Find all the units in M2(Z2)

Thanks guys!!! :)

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- Feb 28th 2006, 03:31 PMsuedenationAlgebra.....units
Find all the units in M2(Z2)

Thanks guys!!! :) - Feb 28th 2006, 05:17 PMThePerfectHackerQuote:

Originally Posted by**suedenation**

$\displaystyle \left( \begin{array}{cc}a&b\\c&d\end{array} \right)$ where $\displaystyle a,b,c,d\in\mathbb{Z}_2$ 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,

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

and,

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

and,

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

and,

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

and,

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

and,

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

Are the only units in this ring. - Feb 28th 2006, 05:53 PMThePerfectHacker
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 $\displaystyle \mathbb{Z}_2$ is not a field.

But let us assume, thus we need that,

$\displaystyle \left| \begin{array}{cc}a&b\\c&d\end{array} \right|\not =0$

Since there are only two elements in the set by elimination we have that,

$\displaystyle \left| \begin{array}{cc}a&b\\c&d\end{array} \right|=1$

Thus, by evaluating the determinant,

$\displaystyle ad-bc=1$

But remember in this ring since it has only two elements, that $\displaystyle -x=x$ for all $\displaystyle x$.

Thus,

$\displaystyle ad+bc=1$

Now, just observe the possibilities for $\displaystyle a,b,c,d$ which make this equation true. Since there are only 16 of them it is easy.

1)If $\displaystyle a=0$ then, $\displaystyle b,c\not=0$ thus, $\displaystyle b,c=1$ and $\displaystyle d$ can be anything (either 1 or 0) this gives us two matrices,

$\displaystyle \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ and $\displaystyle \left(\begin{array}{cc}0&1\\1&1\end{array}\right)$.

2)If $\displaystyle a=1$ then , $\displaystyle d=1$ would mean that either $\displaystyle b$ or $\displaystyle c$ must be zero. Also, if $\displaystyle d=0$ would mean that both $\displaystyle a$ and $\displaystyle b$ cannot be zero (thus they are both one). This gives us four matrices,

$\displaystyle \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ and, $\displaystyle \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$ and,

$\displaystyle \left(\begin{array}{cc}1&0\\1&1\end{array}\right)$ and,$\displaystyle \left(\begin{array}{cc}1&1\\1&0\end{array}\right)$

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

Originally Posted by**ThePerfectHacker**

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

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 $\displaystyle q^n-1$ possibilities. The second row must not the in the subspace generated by the first row, which has dimension one, so there are $\displaystyle q^n-q$ possiblities. The $\displaystyle (r+1)$-th row must not be in the subspace generated by the previous $\displaystyle r$ rows, linearly independent by induction, so of dimension $\displaystyle r$ giving $\displaystyle q^n-q^r$ possibilities. The overal number of possiblities is thus $\displaystyle (q^n-1)(q^n-q)\cdots(q^n-q^r)\cdots(q^n-q^{n-1})$

In the case $\displaystyle q=2, n=2$ this is $\displaystyle (2^2-1)(2^2-2^1) = 6$. - Mar 12th 2006, 09:41 AMThePerfectHackerQuote:

Originally Posted by**rgep**