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Math Help - [SOLVED] Linear transformations

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    MHF Contributor Bruno J.'s Avatar
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    [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.
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    Quote Originally Posted by Bruno J. View Post
    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.
    Last edited by NonCommAlg; May 12th 2010 at 05:50 PM.
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    MHF Contributor Bruno J.'s Avatar
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    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
    Last edited by Bruno J.; May 13th 2010 at 12:37 AM.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    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?
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    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.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Well, what is U\cup G? Anything of importance?
    I don't know... I don't use it in the proof...
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    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!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    You are right about G not being connected!
    Haha, I know. G=\text{det}^{-1}((-\infty,0))\cup\text{det}^{-1}((0,\infty))
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