Solving Clifford Algebra Equation - Need Help!

In summary, Clifford algebra is a mathematical tool used for solving equations involving higher-dimensional objects. The basic principles of solving a Clifford algebra equation include understanding multivector properties, using the distributive property, and applying inverse operations. Geometric interpretation is helpful in simplifying equations and checking solutions. Common challenges include keeping track of multivector properties, non-commutative multiplication, and understanding geometric interpretation. Tips and tricks for solving Clifford algebra equations include breaking down problems, using geometric interpretation, and regular practice.
  • #1
dimension10
371
0
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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  • #2
try k=1,2,3 etc.
 
  • #3
algebrat said:
try k=1,2,3 etc.

Actually, I already know what that expression results in. I was trying to prove it.
 
  • #4
dimension10 said:
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.

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.
 
  • #5
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.
 
Last edited:
  • #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.
 
  • #7
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]
 
  • #8
For k=2 I would have guessed the cross product.

Ah
 
  • #9
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.
 
  • #10
Show us the steps for k=2 and/or 3
 
  • #11
algebrat said:
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.
 
  • #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).
 
  • #13
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!
 
  • #14
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 [Broken]
 

Attachments

  • DET.pdf
    167.2 KB · Views: 214
Last edited by a moderator:

1. What is Clifford algebra and how is it used in solving equations?

Clifford algebra is a mathematical tool that extends classical vector algebra to include higher-dimensional objects such as vectors, bivectors, and multivectors. It is used in solving equations by providing a way to manipulate and simplify complex equations involving multiple variables.

2. What are the basic principles of solving a Clifford algebra equation?

The basic principles of solving a Clifford algebra equation include understanding the properties of multivectors, using the distributive property and geometric interpretation to simplify equations, and applying inverse operations to isolate the variable being solved for.

3. How is geometric interpretation helpful in solving Clifford algebra equations?

Geometric interpretation is helpful in solving Clifford algebra equations because it allows for a visual understanding of the problem, making it easier to manipulate and simplify equations. It also provides a way to check the solutions obtained algebraically.

4. What are some common challenges when solving Clifford algebra equations?

Some common challenges when solving Clifford algebra equations include keeping track of the properties and operations of multivectors, dealing with non-commutative multiplication, and understanding the geometric interpretation of the problem.

5. Are there any tips or tricks for solving Clifford algebra equations?

Some tips and tricks for solving Clifford algebra equations include breaking down the problem into smaller, more manageable parts, using geometric interpretation to simplify equations, and practicing regularly to develop a strong understanding of the principles involved.

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