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Linear Algebra - Column and Null Space

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Ax=b where,

    A = 2 -1
    .....-1 2

    2. Relevant equations

    a) Find Null Space N(A) and Column Space C(A)
    b) For which vectors b does the system Kx=b have a solution?
    c) How many solution x does the system have for any given b?

    3. The attempt at a solution

    a)

    For Null Space, I set Ax=0, and got the zero vector. No problem with that.
    For Column Space, I set Ax=b, and got b = span ( 2 , -1 )
    ..................................................................-1 2
    No problem with Column Space either.
    Those are two vectors, I am rusty with notation on here.

    b) This is simply the Column Space. All vectors b that span R2 allow Kx=b to have a solution. Ni problem with that either.

    c) This is where my issue is. For a given b, how many solutions does x have? No solutions? One? Infinitely many?

    I am having a hard time sorting this out intuitively. I want to say it will be one solution for a given b, but I can't come to that agreement.
     
  2. jcsd
  3. Oct 24, 2009 #2

    Dick

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    Suppose Kx=b has two solutions, Ka1=b and Ka2=b. Subtract the two equations. What can you say about a1-a2 vis a vis a space related to K that it might be in?
     
  4. Oct 25, 2009 #3
    Bit confused about what you mean by vis.
     
  5. Oct 25, 2009 #4

    HallsofIvy

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    "vis a vis" means "in relation to".

    His point was that if Ax= b and Ay= b then Ax- Ay= b- b or A(x-y)= 0. That is, x- y is a vector in the kernel of A. What did you get for the kernel of A?
     
  6. Oct 25, 2009 #5
    Kernel is related to Null Space correct?

    So, since I got the zero vector for Null Space, is the only solution the trivial one?

    A(x-y) = 0, where x-y = 0 = N(A) ?
     
  7. Oct 25, 2009 #6

    Dick

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    Yes. x-y=0 means x=y.
     
  8. Oct 25, 2009 #7

    HallsofIvy

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    "Kernel" is another word for "null space". Sorry, I should have noticed that you used "null space".

    Yes, you are correct that the null space for this operator is {0}. So in order that Ax=b= Ay, which is the same as A(x-y)= 0, x- y must be in the null space: x-y= 0 or x= y. So how many different solutions can Ax= b have?
     
  9. Oct 25, 2009 #8
    Please forgive me, but still a wee bit confused.


    I want to say one solution where x = y = 0.

    They way I am thinking is,

    Let Az = 0, then z = zero vector.

    So, A(x-y) = A(z) = 0

    then, x-y = z = 0, so x = y

    So infinitely many solutions?

    I really want to understand this, because I am a few problems similar so I do not want to just "accept" it. But it feels like I am walking in circles. Maybe it is easier than I what I expect?
     
  10. Oct 25, 2009 #9

    Dick

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    The point is that if A(x)=b and A(y)=b, then A(x-y)=0 (since Ax-Ay=b-b), ok so far? If A(x-y)=0 then x-y is in the null space, still ok? Since you've shown the null space is {0}, then x-y=0. So x=y. So if x and y are both solutions, then x-y. Now, how many solutions to Ax=b?
     
  11. Oct 25, 2009 #10
    ohhhh. Ohhhhh.

    Duh.

    3 solutions. x is a solution, y is a solution, and x-y = 0 is a solution.

    Do I have it now?
     
  12. Oct 25, 2009 #11

    Dick

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    i) 0 isn't a solution to Ax=b (unless b=0, of course). ii) Sure, x is a solution and y is a solution. But x=y!!! You've shown any two solutions are EQUAL. Take a walk around the block and think this over for a bit. Then count them again.
     
  13. Oct 25, 2009 #12

    HallsofIvy

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    NO! For one thing, x- y= 0 is not a solution to Ax= b, which is what you are working with.
    And having said that x- y= 0, it follows that x=y. In other words, x and y are NOT different solutions.
     
  14. Oct 25, 2009 #13
    So that means there is a unique solution for every b.
     
  15. Oct 25, 2009 #14

    Dick

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    Yes. Any problems with that?
     
  16. Oct 25, 2009 #15
    No that is it for me; crystal clear now. Sorry for being so hard-headed.

    I am going to do one more, and I will come back to make sure I understand everything correctly. Thanks for the help! This beats any other sources! :)
     
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