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 $i$ th 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 $i$ th 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 John Armstrong | Algebra, Group Examples, Linear Algebra

1 Comment »

[…] Linear Groups — Generally Monday, we saw that the general linear groups are matrix groups, specifically consisting of those whose columns […]

Pingback by General Linear Groups — Generally « The Unapologetic Mathematician | October 22, 2008 | Reply

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects
Subjects
Archives

October 2008

M T W T F S S

« Sep Nov »

1 2 3 4 5

6 7 8 9 10 11 12

13 14 15 16 17 18 19

20 21 22 23 24 25 26

27 28 29 30 31
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider