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Why is the group $GL(n, \mathbb{R})$ of dimension $n^{2}$?

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    The group $M(n, \mathbb{R})$ of $n\times n$ matrices has dimension $n^2$. Why? It's easy to show that $M(n, \mathbb{R})$ a vector space. Any element $\in M(n, \mathbb{R})$ can be written in terms of a sum of $n^2$ matrices $B_{ij}$ where the entry $i,j)$ of $B_{ij} = 1$, and zero otherwise. You can show that the set $\{B_{ij}\}$ is a basis of $M(n, \mathbb{R})$, and hence $M(n, \mathbb{R})$ has dimension $n^2$. $GL(n,\mathbb{R})$ is a subset of $M(n, \mathbb{R})$ under the determinant map. It has the same dimension by Steve's answer below.2012-03-06

2 Answers 2

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It is an open submanifold of the set of $n$ by $n$ matrices, which is a vector space of dimension $n^2$.

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    Woah, did not expect so many responses. Thanks math.SE!, it makes sense now.2012-03-07
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You should notice that the determinant is a continuous map $f:M(n,\mathbb{R})\equiv\mathbb{R}^{n^2}\to \mathbb{R}$, $f(X)=\det(X)$

Then note that

$GL(n, \mathbb{R})=f^{-1}(\mathbb{R}\setminus\{0\})$

How the pre-image by continuous map of open set is an open set then $GL(n, \mathbb{R})$ is an open set of $\mathbb{R}^{n^2}$ whose dimension is $n^2$.

(I am here using that an open set of $\mathbb{R}^{k}$ has $k$-dimension.)

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    If a manifold as a topological space is locally homeomorphic to $R^n$ then its dimension is $n$. We can also think about the Haussdorf Dimension.2012-11-23