The Unapologetic Mathematician

Mathematics for the interested outsider

The General Linear Groups

Not just any general group \mathrm{GL}(V) for any vector space V, but the particular groups \mathrm{GL}_n(\mathbb{F}). I can’t put LaTeX, or even HTML subscripts in post titles, so this will have to do.

The general linear group \mathrm{GL}_n(\mathbb{F}) is the automorphism group of the vector space \mathbb{F}^n of n-tuples of elements of \mathbb{F}. That is, it’s the group of all invertible linear transformations sending this vector space to itself. The vector space \mathbb{F}^n comes equipped with a basis \{e_i\}, where e_i has a {1} in the ith place, and {0} elsewhere. And so we can write any such transformation as an n\times n matrix.

Let’s look at the matrix of some invertible transformation T:
\displaystyle\begin{pmatrix}t_1^1&t_2^1&\cdots&t_n^1\\t_1^2&t_2^2&\cdots&t_n^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_n^n\end{pmatrix}

How does it act on a basis element? Well, let’s consider its action on e_1:
\displaystyle\begin{pmatrix}t_1^1&t_2^1&\cdots&t_n^1\\t_1^2&t_2^2&\cdots&t_n^2\\\vdots&\vdots&\ddots&\vdots\\t_1^n&t_2^n&\cdots&t_n^n\end{pmatrix}\begin{pmatrix}1\\{0}\\\vdots\\{0}\end{pmatrix}=\begin{pmatrix}t_1^1\\t_1^2\\\vdots\\t_1^n\end{pmatrix}

It just reads off the first column of the matrix of T. Similarly, T(e_i) will read off the ith column of the matrix of T. This works for any linear endomorphism of \mathbb{F}^n: 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 T must form a basis for \mathbb{F}^n.

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 n, and we’ve got n column vectors to consider. If all n are linearly independent, then the column rank of the matrix is n. Then the dimension of the image of T is n, and thus T is surjective.

On the other hand, any vector T(v) in the image of T is a linear combination of the columns of the matrix of T (use the components of v as coefficients). If these columns are linearly independent, then the only combination adding up to the zero vector has all coefficients equal to {0}. And so T(v)=0 implies v=0, and T is injective.

Thus we only need to check that the columns of the matrix of T are linearly independent to know that T is invertible.

Conversely, say we’re given a list of n linearly independent vectors f_i in \mathbb{F}^n. They must be a basis, since any linearly independent set can be completed to a basis, and a basis of \mathbb{F}^n must have exactly n elements, which we already have. Then we can use the f_i as the columns of a matrix. The corresponding transformation T has T(e_i)=f_i, 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 n\times n matrices. They are exactly the ones so that the set of columns is linearly independent.

October 20, 2008 Posted by | Algebra, Group Examples, Linear Algebra | 1 Comment

   

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