Solving Clifford Algebra Equation - Need Help!

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dimension10
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I was trying to solve the following equation:

[tex]\bigwedge\limits_{j=1}^{k}\begin{bmatrix}<br /> a_{1,j}\\ <br /> a_{2,j}\\ <br /> :\\<br /> .\\<br /> a_{k+1,j}<br /> \end{bmatrix}[/tex]

Does anyone know how I can solve it? Thanks in advance.
 
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algebrat said:
try k=1,2,3 etc.

Actually, I already know what that expression results in. I was trying to prove it.
 
dimension10 said:
I was trying to solve the following equation:

[tex]\bigwedge\limits_{j=1}^{k}\begin{bmatrix}<br /> a_{1,j}\\ <br /> a_{2,j}\\ <br /> :\\<br /> .\\<br /> a_{k+1,j}<br /> \end{bmatrix}[/tex]

Does anyone know how I can solve it? Thanks in advance.

There is no equation, just an expression. It is not clear what you are trying to solve.

dimension10 said:
Actually, I already know what that expression results in. I was trying to prove it.

Please show us the result. Please tell us exactly what you are trying to prove.
 
algebrat said:
There is no equation, just an expression. It is not clear what you are trying to solve.

Ya I realized that later. Just made a typo error.
algebrat said:
Please show us the result. Please tell us exactly what you are trying to prove.

[tex]\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }[/tex]

where A_(k+1)Xk is the matrix formed by augmenting the vectors together and the cross_j function means crossing out the jth row of the matrix A_(k+1)Xk.

I'm trying to prove that [itex]\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }[/itex] is the answer, since I found it by finding the special cases where k=1,2,3.
 
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I guess I know very little about clifford algebras, I was expecting a wedge not to be a scalar, which I am guessing is what the root of sum of determinants squared would give.
 
algebrat said:
I guess I know very little about clifford algebras, I was expecting a wedge not to be a scalar, which I am guessing is what the root of sum of determinants squared would give.

Oops! Sorry, you are right. I mean the determinant of the exterior product is equal to [tex]\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }[/tex]
 
For k=2 I would have guessed the cross product.

Ah
 
algebrat said:
For k=2 I would have guessed the cross product.

Ah

No... I think it is the hodge dual of the cross product, actually. The wedge alone would be a bivector in that case.
 
Show us the steps for k=2 and/or 3
 
algebrat said:
Show us the steps for k=2 and/or 3

[tex]\begin{array}{l}<br /> \begin{array}{*{20}{l}}<br /> {\left\| {\left[ {\begin{array}{*{20}{l}}<br /> \alpha \\<br /> \gamma \\<br /> \varepsilon <br /> \end{array}} \right] \wedge \left[ {\begin{array}{*{20}{l}}<br /> \beta \\<br /> \delta \\<br /> \zeta <br /> \end{array}} \right]} \right\| = \left\| {\alpha \delta \left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \alpha \zeta \left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \gamma \beta \left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_1}} \right) + \gamma \zeta \left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + \varepsilon \beta \left( {{{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_1}} \right) + \varepsilon \delta \left( {{{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_2}} \right)} \right\|}\\<br /> { = \left\| {\left( {\alpha \delta - \beta \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \left( {\alpha \zeta - \beta \varepsilon } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \left( {\gamma \zeta - \delta \varepsilon } \right)\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|}\\<br /> { = \left\| {\det \left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta \\<br /> \gamma &\delta <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \det \left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta \\<br /> \varepsilon &\zeta <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \det \left[ {\begin{array}{*{20}{c}}<br /> \gamma &\delta \\<br /> \varepsilon &\zeta <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|}<br /> \end{array}\\<br /> = \sqrt {{{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta \\<br /> \gamma &\delta <br /> \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta \\<br /> \varepsilon &\zeta <br /> \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \gamma &\delta \\<br /> \varepsilon &\zeta <br /> \end{array}} \right]} <br /> \end{array}[/tex]

