Matrix groups

Section 1.3 Matrix groups

Definition 1.3.1. Invertible matrix.

Assume either that \(R\in \{\Z, \Q, \R, \C\}\) or that \(R=\Z/m\Z\) for some \(m\in \Z_{> 0}\text{.}\) Fix a positive integer \(n\text{.}\) The set of all \(n\times n\) matrices with coefficients in \(R\) is denoted \(M_n(R)\text{.}\)

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Given matrices \(A=[a_{ij}], B=[b_{ij}]\in M_n(R)\) we define their product \(AB\) to be the matrix \(C=[c_{ij}]\in M_n(R)\text{,}\) where

\begin{equation} c_{ij}=\sum_{k=1}^na_{ik}b_{kj}\tag{1.3.1} \end{equation}

for all \(1\leq i,j\leq n\text{.}\)

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A matrix \(A\in M_n(R)\) is invertible if there is a matrix \(B\in M_n(R)\) such that

\begin{equation*} AB=BA=I\text{,} \end{equation*}

where \(I\) is the \(n\times n\) identity matrix.

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We denote by \(\GL_n(R)\) the set of all invertible matrices with coefficients in \(R\text{:}\) i.e.,

\begin{equation} \GL_n(R)=\{A\in M_{n}(R)\mid A \text{ is invertible}\}\text{.}\tag{1.3.2} \end{equation}

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Specimen 6. General linear group (generalized).

Assume either that \(R\in \{\Z, \Q, \R, \C\}\) or that \(R=\Z/m\Z\) for some \(m\in \Z_{> 0}\text{.}\) Given a positive integer \(n\text{,}\) the pair \((\GL_n(R),\cdot)\) is a group, where \(\cdot\) is matrix multiplication in \(M_n(R)\text{.}\)

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Remark 1.3.2. Invertible matrices and determinant.

It turns out that matrix algebra in \(M_n(R)\) proceeds essentialy exactly as you are accustomed to, no matter how exotic our ring \(R\) is. In particular, we can define a determinant function

\begin{align*} \det \colon M_n(R) \amp \rightarrow R\\ A \amp \mapsto \det A \end{align*}

using the usual formula, since \(R\) has notions of addition and multiplication. Furthermore, you can show that as usual \(\det\) satisfies

\begin{equation*} \det AB=\det A\det B \end{equation*}

for all \(A,B\in M_n(R)\text{,}\) and that for all \(A\in M_n(R)\) there is a special matrix \(A^\dagger\) called the adjoint matrix of \(A\) that satisfies

\begin{equation*} A^\dagger A=AA^\dagger=(\det A)\, I\text{.} \end{equation*}

From these observations it follows that

\begin{equation*} \GL_n(R)=\{A\in M_{n}(R)\mid \det A\in R^*\}\text{,} \end{equation*}

where \(R^*\) is the group of units of \(R\text{,}\) as defined in Specimen 2 and Specimen 5.

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Of course, that these properties of the determinant and adjoint matrix hold very much requires a proof. We will not do that here, as these will not play a fundamental role in the current course beyond making computation within these groups more convenient. However, the ring theory developed in Math 331-2 will provide a powerful general method that allows us to conclude that properties enjoyed by mathematical objects “defined over” \(\Z\) and \(\C\) are sometimes inherited by the same types of objects “defined over” a more general \(R\text{.}\)

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Example 1.3.3. \(\GL_2(\Z/2\Z)\).

Consider the group \(\GL_2(\Z/2\Z)\text{.}\)

Compute \(\abs{\GL_2(\Z/2\Z)}\) by explicitly enumerating all of its elements. You can use \(0\) and \(1\) to denote the congruence classes \([0]_2, [1]_2\in \Z/2\Z\text{.}\)
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Compute \(\ord A\) for all \(A\in \GL_2(\Z/2\Z)\text{.}\)
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Find \(A,B\in\GL_2(\Z/2\Z)\) such that

\begin{equation} \GL_2(\Z/2\Z)=\{A^iB^j\mid 0\leq i\leq 2, 0\leq j\leq 1\}\text{.}\tag{1.3.3} \end{equation}

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For your choice of \(A, B\) above, express \(BA\) as \(A^iB^j\) as in (1.3.3).
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Use the above to construct a full multiplication table of \(\GL_2(\Z/2\Z)\) in terms of \(A\) and \(B\text{.}\)
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Solution.

Since \((\Z/2\Z)^*=\{1\}\text{,}\) we see that \(A\in M_2(\Z/2\Z)\) to be invertible, we need \(\det A=1\text{.}\) This restriction allows us to easily enumerate the invertible matrices as

\begin{equation*} \GL_2(\Z/2\Z)=\left\{ I, A=\begin{bmatrix} 0\amp 1\\ 1\amp 1 \end{bmatrix}, B=\begin{bmatrix} 0\amp 1\\ 1\amp 0 \end{bmatrix}, \begin{bmatrix} 1\amp 1\\ 1\amp 0 \end{bmatrix}, \begin{bmatrix} 1\amp 0\\ 1\amp 1 \end{bmatrix}, \begin{bmatrix} 1\amp 1\\ 0\amp 1 \end{bmatrix} \right\}\text{.} \end{equation*}

The matrices named \(A\) and \(B\) above satisfy

\begin{align*} A^3 \amp = B^2=I \\ A^2 \amp = \begin{bmatrix} 1\amp 1\\ 1\amp 0 \end{bmatrix}\\ AB \amp =\begin{bmatrix} 1\amp 0\\ 1\amp 1 \end{bmatrix}\\ A^2B \amp = \begin{bmatrix} 1\amp 1\\ 0\amp 1 \end{bmatrix}\text{.} \end{align*}

Thus

\begin{equation*} \GL_2(\Z/2\Z)=\{A^iB^j\mid 0\leq i\leq 2, 0\leq j\leq 1\}\text{.} \end{equation*}

Furthermore, we see that

\begin{equation*} BA=\begin{bmatrix} 1\amp 1\\ 0\amp 1 \end{bmatrix} =A^2B=A^{-1}B\text{.} \end{equation*}

The three relations

\begin{align*} A^3 \amp =I\\ B^2 \amp = I\\ BA \amp =A^2B=A^{-1}B \end{align*}

then easily allow us to compute with the group law: that is as long as we express elements of the group in the form \(A^iB^j\text{.}\) For example, we have

\begin{align*} (A^2B)(AB) \amp = A^2BAB\\ \amp = A^2A^2B^2\\ \amp= A^4 \\ \amp = A \amp (A^3=I)\text{.} \end{align*}

Computations like this allow us to easily compute orders of elements:

\begin{align*} \ord I \amp = 1\\ \ord A =\ord A^2\amp =3\\ \ord B =\ord AB\amp=\ord A^2B=2\text{.} \end{align*}

We leave it to you to compute the entire multiplication table of \(\GL_2(\Z/2/Z)\text{.}\) A good exercise!

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