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Kernal and Range (share zero vector?)

  1. Feb 20, 2007 #1
    True of false: If A is a square matrix, ker A [tex] \cap[/tex] rng A = {0}.

    Now it seems to be like this would be false because if the kernel, aka nullspace, is zero then the matrix will have a non-zero range. Also, the range and kernel have dimensions in different spaces, and they will not have the same bases. I can see how this would work for a function, for example f(x)= x^2 = 0, and then show that f(1) =/= 0.

    I think it ultimately comes down to showing that:

    Ax = 0
    Ax = b = x

    Am I on the right track?
     
  2. jcsd
  3. Feb 21, 2007 #2

    AKG

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    Look at the definition of kernel and range, i.e. look at what it means for something to be in the kernel and what it means to for something to be in the range. Show that 0 satisfies both these definitions, i.e. that it is in both the kernel and range. The equation A(0) = 0 proves both things actually
    No, they have the same dimensions, because A is a square matrix. The range and kernel are thus in the same space, and can be given the same basis.
    What's the relevance of showing f(1) is not 0? And how is f(x) = x2 = 0 a function?
    This doesn't even mean anything.
     
  4. Feb 21, 2007 #3

    matt grime

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    The range and kernel can't 'be given the same basis', AKG. I don't even know what that means: suppose they intersect trivially, what would it mean to give them the same basis now?


    As for the OP.

    You're supposed to think of A as a linear map from V to V. If you didn't it wouldn't even begin to make sense to ask what the intersection of ker and the range are.

    It is very easy to show by example this is false. A can be taken to be 2x2. Just think what it means for something in the range to also be in the kernel. And if that doesn't work, arbitarilty pick a basis, e and f. Suppose that f spans in the kernel (if the kernel weren't 1-dimensional the range and kernel must intersect in 0 - do you see why). Now what can we do?
     
  5. Feb 21, 2007 #4

    HallsofIvy

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    I think AKG just mis-spoke. He was responding to Mindscrape's " the range and kernel have dimensions in different spaces". I'm not sure what is meant by "have dimensions in different spaces" but perhaps Mindsrape was thinking that if L: U--> V then the kernel is a subspace of U and the range (image) is a subspace of V. Since they are in different spaces, they cannot have any vectors in common! Of course, here, L is represented by an n by n matrix so it is really from Rn to Rn. Those are the same and the kernel and image are both in Rn and could, theoretically, have vectors in common. They do NOT necessarily have the same dimension: If the kernel has dimension m, then the image has dimension n- m.

    Obviously, since the kernel and image are both subspaces of Rn and every subspace contains 0, 0 is in the intersection of kernel and image. The problem is to show that no other vector is. That is, either show
    1) If [itex]v\ne 0[/itex] and Lv= 0, then there exist NO u such that Lu= v.
    or show
    2) If there exist u such that Lu= v, then Lv is not 0.
     
  6. Feb 21, 2007 #5

    matt grime

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    Why would anyone want to attempt to prove the impossible (which is what your asking someone to do), Halls?
     
  7. Feb 21, 2007 #6

    HallsofIvy

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    I forgot what the original problem was!!

    The original problem was to determine that decide whether "If A is a square matrix then kernel of A intersect range of A is the 0 vector" was true of false.

    Somehow I got it into my head that the problem was to prove that was true!
     
  8. Feb 21, 2007 #7

    AKG

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    That might have been my fault. I was very tired at the time I posted, and I thought the question asked to determine whether 0 was contained in the intersection of the kernel and the range, not whether {0} is that intersection. So I was treating the question like the answer was 'true'.
     
  9. Feb 21, 2007 #8
    So, you could take the matrix
    [tex] A = \left(
    \begin{array}{ccc}
    1 & -1\\
    1 & -1
    \end{array}
    \right)
    [/tex]

    Then show that the range is {(1,1)^T} and the kernel is {(1,1)^T}, which is not zero?
     
  10. Feb 21, 2007 #9

    AKG

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    The range is not exactly {(1,1)T}, it is the subspace containing every scalar multiple of (1,1)T. So you can write it as:

    Ran(A) = Span{(1,1)T} = {(k,k)T : k is a scalar} = {k(1,1)T : k is a scalar}

    The kernel is the same thing in the example you've given, i.e. Ker(A) = Ran(A). So the intersection of the kernel and the range is the kernel itself, or equivalently, the range itself, i.e. the intersection is:

    Span{(1,1)T}

    so in this case, clearly the intersection is not {0}.
     
  11. Feb 21, 2007 #10
    Er, yeah, I meant to put span before those, but you know what I meant. :p

    Just curious, does anyone know if this is false for an mxn matrix as well? I would actually suspect it would be true for an mxn.
     
  12. Feb 21, 2007 #11

    AKG

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    Is false for an mxn matrix, but for a totally different reason. In fact you hinted to the reason in your first post. The kernel is a subspace of R^n, i.e. it consists of n-tuples. The range is a subspace of R^m, i.e. it consists of m-tuples. No n-tuple is an m-tuple. Even (0,0,...,0) with m zeroes is clearly a different thing from (0,0,...,0) with n zeroes (if m and n are different, of course).
     
  13. Feb 22, 2007 #12

    matt grime

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    I just wanted to ask how the OP came up with that example? The counter example I wanted them to work out wasn't that, but I think it'd be nice if they thought about what it was:

    take e,f a basis, define T(e)=0 and T(f)=e. That is a simple counter example, its matrix with respect to {e,f} is

    [0 1]
    [0 0]

    and every counter example in R^2 is conjugate to it.

    It is a case of the 'just do it'. I.e. construct the (simplest?) possible counter example: write down something with nonzero kernel, and with something mapping into the kernel.
     
  14. Feb 22, 2007 #13
    I knew it would have be a symmetric matrix, and I tried the first one that came to mind, which happened to give a counter example.

    By the way with the stuff about polynomials, I was just trying to think of something in there that would work to. I knew it would probably be an even function that would fulfill. I think you could do f(x)=(x-1)^2 and then f(1) = 0, also that f(0) = 1.

    So you suggest taking two functions T(e)=0 and T(f)=e, and then making a matrix out of the equations to show that their basis would form any counter example? What do you mean by conjugate?
     
  15. Feb 22, 2007 #14

    matt grime

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    Symmetric matrix? What on earth do you meanby that? And who mentioned polynomials?
     
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