# Thread:

1. ## Algebra.....units

Find all the units in M2(Z2)

Thanks guys!!!

2. 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.

3. 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 $a$.

4. 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 $(2^2-1)(2^2-2^1) = 6$.

5. 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!
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