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Homework Help: Understanding nullspace (kernel) of a matrix

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the kernel of the matrix:

    [PLAIN]http://img256.imageshack.us/img256/9015/53369959.jpg [Broken]

    3. The attempt at a solution

    So I row-reduce it and get:

    [PLAIN]http://img812.imageshack.us/img812/1391/97980793.jpg [Broken]

    The system of equations the row-reduced form equals 0.

    So I set x[itex]_{3}[/itex] and x[itex]_{4}[/itex] as the free variables and solve for x[itex]_{2}[/itex]: x[itex]_{2}[/itex] = -x[itex]_{3}[/itex] - x[itex]_{4}[/itex]

    Substitute that into the top equation to get x[itex]_{1}[/itex] -4(x[itex]_{3}[/itex]+x[itex]_{4}[/itex]) +2x[itex]_{3}[/itex] +7x[itex]_{4}[/itex] = 0

    Solve for x[itex]_{1}[/itex]: x[itex]_{1}[/itex] = 2x[itex]_{3}[/itex] - 3x[itex]_{4}[/itex]

    From this we get:

    Vector(2x[itex]_{3}[/itex]-3x[itex]_{4}[/itex], -x[itex]_{3}[/itex]-x[itex]_{4}[/itex], x[itex]_{3}[/itex], x[itex]_{4}[/itex])

    So ker(A) = x = x[itex]_{3}[/itex](2,-1,1,0) + x[itex]_{4}[/itex](-3,-1,0,1)

    What does this mean essentially? I know how to solve it, but I don't really understand what I'm doing or what this is useful for. As far as I understand, the kernel is a subspace of a linear map, so what does this translate to in practical terms?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 3, 2011 #2


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    Science Advisor

    well a linear transformation maps one vector space into another. the kernel is a measure of how many vectors get annihilated by the linear transformation (we lose entire dimensions at a time).

    in this case, we lose 2 dimensions, meaning there is only 2 left for the image space. the matrix A condenses R^4 down to a two dimensional plane in R^3, by sending an entire 2-dimensional plane to the origin.

    it's easier to understand what this means with some really basic examples (we will assume the standard bases are used):

    suppose T:R^3-->R^2 is the map T(x,y,z) = (x,y). we just dump the 3rd dimension. this has the matrix:

    [1 0 0]
    [0 1 0], and it's pretty clear that the nullspace of T is {(0,0,z) : z in R} (the z-axis).

    or suppose L:R^3-->R^3 is the map L(x,y,z) = (x,x,x). this has the matrix

    [1 0 0]
    [1 0 0]
    [1 0 0], and has nullspace {(0,y,z) : y,z in R}. here our image space is just the line x(1,1,1), anything in the yz-plane maps to the origin.

    the nullspace tells you how much information you are losing because of a linear transformation T. if a nullspace is just {0}, you know that T is preserving a faithful copy of the original space (although it may "stretch" or "rotate" vectors, we can still trade coordinates one-for-one). a matrix for such a T just sends a basis straight over to some other basis.

    in terms of solving systems of linear equations, finding x for which Ax = 0, and finding nullspace(A), are the same problem.
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