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Homework Help: Basis, Nullspace, Linear transformtion

  1. Oct 29, 2012 #1
    The problem is attached, I did parts 1-3, but I am having trouble with part 4.

    This is what i was planning on doing for part 4 (my teacher said this wasn't the correct method):

    set T(v)=0
    and solve the augmented matrix
    1 0 -1 1 0
    2 1 -2 4 0
    3 1 -1 7 0

    rref gives
    1 0 0 2 0
    0 1 0 2 0
    0 0 1 1 0

    So N(T)=span{[-2 -2 -1 1]^t}
    So shouldn't a basis for N(T) be [-2 -2 -1 1]^t?

    I don't understand how this is wrong.

    Attached Files:

    Last edited: Oct 29, 2012
  2. jcsd
  3. Oct 30, 2012 #2


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    I don't see anything wrong with it either. What you're doing is to solve the equation
    $$T\begin{bmatrix}a\\ b\\ c\\ d\end{bmatrix}=0.$$ Maybe your teacher just thought that this was a bad idea compared to solving
    $$T\begin{bmatrix}2b-2c\\ b\\ c\\ -c\end{bmatrix}=0,$$ where two of the unknowns have already been eliminated because we're using what we know about the null space of A. But I get the same result with both methods, so I can't say that the second one is significantly better.
  4. Oct 30, 2012 #3
    I talked to him again and he said "the way I did it, I was finding all of the vectors in R4 such that T(x)=0. The question is asking to find all vectors in V such that T(x)=0 which is a different question."

    The basis found in part 1 is {[2 1 0 0] [2 0 -1 1]}.
    The linear transformation of these two vectors is
    [2 5 7]^t and [4 10 14]^t respectively.

    Te matrix for this is
    2 4
    5 10
    7 14

    Rref is
    1 2
    0 0
    0 0

    The coordinate vector that spans the null space is [2 -1]^t

    So the basis is 2[2 1 0 0] + 1[2 0 -1 1]= [-2 -2 -1 1]

    Wow that got really tricky a d conplicated
  5. Oct 31, 2012 #4


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    Right, that's what I meant. The thing is, both methods give us the same result, so what you did can't be completely wrong. I think the only problem is this: When you do it your way, you don't automatically know that the vector you find is actually a member of V (= the null space of A), so you also have to verify that it is.
  6. Oct 31, 2012 #5


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    I don't understand exactly what you're doing here. I think it may be an advantage for me that I don't remember the fancy ways of doing these things. :smile: All I did was this: You know from part 1 that any member of V can be written as a linear combination of the two basis vectors you found: a[2 1 0 0]T+b[2 0 -1 1]T =[2a+2b a -b b]T So now you just need to solve T[2a+2b a -b b]T=0. This gives you a relationship between a and b. Use that relationship and set b=1, or whatever is convenient.
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