## Question on Clifford Algebra

I was trying to solve the following equation:

$$\bigwedge\limits_{j=1}^{k}\begin{bmatrix} a_{1,j}\\ a_{2,j}\\ :\\ .\\ a_{k+1,j} \end{bmatrix}$$

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

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 try k=1,2,3 etc.

 Quote by algebrat try k=1,2,3 etc.
Actually, I already know what that expression results in. I was trying to prove it.

## Question on Clifford Algebra

 Quote by dimension10 I was trying to solve the following equation: $$\bigwedge\limits_{j=1}^{k}\begin{bmatrix} a_{1,j}\\ a_{2,j}\\ :\\ .\\ a_{k+1,j} \end{bmatrix}$$ 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 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.

 Quote by algebrat 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 Please show us the result. Please tell us exactly what you are trying to prove.
$$\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }$$

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 $\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }$ is the answer, since I found it by finding the special cases where k=1,2,3.

 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.

 Quote by algebrat 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 $$\sqrt {\sum\limits_{j = 1}^{k + 1} {{{\det }^2}\left( {{{{\mathop{\rm cross}\nolimits} }_j}\left( {\mathop {\bf{A}}\limits_{(k + 1) \times k} } \right)} \right)} }$$

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

 Quote by algebrat 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

 Quote by algebrat Show us the steps for k=2 and/or 3
$$\begin{array}{l} \begin{array}{*{20}{l}} {\left\| {\left[ {\begin{array}{*{20}{l}} \alpha \\ \gamma \\ \varepsilon \end{array}} \right] \wedge \left[ {\begin{array}{*{20}{l}} \beta \\ \delta \\ \zeta \end{array}} \right]} \right\| = \left\| {\alpha \delta \left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \alpha \zeta \left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \gamma \beta \left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_1}} \right) + \gamma \zeta \left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right) + \varepsilon \beta \left( {{{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_1}} \right) + \varepsilon \delta \left( {{{{\bf{\hat e}}}_3} \wedge {{{\bf{\hat e}}}_2}} \right)} \right\|}\\ { = \left\| {\left( {\alpha \delta - \beta \gamma } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \left( {\alpha \zeta - \beta \varepsilon } \right)\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \left( {\gamma \zeta - \delta \varepsilon } \right)\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|}\\ { = \left\| {\det \left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \gamma &\delta \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_2}} \right) + \det \left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \varepsilon &\zeta \end{array}} \right]\left( {{{{\bf{\hat e}}}_1} \wedge {{{\bf{\hat e}}}_3}} \right) + \det \left[ {\begin{array}{*{20}{c}} \gamma &\delta \\ \varepsilon &\zeta \end{array}} \right]\left( {{{{\bf{\hat e}}}_2} \wedge {{{\bf{\hat e}}}_3}} \right)} \right\|} \end{array}\\ = \sqrt {{{\det }^2}\left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \gamma &\delta \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \varepsilon &\zeta \end{array}} \right] + {{\det }^2}\left[ {\begin{array}{*{20}{c}} \gamma &\delta \\ \varepsilon &\zeta \end{array}} \right]} \end{array}$$

$$\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}$$

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).

 Quote by algebrat 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!