[tex]\begin{array}{l}<br /> \left\| {\left[ \begin{array}{l}<br /> \alpha \\<br /> \delta \\<br /> \eta \\<br /> \kappa <br /> \end{array} \right] \wedge \left[ \begin{array}{l}<br /> \beta \\<br /> \varepsilon \\<br /> \theta \\<br /> \lambda <br /> \end{array} \right] \wedge \left[ \begin{array}{l}<br /> \gamma \\<br /> \zeta \\<br /> \iota \\<br /> \mu <br /> \end{array} \right]} \right\| = \left\| {\left( {\alpha {{{\bf{\hat e}}}_1} + \delta {{{\bf{\hat e}}}_2} + \eta {{{\bf{\hat e}}}_3} + \kappa {{{\bf{\hat e}}}_4}} \right) \wedge \left( {\beta {{{\bf{\hat e}}}_1} + \varepsilon {{{\bf{\hat e}}}_2} + \theta {{{\bf{\hat e}}}_3} + \lambda {{{\bf{\hat e}}}_4}} \right) \wedge \left( {\gamma {{{\bf{\hat e}}}_1} + \zeta {{{\bf{\hat e}}}_2} + \iota {{{\bf{\hat e}}}_3} + \mu {{{\bf{\hat e}}}_4}} \right)} \right\|\\<br /> {\rm{ }} = \left| {\left| {\left( {\alpha \varepsilon \iota - \delta \beta \iota - \alpha \theta \zeta + \eta \beta \zeta + \delta \theta \gamma - \eta \varepsilon \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + } \right.} \right.\\<br /> {\rm{ }}\left( {\alpha \varepsilon \mu - \delta \beta \mu - \alpha \lambda \zeta + \kappa \beta \zeta - \delta \lambda \gamma + \kappa \varepsilon \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_4}} \right) + \\<br /> {\rm{ }}\left( {\alpha \theta \mu - \eta \beta \mu - \alpha \lambda \iota + \kappa \beta \iota - \eta \lambda \gamma + \kappa \theta \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right) + \\<br /> \left. {\left. {{\rm{ }}\left( {\delta \theta \mu - \eta \varepsilon \mu - \delta \lambda \iota + \kappa \varepsilon \iota - \eta \lambda \zeta + \kappa \theta \zeta } \right)\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right)} \right|} \right|\\<br /> {\rm{ }} = \left| {\left| {\det \left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta &\gamma \\<br /> \delta &\varepsilon &\zeta \\<br /> \eta &\theta &\iota <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + } \right.} \right.\\<br /> {\rm{ }}\det \left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta &\gamma \\<br /> \delta &\varepsilon &\zeta \\<br /> \kappa &\lambda &\mu <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_4}} \right) + \\<br /> {\rm{ }}\det \left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta &\gamma \\<br /> \eta &\theta &\iota \\<br /> \kappa &\lambda &\mu <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right) + \\<br /> \left. {\left. {{\rm{ }}\det \left[ {\begin{array}{*{20}{c}}<br /> \delta &\varepsilon &\zeta \\<br /> \eta &\theta &\iota \\<br /> \kappa &\lambda &\mu <br /> \end{array}} \right]\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_4}} \right)} \right|} \right|\\<br /> {\rm{ }} = \sqrt {{{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta &\gamma \\<br /> \delta &\varepsilon &\zeta \\<br /> \eta &\theta &\iota <br /> \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta &\gamma \\<br /> \delta &\varepsilon &\zeta \\<br /> \kappa &\lambda &\mu <br /> \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \alpha &\beta &\gamma \\<br /> \eta &\theta &\iota \\<br /> \kappa &\lambda &\mu <br /> \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}<br /> \delta &\varepsilon &\zeta \\<br /> \eta &\theta &\iota \\<br /> \kappa &\lambda &\mu <br /> \end{array}} \right]} <br /> \end{array}[/tex]

But when I tried it for the general case, it was not possible.
 
maybe you could somehow argue that the coefficient of e_1^...^e_{j-1}^e_{j+1}^...^e_{k+1} would be the det of cross_j(A).
 
algebrat said:
maybe you could somehow argue that the coefficient of e_1^...^e_{j-1}^e_{j+1}^...^e_{k+1} would be the det of cross_j(A).

Thanks a lot! I think that you're right! Thanks again!
 
Basic Geometric Algebra

See attached file DET.pdf. The thing to note is that in the case of an orthogonal basis e1,...,en we have e1^...^en = e1...en (wedge product is the same as the geometric/Clifford product).

Detailed note based on "Geometric Algebra for Physicists" by Doran and Laseby is at

https://github.com/brombo/GA
 

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