[SOLVED] Linear transformations

Bruno J.

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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

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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.

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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

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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.

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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

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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
Jun 2009
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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

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