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

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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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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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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