Understanding Exterior Algebra: Determinant from Wedge Product

[Matthollyw00d]

Can someone please thoroughly explain how the determinant comes from the wedge product? I'm only in Cal 3 and Linear at the moment. I'm somewhat trying to learn more about the Wedge Product in Exterior Algebra to understand the determinant on a more fundamental basis. A thorough website or book would be of great help also.

 

[Peeter]

The wedge product required for this is that it changes sign on alternation, and is linear in either term

ie:
$$b \wedge a = - a \wedge b$$

and
$$(a + b) \wedge c = a \wedge c + b \wedge c$$

and

$$a \wedge (b+c) = a \wedge b + a \wedge c$$(the first assumes one is talking about what is referred to as a simple element, one that can be built up of wedge products of other simple elements).

Note that a consequence of the first is that a ^ a = 0 for any simple element a (for example a vector in the two element case).

As an illustration of how a determinant enters the picture consider the two variable case. The wedge products of two vectors in a plane

$$\begin{align*}<br /> (a_1 e_1 + a_2 e_2 ) \wedge (b_1 e_1 + b_2 e_2 )<br /> &= <br /> a_1 b_1 e_1 \wedge e_1<br /> +a_1 b_2 e_1 \wedge e_2<br /> +a_2 b_1 e_2 \wedge e_1<br /> +a_2 b_2 e_2 \wedge e_2 \\<br /> &= <br /> a_1 b_2 e_1 \wedge e_2<br /> +a_2 b_1 e_2 \wedge e_1 \\<br /> &= <br /> a_1 b_2 e_1 \wedge e_2<br /> -a_2 b_1 e_1 \wedge e_2 \\<br /> &= <br /> (a_1 b_2 -a_2 b_1) e_1 \wedge e_2<br /> \end{align*}$$

Observe the determinant above as a factor of the wedge product.

edit. As for books I'd personally recommend Geometric Algebra for Computer Science, and New Foundations of Classical Mechanics (but be warned that neither of these exclusively treat the wedge product and exterior algebra nor are about differential forms if that is what you are looking for).

 

[Matthollyw00d]

The wedge product required for this is that it changes sign on alternation, and is linear in either term

ie:
$$b \wedge a = - a \wedge b$$

and
$$(a + b) \wedge c = a \wedge c + b \wedge c$$

and

$$a \wedge (b+c) = a \wedge b + a \wedge c$$


(the first assumes one is talking about what is referred to as a simple element, one that can be built up of wedge products of other simple elements).

Note that a consequence of the first is that a ^ a = 0 for any simple element a (for example a vector in the two element case).

As an illustration of how a determinant enters the picture consider the two variable case. The wedge products of two vectors in a plane

$$\begin{align*}<br /> (a_1 e_1 + a_2 e_2 ) \wedge (b_1 e_1 + b_2 e_2 )<br /> &= <br /> a_1 b_1 e_1 \wedge e_1<br /> +a_1 b_2 e_1 \wedge e_2<br /> +a_2 b_1 e_2 \wedge e_1<br /> +a_2 b_2 e_2 \wedge e_2 \\<br /> &= <br /> a_1 b_2 e_1 \wedge e_2<br /> +a_2 b_1 e_2 \wedge e_1 \\<br /> &= <br /> a_1 b_2 e_1 \wedge e_2<br /> -a_2 b_1 e_1 \wedge e_2 \\<br /> &= <br /> (a_1 b_2 -a_2 b_1) e_1 \wedge e_2<br /> \end{align*}$$

Observe the determinant above as a factor of the wedge product.

edit. As for books I'd personally recommend Geometric Algebra for Computer Science, and New Foundations of Classical Mechanics (but be warned that neither of these exclusively treat the wedge product and exterior algebra nor are about differential forms if that is what you are looking for).

I'm also somewhat confused on notation. The 2 vectors you're using in your example are A= [a1 a2] and B= [b1 b2], correct? Then are the e1 and e2, unit vectors, like i, j, and k? And then what does that final e1 wedge e2 do to vanish when dealing with determinants?

 

[Peeter]

e1 and e2 are two vectors that aren't colinear, but they can be i and j if you like. If you use a = a_1 i + a_2 j + a_3 k, and b = b_1 i + ...
then you'll get something like:

$$a \wedge b = <br /> \begin{vmatrix}<br /> a_1 & a_2 \\<br /> b_1 & b_2 <br /> \end{vmatrix}<br /> i \wedge j<br /> +<br /> \begin{vmatrix}<br /> a_2 & a_3 \\<br /> b_2 & b_3 <br /> \end{vmatrix}<br /> j \wedge k<br /> +<br /> \begin{vmatrix}<br /> a_1 & a_3 \\<br /> b_1 & b_3 <br /> \end{vmatrix}<br /> i \wedge k$$

The final wedges do not vanish unless you'll chosen i, j, k to be linearly dependent (which wouldn't be the case if this is your standard orthonormal basis for R^3). The wedge products i^j, j^k, i^k can be thought of as forming a basis in a "wedge product" space in their own right.