Not just any general group for any vector space , but the particular groups . I can’t put LaTeX, or even HTML subscripts in post titles, so this will have to do.
The general linear group is the automorphism group of the vector space of -tuples of elements of . That is, it’s the group of all invertible linear transformations sending this vector space to itself. The vector space comes equipped with a basis , where has a in the th place, and elsewhere. And so we can write any such transformation as an matrix.
Let’s look at the matrix of some invertible transformation :
How does it act on a basis element? Well, let’s consider its action on :
It just reads off the first column of the matrix of . Similarly, will read off the th column of the matrix of . This works for any linear endomorphism of : its columns are the images of the standard basis vectors. But as we said last time, an invertible transformation must send a basis to another basis. So the columns of the matrix of must form a basis for .
Checking that they’re a basis turns out to be made a little easier by the special case we’re in. The vector space has dimension , and we’ve got column vectors to consider. If all are linearly independent, then the column rank of the matrix is . Then the dimension of the image of is , and thus is surjective.
On the other hand, any vector in the image of is a linear combination of the columns of the matrix of (use the components of as coefficients). If these columns are linearly independent, then the only combination adding up to the zero vector has all coefficients equal to . And so implies , and is injective.
Thus we only need to check that the columns of the matrix of are linearly independent to know that is invertible.
Conversely, say we’re given a list of linearly independent vectors in . They must be a basis, since any linearly independent set can be completed to a basis, and a basis of must have exactly elements, which we already have. Then we can use the as the columns of a matrix. The corresponding transformation has , and extends from there by linearity. It sends a basis to a basis, and so must be invertible.
The upshot is that we can consider this group as a group of matrices. They are exactly the ones so that the set of columns is linearly independent.