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Nullspace Proof

  1. Nov 12, 2008 #1

    danago

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    Hey. Im doing some last minute exam study and came across this question:

    let A and B be nxn matrices. Prove that [tex]Null(AB) \geq Null(B)[/tex]

    Now i think i did it correctly, but i wasn't sure if i perhaps overlooked something and there is a flaw in my proof.

    I began by stating that if any vector is in the nullspace of B, then it MUST also be in the nullspace of AB. Let x be in the nullspace of B, then:

    Bx = 0

    Then:

    ABx = A(Bx) = A(0) = 0

    Hence any vector in the nullspace of B is also in the nullspace of AB i.e. the nullspace of B is a subset of the nullspace of AB, hence the dimension (nullility) of the nullspace of AB will be greater (or equal).

    That look ok?

    The second part of the question was:

    When is the Null(AB) = Null(B)?

    I said when A is invertible, since:

    ABx = 0 imples that Bx = 0 if the inverse of A exists. Hence, any vector in the nullspace of AB will also be in the nullspace of B. Since every element of the nullspace of AB is also in nullspace of B, and every element of the nullspace of B is also in the nullspace of AB (from the first part of the question), the two subspaces are equal and hence have the same dimension.


    Do they look alright?

    Thanks in advance,
    Dan.

    PS. I am so sick of typing the word nullspace
     
  2. jcsd
  3. Nov 12, 2008 #2

    Dick

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    It looks fine to me. Why don't you make up a notation like 'N(A) means nullspace of A'? That way you won't have to type the word over and over.
     
  4. Nov 12, 2008 #3

    danago

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    Thanks, and glad to hear that its ok :smile:

    And haha yea i did think about doing that, but only after i had posted. Im in a bit of a crazy pre-exam mood at the moment because i have an accounting exam tomorrow and pretty much hate the unit with a passion and not so confident about it :P
     
  5. Nov 12, 2008 #4

    Office_Shredder

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    I don't think A needs to be invertible. What you need is [tex]im(B) \cap ker(A) = {0}[/tex]
     
  6. Nov 12, 2008 #5

    Dick

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    Actually, there is maybe an oversight. It's certainly true that if A is invertible N(AB)=N(B). But there is a weaker possibility. Can you show that if N(A) is contained in N(B) then it's also true that N(AB)=N(B)?
     
  7. Nov 13, 2008 #6

    danago

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    Hmm im not quite sure. I dont know if im completely missing something, but i cant see how to show that. From the first part ive shown that Bx=0 ==> ABx=0, so i guess i need to use the fact that N(A) is in N(B) to show that ABx=0 ==> Bx=0 and that will conclude the proof?

    Any hints? :smile:


    Also, Office_Shredder, is the kernal of a matrix the same as its nullspace? And what does im(B) mean?
     
  8. Nov 13, 2008 #7

    Office_Shredder

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    kernel of the matrix is its nullspace, the image of the matrix (im(B)) is the set of points that the matrix maps to... so for example, the matrix

    [1 0]
    [0 0]

    maps the vector (a,b) to (a,0) so the image of this matrix is just the set of all points (a,0).
     
  9. Nov 13, 2008 #8

    Dick

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    Office_Shredder said it much more accurately than I did. The kernel is the same as the nullspace. im(B) is the set of all set of all y such that y=B(x) for some x. So the statement if really just saying if B(x) is not zero then A(B(x)) is not zero. Which is exactly what you need.
     
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