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Finding basis for kernal of linear map

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A = 1 3 2 2
    1 1 0 -2
    0 1 1 2

    Viewing A as a linear map from M_(4x1) to M_(3x1) find a basis for the kernal of A and verify directly that these basis vectors are indeed linearly independant.


    3. The attempt at a solution

    Ok so first i found the reduced row echelon form of A. This equals:

    rref(A) =

    1 0 -1 -4
    0 1 1 2
    0 0 0 0

    So i found the kernal of this by-

    1 0 -1 -4
    0 1 1 2
    0 0 0 0

    Multiplied by

    x_1
    x_2
    x_3
    x_4

    Equals

    0
    0
    0
    0.



    x_1 = x_3 + x_4
    x_2 = -x_3-2x_4
    x_3 = x_3
    x_4 = x_4

    Therefore kernal...

    =x_3 {1, -1, 1, 0} + x_4 {1, -2, 0, 1}

    So i thought this meant the basis equalled

    Basis of kernal = (1,-1,1,0),(1,-2,0,1)


    I have idea what to do now though. I have no idea if what i have done is vaguely right and am not sure if it is how to fulfill the rest of the question. The problem is i have not really incoperated the fact that in the question it states that Viewing A as a linear map from M_(4x1) to M_(3x1). I do not understand this terminology, what does it mean exactly?

    (ps - i appologise for the bad formatting)


    Thanks
     
  2. jcsd
  3. May 22, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi PhyStan7! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Nooo :redface:4x4 :wink:
    I assume M4x1 is the 4x1 matrices or column vectors.

    So A is a function from the 4-column vectors to the 3-column vectors.

    Any (constant) matrix is linear, so it's a linear function (linear map).

    Only functions (maps) have kernels, so you have to view the matrix as a map to talk about a kernel. :smile:
     
  4. May 22, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You want to solve

    [tex]\begin{bmatrix}1 & 3 & 2 & 2 \\ 1 & 1 & 0 & -2 \\ 0 & 1 & 1 & 1\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}[/tex].



    Which is the same as the three equations w+ 3x+ 2y+ 2z= 0, w+ x- 2z= 0, x +y+ z= 0.

    Adding the first two equations eliminates z: 2w+ 4x+ 2y= 0. Multiplying the third equation by 2 and adding to the second equation also eliminates z: w+ 3x+ 2y= 0.

    Subtracting the second of those from the first eliminaes y: w+ x= 0 so x= -w.

    Putting that back into the previous equations will allow you to write each of x, y, and z in terms of w. The kernel is one-dimensional, not two-dimensional.

    A "linear map" is a "linear" function from one vector space to another. If f is a linear map then f(au+ bv)= af(u)+ bf(v) fpr any vectors u and v in the domain, any scalars a and b.

    You can think of an "m by n" matrix as a linear map from [itex]R^m[/itex] to [itex]R^n[/itex]. Conversely, any linear map from from m-dimensional U to n dimensional V can be written as an m by n matrix for specific bases for U and V.
     
    Last edited: May 23, 2010
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