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Subspace theorem problem (matrix)

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose A is a fixed matrix in M. Apply the subspace theorem to show that

    S = {x [itex]\in[/itex] ℝ : Ax = 0}

    3. The attempt at a solution

    Zero vector for x = <0,0,0>

    A*<0,0,0> = 0

    Therefore zero vector is in ℝ and S is non-empty.


    For u & v [itex]\in[/itex] S

    u + v = 0 + 0 = 0

    A*(u+v) = 0 => A(0) = 0

    S is closed under addition


    λ [itex]\in[/itex] R

    λ*A*u = λ*0 = 0

    Therefore closed under multiplication and by subspace theorem, if a subspace of S.

    Is this correct?
  2. jcsd
  3. Jan 18, 2012 #2


    Staff: Mentor

    I can't tell what space x is in. It just shows up as a box in my browser. Is it R3?

    Also, what is it you're supposed to show. All you have above is a definition for S. Are you supposed to show that S is a subspace of R3 (or whatever the space is)?
    No, you're not given any information that allows you to conclude that u and v are zero vectors or that they add to 0.
    No, for the reason given above. You need to use the properties of matrix multiplication.
    No, because you can't say that u is necessarily equal to 0.
  4. Jan 18, 2012 #3


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    I assume that instead of ℝ you mean Rn (Mark: it shows up in my browser as a blackboard [itex]\mathbb{R}[/itex]).

    It looks almost correct to me, except for this line:
    u + v = 0 + 0 = 0
    You are taking u and v arbitrarily in S then they are not necessarily equal to zero. You may want to use the linearity of matrix multiplication here.

    Actually he used that Au = 0 there, so that is correct.
  5. Jan 18, 2012 #4


    Staff: Mentor

    The OP is using vectors in R3, though.
    Right. I mistakenly thought the OP replaced u with 0, but it was Au being replaced. I didn't catch that.
  6. Jan 18, 2012 #5


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    You mean "show that S is a subspace of"- I guess R3 although you said ℝ.

    Yes, that is correct.

    No. What you have written here is that u= v= 0 and that is not necessarily true.
    However, what you may have meant to write may have been

    Not quite. You need A(u+ v)= Au+ Av= 0+ 0= 0.

    Again, not quite. You want [itex]A(\lambda u)= \lambda A(u)= \lambda(0)= 0[/itex]

  7. Jan 19, 2012 #6
    That should have been an R^n yes! I mistakenly left that out, sorry about the confusion!
  8. Jan 19, 2012 #7
    Thanks HallsofIvy, I guess I need to expand and be more explicit in each step of the way. I just assumed I didn't have to show that commutative, or was it associative law of multiplication. But I understand what you mean!
  9. Jan 19, 2012 #8
    I'm not too sure what's wrong with the addition aspect.

    Given that I want to essentially prove that the matrix A multiplied by an addition of vectors would equal zero A*(u + v) = 0

    So would it suffice to say that the above statement is true, and u + v = 0 but that neither u or v has to equal 0?
  10. Jan 19, 2012 #9


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    But it isn't true that u+ v= 0. What is true is that Au= 0 and Av= 0. That's completely different.
  11. Jan 19, 2012 #10


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    for emphasis, u in S does NOT mean u is 0. it just means that Au = 0.

    of course, the 0-vector 0 is ONE of the vectors in S, but an arbitrary vector in S need not be the 0-vector (it's hard to say for sure without knowing more about the matrix (or linear map) A).

    what you need to show, is that:

    if u,v is in S, so Au = 0 and Av = 0,

    then A(u+v) = 0.

    next, you need to show that if c is in R, and u is in S, that cu is also in S

    (which entails showing that A(cu) = 0).
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