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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?)

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

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up vote 1 down vote accepted

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


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

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I did try this...guess I didn't stick to it long enough! Thank you. :) – Siyanda Nov 17 '12 at 8:52
@Siyanda: You’re welcome. – Brian M. Scott Nov 17 '12 at 8:53

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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I actually did...and I didn't get very far before realizing that equating the two would actually be a bit of a stretch – Siyanda Nov 17 '12 at 8:39
Evidently, you were wrong. – Gerry Myerson Nov 17 '12 at 8:55
Actually was (laughing) just about to delete this comment out of embarrassment...sorry – Siyanda Nov 17 '12 at 9:03

$$AB = \begin{pmatrix} a_1b_1+a_2b_3 & a_1b_2+a_2b_4 \\ a_3b_1+a_4b_3 & a_3b_2+a_4b_4 \\ \end{pmatrix}$$

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

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

$$= a_1a_4b_1b_4+a_2a_3b_2b_3-a_1a_4b_2b_3-a_2a_3b_1b_4 $$

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

$$= \det A \det B.$$

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