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Show that this map is a group homomorphism and find its kernel:

$\theta: GL_2 (\Bbb Q) \rightarrow \Bbb Q\setminus \{0\}$ given by $\theta(A) = \det A.$

My attempt:

Let $A = \begin{pmatrix} a_1 & a_2 \\ a_3 & a_4 \\ \end{pmatrix}$ then $ \theta (A) =\det A = a_1a_4 - a_2a_3$ And let $B \in GL_2 (\Bbb Q)$ such that B = \begin{pmatrix} b_1 & b_2 \\ b_3 & b_4 \\ \end{pmatrix} and $\theta(B) = \det B = b_1b_4 - b_2b_3$

Then checking for homomorphism...

$ \begin{align} \theta(A)\theta(B)= \det A \det B & = \ (a_1a_2-a_3a_4)(b_1b_4 - b_2b_3) \\ & = a_1a_2b_1b_2 - a_3a_4b_1b_4 - a_1a_2b_2b_3 + a_3a_4b_3b_4\\ & = \det(AB) = \theta(AB) \end{align}$

(to be honest I couldn't actually figure out how $\det A\det B$ became $\det AB$ with the method I used. i.e. the expansions were just not working out. Is there a better way of doing this? And am I horrificaly wrong?)

$\ker \theta = A: \det A =1$

3 Answers 3

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Start working from the other end. It’s usually better either to work from the more complicated end or to work on both ends of the calculation simultaneously.

You know that $AB=\pmatrix{a_1&a_2\\a_3&a_4}\pmatrix{b_1&b_2\\b_3&b_4}=\pmatrix{a_1b_1+a_2b_3&a_1b_2+a_2b_4\\a_3b_1+a_4b_3&a_3b_2+a_4b_4}\;,$

so

$\begin{align*} \det AB&=(a_1b_1+a_2b_3)(a_3b_2+a_4b_4)-(a_1b_2+a_2b_4)(a_3b_1+a_4b_3)\\ &=\color{red}{a_1b_1a_3b_2}+a_1b_1a_4b_4+a_2b_3a_3b_2+\color{blue}{a_2b_3a_4b_4}\\ &\qquad-\color{red}{a_1b_2a_3b_1}-a_1b_2a_4b_3-a_2b_4a_3b_1-\color{blue}{a_2b_4a_4b_3}\\ &=a_1b_1a_4b_4+a_2b_3a_3b_2-a_1b_2a_4b_3-a_2b_4a_3b_1\\ &=a_1a_4b_1b_4-a_1a_4b_2b_3+a_2a_3b_2b_3-a_2a_3b_1b_4\\ &=a_1a_4(b_1b_4-b_2b_3)-a_2a_3(b_1b_4-b_2b_3)\\ &=(a_1a_4-a_2a_3)(b_1b_4-b_2b_3)\\ &=\det A\det B\;. \end{align*}$

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    @Siyanda: You’re welcome.2012-11-17
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You haven't multiplied out $AB$ ---- you have to do that, then you can compute $\det(AB)$ and see whether it equals $\det A\det B$.

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    Actually was (laughing) just about to delete this comment out of embarrassment...sorry2012-11-17
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$AB = \begin{pmatrix} a_1b_1+a_2b_3 & a_1b_2+a_2b_4 \\ a_3b_1+a_4b_3 & a_3b_2+a_4b_4 \\ \end{pmatrix}$

So $\det AB = (a_1b_1+a_2b_3)(a_3b_2+a_4b_4) - (a_1b_2 + a_2b_4)(a_3b_1+a_4b_3)$

$= (a_1a_3b_1b_2 + a_1a_4b_1b_4 + a_2a_3b_2b_3 + a_2a_4b_3b_4) - (a_1a_3b_1b_2+a_1a_4b_2b_3+ a_2a_3b_1b_4+ a_2a_4b_3b_4)$

$= a_1a_4b_1b_4+a_2a_3b_2b_3-a_1a_4b_2b_3-a_2a_3b_1b_4 $

$= a_1a_4(b_1b_4-b_2b_3) -a_2a_3(b_1b_4-b_2b_3) $ $= (a_1a_4-a_2a_3)(b_1b_4-b_2b_3) $

$= \det A \det B.$