbb{R}^{n^2}$ and thus connected and aren't $U,G$ open and their union is $M_{n\times n}$ and so if they were disjoint this would contradict the assumption that $M_{n\times n}$ is connected?

5. Originally Posted by Drexel28
So, I'm not super knowledgeable about this kind of stuff, but isn't $M_{n\times n}\approx\mathbb{R}^{n^2}$ and thus connected and aren't $U,G$ open and their union is $M_{n\times n}$ and so if they were disjoint this would contradict the assumption that $M_{n\times n}$ is connected?
It's not true that $M=U \cup G$ necessarily. For instance if $T=0$ then $U=G$, but $M \neq G$.

6. Originally Posted by Bruno J.
It's not true that $M=U \cup G$ necessarily. For instance if $T=0$ then $U=G$, but $M \neq G$.
Well, what is $U\cup G$? Anything of importance? I feel like a connectedness argument should work here. I know that $G$'s not connected? I'm grasping for straws and not really putting effort into this. Ignore me if you want.

7. Originally Posted by Drexel28
Well, what is $U\cup G$? Anything of importance?
I don't know... I don't use it in the proof...

8. Originally Posted by Drexel28
Well, what is $U\cup G$? Anything of importance? I feel like a connectedness argument should work here. I know that $G$'s not connected? I'm grasping for straws and not really putting effort into this. Ignore me if you want.
You are right about $G$ not being connected!

9. Originally Posted by Bruno J.
You are right about $G$ not being connected!
Haha, I know. $G=\text{det}^{-1}((-\infty,0))\cup\text{det}^{-1}((0,\infty))$