# Algebra.....units

• Feb 28th 2006, 03:31 PM
suedenation
Algebra.....units
Find all the units in M2(Z2)

Thanks guys!!! :)
• Feb 28th 2006, 05:17 PM
ThePerfectHacker
Quote:

Originally Posted by suedenation
Find all the units in M2(Z2)

Thanks guys!!! :)

You need to find all the matrices, of form
$\left( \begin{array}{cc}a&b\\c&d\end{array} \right)$ where $a,b,c,d\in\mathbb{Z}_2$ such as it has an inverse.
There are 16 possible such matrices, check each one if it is invertible.

If I am not making a mistake, then I think that,
$\left( \begin{array}{cc}0&1\\1&0\end{array} \right)$
and,
$\left( \begin{array}{cc}1&0\\0&1\end{array} \right)$
and,
$\left( \begin{array}{cc}0&1\\1&1\end{array} \right)$
and,
$\left( \begin{array}{cc}1&1\\1&0\end{array} \right)$
and,
$\left( \begin{array}{cc}1&1\\0&1\end{array} \right)$
and,
$\left( \begin{array}{cc}1&0\\1&1\end{array} \right)$
Are the only units in this ring.
• Feb 28th 2006, 05:53 PM
ThePerfectHacker
I was thinking of a more elegant way of solving this. There is a theorem in Linear Algebra which states that a square matrix is invertible (a unit) if and only if its determinant is non-zero.
The problem is I do not know if this is going to work here, possibly because $\mathbb{Z}_2$ is not a field.

But let us assume, thus we need that,
$\left| \begin{array}{cc}a&b\\c&d\end{array} \right|\not =0$
Since there are only two elements in the set by elimination we have that,
$\left| \begin{array}{cc}a&b\\c&d\end{array} \right|=1$
Thus, by evaluating the determinant,
$ad-bc=1$
But remember in this ring since it has only two elements, that $-x=x$ for all $x$.
Thus,
$ad+bc=1$
Now, just observe the possibilities for $a,b,c,d$ which make this equation true. Since there are only 16 of them it is easy.

1)If $a=0$ then, $b,c\not=0$ thus, $b,c=1$ and $d$ can be anything (either 1 or 0) this gives us two matrices,
$\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ and $\left(\begin{array}{cc}0&1\\1&1\end{array}\right)$.

2)If $a=1$ then , $d=1$ would mean that either $b$ or $c$ must be zero. Also, if $d=0$ would mean that both $a$ and $b$ cannot be zero (thus they are both one). This gives us four matrices,
$\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ and, $\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$ and,
$\left(\begin{array}{cc}1&0\\1&1\end{array}\right)$ and, $\left(\begin{array}{cc}1&1\\1&0\end{array}\right)$

Now there is no need to observe what happens with $b,c,d$ because we know what matrix would it be depending on the value of $a$.
• Mar 12th 2006, 02:42 AM
rgep
Quote:

Originally Posted by ThePerfectHacker
There is a theorem in Linear Algebra which states that a square matrix is invertible (a unit) if and only if its determinant is non-zero.
The problem is I do not know if this is going to work here, possibly because $\mathbb{Z}_2$ is not a field.

It works fine, because $\mathbb{Z}_2$ actually is a field.

Here's another way which generalises to any finite field $GF(q)$ and any dimension $n$.

A square matrix is invertible if and only if its rows are linearly independent. Take each row in turn. The first row must be non-zero so there are $q^n-1$ possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are $q^n-q$ possiblities. The $(r+1)$-th row must not be in the subspace generated by the previous $r$ rows, linearly independent by induction, so of dimension $r$ giving $q^n-q^r$ possibilities. The overal number of possiblities is thus $(q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})$

In the case $q=2, n=2$ this is $(2^2-1)(2^2-2^1) = 6$.
• Mar 12th 2006, 09:41 AM
ThePerfectHacker
Quote:

Originally Posted by rgep
It works fine, because $\mathbb{Z}_2$ actually is a field.

Here's another way which generalises to any finite field $GF(q)$ and any dimension $n$.

A square matrix is invertible if and only if its rows are linearly independent. Take each row in turn. The first row must be non-zero so there are $q^n-1$ possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are $q^n-q$ possiblities. The $(r+1)$-th row must not be in the subspace generated by the previous $r$ rows, linearly independent by induction, so of dimension $r$ giving $q^n-q^r$ possibilities. The overal number of possiblities is thus $(q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})$

In the case $q=2, n=2$ this is $(2^2-1)(2^2-2^1) = 6$.

Thanks for the tip!