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Thanks for your time and effort. I appreciate your help.

I'm new to geometric algebra and I get that it supersedes linear algebra.

I was wondering though how I could learn to take an operator in linear algebra and express it in terms of geometric algebra?

For example,

Suppose I had the matrix operator:

\begin{vmatrix} \mathbf{1} & \mathbf{ 1} & \mathbf{-1} \\ \mathbf{0} & \mathbf{ 1} & \mathbf{-1} \\ \mathbf{0} & \mathbf{-1} & \mathbf{ 1} \ \end{vmatrix}

How would I express this same function in the language of geometric algebra?

Thanks for your help.

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The standard (usual) model of geometric algebra provides powerful expressions for some of the most important kinds of linear transformations: projections, rotations, reflections, orthogonal, skew symmetric (but not symmetric). You can find all this in my book Vector and Geometric Algebra: faculty.luther.edu/~macdonal/laga (Pardon the self-promotion.) –  Alan Macdonald Aug 24 '13 at 1:07
    
It's silly to say it supersedes linear algebra. It is just an alternative way that offers different insights about a proper subset of linear algebra, at best. –  rschwieb Mar 23 at 21:00
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@Alan: I've changed your answer to a comment. This isn't because of the self-promotion, but because we want answers to be self-contained. If you were to link to your book and also answer the question, it would be just fine. –  mixedmath Mar 23 at 21:07
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2 Answers 2

You could express it as a function. Let your operator be $\underline T$. It could be described by

$$\begin{align*}\underline T(e_1) &= e_1 \\ \underline T(e_2) &= e_1 + e_2 - e_3 \\ \underline T(e_3) &= -e_1 -e_2 + e_3\end{align*}$$

You could instead use dot products to combine this into a single expression. Let $a$ be an arbitrary vector, and you have

$$\underline T(a) = (a \cdot e_1) e_1 + (a \cdot e_2) (e_1 + e_2 - e_3) + (a \cdot e_3) (-e_1 - e_2 + e_3)$$

In particular, notice that the last column is just the negative of the second column, so the expression simplifies to

$$\underline T(a) = (a \cdot e_1) e_1 + (a \cdot e_2 - a \cdot e_3)(e_1 + e_2 - e_3)$$

There is (so far) nothing inherently GA-like to expressing a linear operator this way, but it is a bit more amenable to some of the operations you might be asked to perform that come from GA.


Edit: about tensor products. The GA is sometimes referred to as a quotient of the tensor algebra. That is, there is some loss of information in using geometric products instead of tensor products. Let me illustrate: let $a$ and $b$ be two vectors. The tensor product is

$$a \otimes b =\sum_{i} (a^i b^i)(e_i \otimes e_i) + \sum_{i \neq j} (a^i b^j) (e_i \otimes e_j)$$

In GA, the corresponding use of the geometric product gives

$$a b = a \cdot b + a \wedge b = \sum_i a^i b^i + \sum_{i \neq j} a^i b^j (e_i \wedge e_j)$$

The tensor product keeps $e_1 \otimes e_1$ separate from $e_2 \otimes e_2$ and so on. Thus, the GA can't keep information about those terms distinct. It doesn't need to because these are all scalar terms when interpreted in the context of geometry.

When using tensor products to build up linear maps, however, GA can replicate this rather well. You should be able to see that $a \otimes b$ interpreted as a map becomes $\underline M(c) = a (b \cdot c)$.

A tensor product of 2 matrices can probably be interpreted in some similar way.

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Thanks for your help. Is there anything different if the elements are the result of a tensor product? For example, if $$\begin{align*}\underline T(e_1) &= e_1 \\ \underline T(e_2) &= e_1 + e_2 - e_3 \\ \underline T(e_3) &= -e_1 -e_2 + e_3\end{align*}$$ You could instead use dot products to combine this into a single expression. Let $a$ be an arbitrary vector, and you have $$\underline T(a) = (a \cdot e_1) e_1 + (a \cdot e_2) (e_1 + e_2 - e_3) + (a \cdot e_3) (-e_1 - e_2 + e_3)$$ –  New-to-GA Aug 24 '13 at 12:44
    
