# [SOLVED] Linear transformations

#### Bruno J.

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

#### NonCommAlg

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

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

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• Bruno J. and Chris L T521

#### Bruno J.

MHF Hall of Honor
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 $$\displaystyle U \cap G \not=\emptyset$$. $$\displaystyle \ \ \ \ \ \square$$

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#### Drexel28

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

#### Bruno J.

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

#### Drexel28

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

#### Bruno J.

MHF Hall of Honor
Well, what is $$\displaystyle U\cup G$$? Anything of importance?
I don't know... I don't use it in the proof...

#### Bruno J.

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

#### Drexel28

MHF Hall of Honor
You are right about $$\displaystyle G$$ not being connected!
Haha, I know. $$\displaystyle G=\text{det}^{-1}((-\infty,0))\cup\text{det}^{-1}((0,\infty))$$