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Thread: [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 $\displaystyle T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is the sum of two invertible linear transformations $\displaystyle T_1, T_2$.
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    Quote Originally Posted by Bruno J. View Post
    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 $\displaystyle A=\lambda I + B.$
    Last edited by NonCommAlg; May 12th 2010 at 04:50 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 $\displaystyle M=\mathbb{R}_{n\times n}$ 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 $. $\displaystyle \ \ \ \ \ \square$
    Last edited by Bruno J.; May 12th 2010 at 11:37 PM.
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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 $\displaystyle M=\mathbb{R}_{n\times n}$ 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?
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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 $\displaystyle M_{n\times n}\approx\mathbb{R}^{n^2}$ 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 $\displaystyle M \neq G$.
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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 $\displaystyle M=U \cup G$ 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.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Well, what is $\displaystyle U\cup G$? 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 $\displaystyle U\cup G$? 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!
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    You are right about $\displaystyle G$ not being connected!
    Haha, I know. $\displaystyle G=\text{det}^{-1}((-\infty,0))\cup\text{det}^{-1}((0,\infty))$
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