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Find basis for the kernel of linear map

  1. Apr 7, 2007 #1
    I need help. For this problem, you have to use the Gram-Schmidt process to make it orthogonal.

    My trouble is finding the bais for the kernel of the linear map

    L: R4 -> R1 defined by L([a,b,c,d)]=a-b-2c+d

    I know the dimension of the kernel is 3, but how?
    I have tried setting it against the standard basis and that's not right.
    I tried solving it by using four vectors with different values, and that keeps giving me a linear dependent vector.


    Am I missing something? I can row reduce and pull out the constants, but I have no idea how to get to the matrix.
  2. jcsd
  3. Apr 7, 2007 #2
    It's a linear map! L(x1+x2+x3+x4)=L(x1)+L(x2)+L(x3)+L(x4).

    Can you see how this helps you?

    You mave already solved it. Suppose you had a set of linearly independent vectors x1,x,2,x3,x4,...,xk that spans R^n (k necessarily greater than n for our purposes). Then you can take away one of the vectors and see if they still span and if they're linearly independent or not. You can repeat this process until you find a linearly independent set of vectors that span, which of course is going to be a basis.
    Last edited: Apr 7, 2007
  4. Apr 7, 2007 #3
    I think I have it...if there is constants for artibary numbers then those constants make up the kernal and the other is the range, correct?

    SO if the the dim is 4 and the kernel is three, then the range is one...

    if I solve it for a=b+2c-d, then the kernal is the constants, b, c, d times the vectors of (a, b, c, d)
  5. Apr 8, 2007 #4


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    The kernel is {(b + 2c - d, b, c, d) : b, c, d in R}. Do you know how to find the basis from here?
  6. Apr 8, 2007 #5


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    Since you have a single equation, a- b- 2c+ d= 0, you can solve for one variable: for example, d= b+ 2c- a. A standard way of finding a basis is to set each of the variables on the right equal to 1 and the other 0 in turn: If a= 1, b= c= 0, then d= -1; a basis vector is [1, 0, 0, -1]. If b= 1, a= c= 0, then d= 1; a basis vector is [0, 1, 0, 1]. If c= 1, a= b= 0, then d= 2; a basis vector is [0, 0, 1, 2]. Now use Gram-Schmidt to find an orthonormal basis.
  7. Apr 8, 2007 #6


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    Just to make sure -- you know that the kernel is the solution space to the equation
    L(v) = 0,​
  8. Apr 8, 2007 #7
    Got it...

    I was making it harder than it was...

    I solve for one of the variables, say a....


    Then I plug this value into L.

    [b+2c-d, b,c,d]

    I could set up in aug matrix, but I can pull the constants b, c, d right out to get the kernel basis.

    Then my vectors are
    u2= [2,0,1,0]
    u3= [-1,0,01]

    from here orthogonalize with the GS process.

    I do know to be in the kernel it has to be in the zero vector, L(v) = 0, or at least that is how my instructor describes it.

    I think I have it from here. Thanks to everyone for their help.
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