`Forgotten' linear algebra

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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

[itex]A \mathbf{x} =\mathbf{0}[/itex].

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

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

Now suppose the system is inhomogeneous

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

[itex]\mathbf{x} = A^{-1}\mathbf{b}[/itex].

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

Answers and Replies

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Now how do we actually solve this for [itex]\mathbf{x}[/itex], 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

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

[itex]\mathbf{x} = A^{-1}\mathbf{b}[/itex].

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.
 
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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
 
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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/Geometric_algebra#Cramer.27s_rule.2C_determinants.2C_and_matrix_inversion_can_be_naturally_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;)
 
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