Dimension of $GL(n, \mathbb{R})$ Why is the group $GL(n, \mathbb{R})$ of dimension $n^{2}$?
 A: I'm sorry people misunderstood your question. You are clearly asking about dimension in terms of linear algebra and not in terms of manifolds.
The set $GL(n)$ is not a subspace. To see this, take any invertible matrix $A$, then $-A$ is also invertible. But $A-A=0$ is obviously not invertible, thus $GL(n)$ is not a vector subspace for any $n$ so it makes no sense to talk about it's dimension in terms of independent vectors.
However, GL(n) is what is known as a submanifold which means that although $GL(n)$ is not a vector space, you can "locally" (i.e. around some neighborhood around the set) view it a vector space structure. But this is something you shouldn't worry about in linear algebra.
A: It is an open submanifold of the set of $n$ by $n$ matrices, which is a vector space of dimension $n^2$.
A: 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.)
