Linear Algebra: Solving Systems of Equations with Wedge Product?

[jdstokes]

Hi all,

I learned this stuff years ago and wasn't brilliant at it even then so I think a refresher is in order.

Suppose I have n distinct homogeneous equations in n unknowns. I want to find the solution so I write down the matrix of coefficients multiplying my vector of variables as follows

$A \mathbf{x} =\mathbf{0}$.

Now, we don't want $\deta A \neq 0$ to happen otherwise the columns of A are linearly independent so the only solution to $A \mathbf{x} = \mathbf{C}_1 x_1 + \cdots \mathbf{C}_n x_n = \mathbf{0}$ is $\mathbf{0}$.

Now how do we actually solve this for $\mathbf{x}$, do we just do Gaussian elimination followed by back-substitution? Is the solution unique in this case?

Now suppose the system is inhomogeneous

$A\mathbf{x} = \mathbf{b}$ where $\mathbf{b}\neq 0$. In this case we actually want $\det A \neq 0$ because then we can instantly write down the unique solution

$\mathbf{x} = A^{-1}\mathbf{b}$.

Have I gotten the solution to square systems about right? If yes, I'll try to figure out the non-square case.

 

[slider142]

Now how do we actually solve this for $\mathbf{x}$, do we just do Gaussian elimination followed by back-substitution? Is the solution unique in this case?

If the null space of A is not empty, then it is a subspace, so the solution is an entire subspace of the space you're working with, not just a single vector. The subspace containing only the zero vector is the only degenerate subspace that does consist of a single vector, and it is always in the null space.

Now suppose the system is inhomogeneous

$A\mathbf{x} = \mathbf{b}$ where $\mathbf{b}\neq 0$. In this case we actually want $\det A \neq 0$ because then we can instantly write down the unique solution

$\mathbf{x} = A^{-1}\mathbf{b}$.

Have I gotten the solution to square systems about right? If yes, I'll try to figure out the non-square case.

Yep, that's right.

 

[transgalactic]

i suggest that after you write your matrix
just make a row reduction
and you are not supposed to write a column of zeros in the end
the last column depends on the last number after the "=" sign

 

[Peeter]

You can also solve systems of equations of this form with the wedge product (wedging the column vectors). I'd put an example of this in the wiki Geometric Algebra page a while back when I started learning the subject:

http://en.wikipedia.org/wiki/Geomet...rally_expressed_in_terms_of_the_wedge_product.

Looking at the example now, I don't think it's the greatest. It should also probably be in a wedge product page instead of GA ... but that was the context that I learned about it first (I chose to use the mostly empty wiki page to dump down my initial notes on the subject as I started learning it;)