The definition of $GL(n)$

Let $F$ denote a field and let $GL(n,F)$ denote the set of all $n\times n$ matrices with entries in $F$ having non-zero determinant. This is called the general linear group of degree $n$ over $F$ . Each element $g\in GL(n,F)$ defined a function $g:F^n\rightarrow F^n$ by (5.1).

Proposition 5.5.4   $GL(n,F)$ is a group under ordinary matrix multiplication.

proof: The identity matrix is the identity element. Associativity is a property of matrix multiplication. (Left to the reader as an exercise.) The set $GL(n,F)$ is closed under multiplication by Lemma 5.5.3. Inverses exist by Lemma 5.5.2. $\Box$

Let

$$A= \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$

be a $2\times 2$ matrix with $a,b,c,d\in \mathbb{Z}$ . Suppose that the determinant ^5.3 of $A$ is $1$ , $ad-bc=1$ . Thanks to Theorem 1.4.3, this forces $gcd(a,b)=1$ (why?). Similarly, we must have $gcd(a,c)=1$ , $gcd(b,d)=1$ , and $gcd(c,d)=1$ . On other words, to each $2\times 2$ integer matrix with determinant $1$ , is associated several pairs of integers with no common factor.

Conversely, if $a,b\in \mathbb{Z}$ have no common factor then by Theorem 1.4.3 there are $m,n\in \mathbb{Z}$ such that $am+bn=1$ . This means that the matrix

$$A= \left( \begin{array}{cc} a & b \\ -n & m \end{array} \right)$$

has determinant $1$ .

Exercise 5.5.5   Let $SL(n,F)$ be the subset of all $g\in GL(n,F)$ such that $\det(g)=1$ . Show that $SL(n,F)$ is a group.

Exercise 5.5.6   Let $SL(2,\mathbb{Z})$ be the subset of all $g\in GL(2,\mathbb{Z})$ such that $\det(g)=1$ . Show that $SL(2,\mathbb{Z})$ is a group.

Exercise 5.5.7   Find two matrices of the form

$$\left( \begin{array}{cc} 5 & 4 \\ * & * \end{array} \right)$$

having determinant $1$ .

Exercise 5.5.8   Find a matrix of the form

$$\left( \begin{array}{cc} * & 7 \\ * & 9 \end{array} \right)$$

having determinant $1$ .

Exercise 5.5.9   Find a matrix of the form

$$\left( \begin{array}{cc} * & * \\ 11 & 13 \end{array} \right)$$

having determinant $1$ .

Exercise 5.5.10   Find a matrix of the form

$$\left( \begin{array}{cc} 15 & * \\ 11 & * \end{array} \right)$$

having determinant $1$ .

Exercise 5.5.11   Compute

$$\det\left( \begin{array}{cccc} 1&2&3&4\\ 5&6&7&8\\ 9&0&0&9\\ 1&1&1&1 \end{array} \right)$$

using the Lagrange expansion

Exercise 5.5.12   Let

$$A= \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$

have determinant $1$ . Show that there is a matrix

$$B=\left( \begin{array}{cc} a' & b' \\ c' & d' \end{array} \right)$$

with $a',b',c',d'\in \mathbb{Z}$ and having determinant $1$ , such that $AB$ has the form

$$\left( \begin{array}{cc} x & y \\ 0 & 1 \end{array} \right)$$

for some $x,y\in \mathbb{Z}$ . (Hint: Take $a'=d$ , $c'=-c$ , and use Theorem 1.4.3 to determine $b',d'$ .)

Exercise 5.5.13   Imagine a chessboard in front of you. You can place at most $8$ non-attacking rooks on the chessboard. (Rooks move only horizontally and vertically.) Now imagine you have done this and let $A=(a_{ij})$ be the $8\times 8$ matrix of 0 's and $1$ 's (called a $(0,1)$ -matrix) where $a_{ij}=1$ if there is a rook on the square belonging to the $i^{th}$ horizontal down and the $j^{th}$ vertical from the left. Call such a matrix a rook matrix. If there are exactly 8 $1$ 's in $A$ then we shall call $A$ a full rook matrix. For example,

$$\left( \begin{array}{cccccccc} 0&0&0&0&0&0&1&0\\ 0&0&0&... ...\ 0&0&0&1&0&0&0&0\\ 0&1&0&0&0&0&0&0 \end{array} \right)$$

is a full rook matrix. Show that

(a) a full rook matrix is an $8\times 8$ permutation matrix,

(b) any full rook matrix is invertible,

(c) the product of any two rook matrices is a rook matrix.