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Question on Clifford Algebra

 
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May27-12, 11:44 PM   #1
 

Question on Clifford Algebra


I was trying to solve the following equation:

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

Does anyone know how I can solve it? Thanks in advance.
 
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May28-12, 12:09 AM   #2
 
try k=1,2,3 etc.
 
May28-12, 03:04 AM   #3
 
Quote by algebrat View Post
try k=1,2,3 etc.
Actually, I already know what that expression results in. I was trying to prove it.
 
May28-12, 05:15 AM   #4
 

Question on Clifford Algebra


Quote by dimension10 View Post
I was trying to solve the following equation:

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

Quote by dimension10 View Post
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.
 
May28-12, 05:50 AM   #5
 
Quote by algebrat View Post
There is no equation, just an expression. It is not clear what you are trying to solve.
Ya I realised that later. Just made a typo error.



Quote by algebrat View Post
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.
 
May28-12, 06:11 AM   #6
 
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.
 
May28-12, 06:13 AM   #7
 
Quote by algebrat View Post
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]
 
May28-12, 06:13 AM   #8
 
For k=2 I would have guessed the cross product.

Ah
 
May28-12, 06:14 AM   #9
 
Quote by algebrat View Post
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.
 
May28-12, 06:16 AM   #10
 
Show us the steps for k=2 and/or 3
 
May28-12, 06:20 AM   #11
 
Quote by algebrat View Post
Show us the steps for k=2 and/or 3
[tex]\begin{array}{l}
\begin{array}{*{20}{l}}
{\left\| {\left[ {\begin{array}{*{20}{l}}
\alpha \\
\gamma \\
\varepsilon
\end{array}} \right] \wedge \left[ {\begin{array}{*{20}{l}}
\beta \\
\delta \\
\zeta
\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\|}\\
{ = \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\|}\\
{ = \left\| {\det \left[ {\begin{array}{*{20}{c}}
\alpha &\beta \\
\gamma &\delta
\end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \det \left[ {\begin{array}{*{20}{c}}
\alpha &\beta \\
\varepsilon &\zeta
\end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \det \left[ {\begin{array}{*{20}{c}}
\gamma &\delta \\
\varepsilon &\zeta
\end{array}} \right]\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|}
\end{array}\\
= \sqrt {{{\det }^2}\left[ {\begin{array}{*{20}{c}}
\alpha &\beta \\
\gamma &\delta
\end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
\alpha &\beta \\
\varepsilon &\zeta
\end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}}
\gamma &\delta \\
\varepsilon &\zeta
\end{array}} \right]}
\end{array}[/tex]

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

But when I tried it for the general case, it was not possible.
 
May28-12, 06:29 AM   #12
 
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).
 
May28-12, 06:33 AM   #13
 
Quote by algebrat View Post
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!
 
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