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Non-square linear systems with exterior product

  1. Sep 11, 2014 #1
    Hi, how can I compute the general solution of a system of linear equations? Non-square systems for example. I have the book Linear Algebra via exterior products, but it is the worst book in the history of math books, I think I'll burn it somehow, whatever. I can calculate the solution with the exterior product easily with Cramer's rule. For non-square systems, or systems with infinite many solutions Winitzki talks gibberish about a maximal non-zero exterior product between the vectors of the matrix and then does some magic with it to calculate the homogeneous solution. Can someone explain this to me in a clear way? I think one should calculate it by first finding the non-zero exterior product between the vectors of the matrix (which should be a subspace) and then somehow express the remaining vectors in terms of the non-zero set? But that doesn't make sense. hm
     
    Last edited by a moderator: Sep 29, 2014
  2. jcsd
  3. Sep 12, 2014 #2
    Can someone help me with this? I give an example of the book:

    Let's say the vectors a,b,c are not a basis in the vectorspace V, then there exists a maximal nonzero exterior product (which should also tell you what the rank is) of a linear independent subset of a,b,c Take an example

    [itex]2x+y=1[/itex]

    [itex]2x+2y+z=4[/itex]

    [itex]y+z=3[/itex]

    Now
    [itex]a=(2,2,0)[/itex],
    [itex]b=(1,2,1)[/itex],
    [itex]c=(0,1,1)[/itex],
    [itex]p=(1,4,3)[/itex]

    We see that [itex]a \wedge b \wedge c=0[/itex]. And the maximal nonzero exterior product can be written as [itex]\omega =a \wedge b[/itex] (which is not equal to zero.
    Now we can check if [itex]p[/itex] is a subset of the span [itex]{a,b}[/itex] with [itex]\omega \wedge p=0[/itex] so [itex]p[/itex] can be expressed with [itex]a,b[/itex]. We can find the coefficients with Cramer's rule

    [itex]\alpha = \frac{p\wedge b}{a\wedge b}=-1[/itex]

    [itex]\beta = \frac{a\wedge p}{a\wedge b}= 3[/itex]

    Therefore [itex]p=-a+3b[/itex] so the inhomogeneous solution is [itex]x^{1}= (-1,3,0)[/itex] Now to determine the space of homogeneous solutions, the vector [itex]c[/itex] get's decomposed into a linear combination of [itex]a[/itex] and [itex]b[/itex] again by Cramer's rule. This gives [itex]c=-\frac{1}{2}a+b[/itex] And the space of homogeneous solutions is given by the span of [itex]x_{i}^{(0)(1)}=(-\frac{1}{2},1,-1)[/itex]. So the general solution is

    [itex]x_{i}=x_{i}^{1}+ \beta x_{i}^{(0)(1)}= (-1-\frac{1}{2}\beta, 3+\beta, -\beta)[/itex]

    Then he gives another example with a non-square system

    [itex]x+y=1[/itex]

    [itex]y+z=1[/itex]

    And the general solution is [itex]x_{i}=(1,0,1)+\alpha (1,-1,1)[/itex] (with no explanation).

    I don't see a pattern here. What is going on?
     
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