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Linear Algebra Proof

  1. Oct 4, 2013 #1
    1. The problem statement, all variables and given/known data

    SHow that the set of solutions to a homogenous system of m linear equations in n variabes is a subspace of [itex]ℝ^{n}[/itex] (Show that this set satisfies the definition of a subspace)

    2. Relevant equations



    3. The attempt at a solution
    If {V1,...Vk}=[itex]ℝ^{n}[/itex] then every vector [itex]\vec{q}[/itex][itex]\in[/itex]ℝ can be written as a linear combination of the set
    c1V1+....+ckVk=[itex]\vec{q}[/itex]
    This system of linear equations must have a solution for every [itex]\vec{q}[/itex][itex]\in[/itex]ℝ and therefore the rank of the coefficient matrix = n
    If the rank of the coefficient matrix of a system
    c1V1+...+ckVk=v
    is n, then the system is consistent for all V[itex]\in[/itex]ℝ
    ∴ {V1,....,Vk}=[itex]ℝ^{n}[/itex]


    I thought I was on the right track, but a theorem in my textbook says
    " Let [A|[itex]\vec{b}[/itex]] be a system of m linear equations in n variables. Then [A|[itex]\vec{b}[/itex]] is consistent for all [itex]\vec{b}[/itex]=[itex]ℝ^{n}[/itex] if and only if rank(A)=m"

    Does the requirement change is they are homogenous? Am I even on the right track?
     
  2. jcsd
  3. Oct 4, 2013 #2

    Ray Vickson

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    You are on the wrong track. You need to worry about the rank of the matrix. For example, what is the dimensionality of the solutions to the following linear system?
    [tex] x_1 + x_2 + x_3+ x_4 = 0\\
    2x_1 + 2x_2 + 2x_3 + 2x_4 = 0
    [/tex]
     
  4. Oct 4, 2013 #3

    jbunniii

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    I think you're making this much harder than it needs to be. You don't need to know the rank of the system.

    A homogeneous system can be written as follows: ##Ax = 0##, where ##A## is some matrix.

    A vector ##x## is a solution to the system if and only if ##Ax = 0##. So the set of solutions is precisely the set of vectors which satisfy the equation.

    What are the requirements to show that a set is a subspace?
     
  5. Oct 4, 2013 #4
    Not sure what you mean by dimentionality of the system
    but the RREF form yields the system
    [tex] x_1 + x_2 + x_3+ x_4 = 0
    [/tex]
     
  6. Oct 4, 2013 #5
    Set is a subspace if it is closed under addition and scalar multiplication (and therefore include the zero vector)
     
  7. Oct 4, 2013 #6

    Ray Vickson

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    Yes, and that is my point exactly: the matrix has rank 1, so the dimensionality of the solution set is 4 - 1 = 3. This counters your statement in the OP that the rank of A is n.

    However, as has already been pointed out, you don't even need to know this to do the question; you would need to know it if you were asked about the dimension of the subspace.
     
    Last edited: Oct 4, 2013
  8. Oct 4, 2013 #7

    jbunniii

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    OK, let's start with addition. If ##x## and ##y## are solutions, what can you say about ##x+y##?
     
  9. Oct 4, 2013 #8
    That x+y is also a solution? It should be if it's closed under addition
     
  10. Oct 4, 2013 #9

    jbunniii

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    Are you asking me or telling me? You need ##A(x+y) = 0##. Is it true? If so, why?
     
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