# The Unapologetic Mathematician

## 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.