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I have been reading Alan Macdonald's "Linear and Geometric Algebra" and got stuck on the example on pg 149 on the determinant of a two by two matrix. I have visited this question and am convinced this is not a duplicate because I am asking about the specific derivation in the book. The example is as follows:

Consider the general linear transformation on $\mathbb{R}^2:f(\mathbf{e}_1) = a\mathbf{e}_1 + c\mathbf{e}_2, f(\mathbf{e}_2) = b\mathbf{e}_1 + d\mathbf{e}_2$. Hence,

$$[f] = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $$ Then $\det(f) = ad-bc$: $f(\mathbf{i}) = f(\mathbf{e}_1 \wedge \mathbf{e}_2) = f(\mathbf{e}_1)\wedge f(\mathbf{e_2}) = (a\mathbf{e}_1 + c\mathbf{e}_2) \wedge (b\mathbf{e}_1 + d\mathbf{e}_2) = (ad-bc)(\mathbf{e}_1 \wedge \mathbf{e}_2) = (ad-bc)\mathbf{i}$.

My question is, why is $(a\mathbf{e}_1 + c\mathbf{e}_2) \wedge (b\mathbf{e}_1 + d\mathbf{e}_2) = (ad-bc)(\mathbf{e}_1 \wedge \mathbf{e}_2)$ true? No justification is given in this crucial step. I have been trying to simplify using the outer product space rules, but didn't get anywhere. Any help would be much appreciated.

Linus Rastegar
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  • It’s an application of the distributive law of $\wedge$ over $+$, i.e., of bilinearity of $\wedge$. – amd Dec 26 '17 at 21:00

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$(x,y)\rightarrow x\wedge y$ is bilinear and alternated. This implies that

$(ae_1+ce_2)\wedge (be_1+de_2)=ad e_1\wedge e_2+bc e_2\wedge e_1=(ad-bc)e_1\wedge e_2$ since $e_1\wedge e_1=e_2\wedge e_2=0$ and $e_2\wedge e_1=-e_1\wedge e_2$.

Tsemo Aristide
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  • Thanks for this explanation! How do we know that $x \wedge y$ is alternating? Also, is it acceptable convention to write: $ad e_1 \wedge e_2$ instead of $ad(e_1 \wedge e_2)$. I'm just wondering to make my own notation needs easier. – Linus Rastegar Dec 26 '17 at 16:35
  • By definition, $\wedge$ is an alternating product. You should look at your how your book explains what it is! Or look at some other resources that define it. – Osama Ghani Dec 26 '17 at 16:59