Sorry about that, I started typing and my computer crashed, please ignore the aborted comment. Is there anything different if the elements are a result of a tensor product? $$\\e_{ij} \otimes e_{kl}$$ = \begin{vmatrix} \mathbf{1} & \mathbf{ 1} & \mathbf{-1} \\ \mathbf{0} & \mathbf{ 1} & \mathbf{-1} \\ \mathbf{0} & \mathbf{-1} & \mathbf{ 1} \ \end{vmatrix} –  New-to-GA Aug 24 '13 at 13:13
    
Sorry browser crashed. If we had a tensor with elements from a tensor product of two 2x2 tensors:\begin{vmatrix} \mathbf{e_{11}f_{11}} & \mathbf{e_{11}f_{21}} & \mathbf{e_{21}f_{11}} & \mathbf{e_{21}f_{21}}\\ \mathbf{e_{11}f_{12}} & \mathbf{e_{11}f_{22}} & \mathbf{e_{21}f_{12}} & \mathbf{e_{21}f_{22}}\\ \mathbf{e_{12}f_{11}} & \mathbf{e_{12}f_{21}} & \mathbf{e_{22}f_{11}} & \mathbf{e_{22}f_{21}}\\ \mathbf{e_{12}f_{12}} & \mathbf{e_{12}f_{22}} & \mathbf{e_{22}f_{12}} & \mathbf{e_{22}f_{22}}\\ \end{vmatrix} –  New-to-GA Aug 24 '13 at 13:32
    
Sorry browser crashed. If we had a tensor with elements from a tensor product of two 2x2 tensors:\begin{vmatrix} \mathbf{e_{11}f_{11}} & \mathbf{e_{11}f_{21}} & \mathbf{e_{21}f_{11}} & \mathbf{e_{21}f_{21}}\\ \mathbf{e_{11}f_{12}} & \mathbf{e_{11}f_{22}} & \mathbf{e_{21}f_{12}} & \mathbf{e_{21}f_{22}}\\ \mathbf{e_{12}f_{11}} & \mathbf{e_{12}f_{21}} & \mathbf{e_{22}f_{11}} & \mathbf{e_{22}f_{21}}\\ \mathbf{e_{12}f_{12}} & \mathbf{e_{12}f_{22}} & \mathbf{e_{22}f_{12}} & \mathbf{e_{22}f_{22}}\\ \end{vmatrix} would anything more GA like be in the result or is it more like your above example? –  New-to-GA Aug 24 '13 at 13:40
    
I've added a section on tensor products. –  Muphrid Aug 24 '13 at 15:57
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The key idea of Geometric Algebra is that it converts abstract algebraic concepts into concrete spatial shapes, and the shapes are meaningful to the concept represented.

In the case of your matrix, that can be viewed as three spatial vectors of (x,y,z) each, sticking out of the origin in three directions. You can now multiply this structure with a scalar, a vector, bivector, or another trivector and it will perform a logically reasonable spatial transformation on its operand. The shape of the structure is meaningful. For example the wedge product between the three vectors represents the volume spanned by them, and that turns out to be equal to the determinant of the matrix, and its volume goes to zero as the three vectors become parallel, as explained on the Wiki page on Exterior Algebra

In general, I find the association of familiar algebraic concepts with novel spatial concepts to be the most interesting. If you multiply a vector by -1 it points in the opposite direction: Negation is Reflection! Rotation by 90 degrees == x sqrt(-1), because if you do it twice, you get to 180 degrees. Clifford multiplication between vectors produces a scaling and a rotation. The rotation also works like reflection. Vectors a*b = bivector ab (plus a scalar), which represents a twist from the direction of a to the direction of b. If you multiply a third vector c*a*b it will rotate c through the same angle as the one from a to b.

It is imagining the spatial interpretation of the algebraic expression that is the most interesting aspect of Clifford Algebra. Remember when you learned parallel lines crossed by an angled line, =/= your teacher showed you which angles are equal, and which are complementary, etc. but he never had to prove it, he merely had to point it out to you, because its truth is self-evident by inspection. Clifford Algebra does that for algebra. Instead of manipulating abstract symbols back and forth across the =equals= sign, in Clifford Algebra you imagine reflections and rotations of vectors about the origin.

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I checked three times to be sure, and I can conclusively say that none of these random anecdotes about GA seem to come close to addressing the poster's question about linear transformations. –  rschwieb Mar 23 at 22:29
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