[SOLVED] Linear transformations

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May 12th 2010, 12:31 PM
Bruno J.

[SOLVED] Linear transformations

Show that every linear transformation $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is the sum of two invertible linear transformations $T_1, T_2$ .
May 12th 2010, 01:05 PM
NonCommAlg

Quote:

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 $A=\lambda I + B.$
May 12th 2010, 11:22 PM
Bruno J.

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 $M=\mathbb{R}_{n\times n}$ 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 $U \cap G \not=\emptyset$ . $\ \ \ \ \ \square$
May 12th 2010, 11:35 PM
Drexel28

Quote:

Originally Posted by Bruno J.

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 $M=\mathbb{R}_{n\times n}$ 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?
May 12th 2010, 11:42 PM
Bruno J.

Quote:

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 $M \neq G$ .
May 12th 2010, 11:44 PM
Drexel28

Quote:

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.
May 12th 2010, 11:47 PM
Bruno J.

Quote:

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...
May 12th 2010, 11:48 PM
Bruno J.

Quote:

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!
May 12th 2010, 11:50 PM
Drexel28

Quote:

Originally Posted by Bruno J.

You are right about $G$ not being connected!

Haha, I know. $G=\text{det}^{-1}((-\infty,0))\cup\text{det}^{-1}((0,\infty))$