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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- May 12th 2010, 12:31 PMBruno J.[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 $\displaystyle T_1, T_2$.

- May 12th 2010, 01:05 PMNonCommAlg
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 $\displaystyle A=\lambda I + B.$ - May 12th 2010, 11:22 PMBruno 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 $\displaystyle M=\mathbb{R}_{n\times n}$ 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 $. $\displaystyle \ \ \ \ \ \square$ - May 12th 2010, 11:35 PMDrexel28
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?

- May 12th 2010, 11:42 PMBruno J.
- May 12th 2010, 11:44 PMDrexel28
- May 12th 2010, 11:47 PMBruno J.
- May 12th 2010, 11:48 PMBruno J.
- May 12th 2010, 11:50 PMDrexel28