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Confused about dimensions

  1. Oct 30, 2012 #1
    I am a little confused about dimensions, if the nullspace of a matrix is spanned by the 0 vector, does that mean the dimension of the nullspace of this matrix is 0?

    In the problems I attached, both A and B reduced to the identity matrix.
    Note (2) is supposed to be dim(N(B)) and dim(col(B)) instead of dim(N(A)) and dim(col(A)).


    Since they both reduce to I, that means a pivot is in every column, which means that the dimension of N(A) and N(B) is 0 right?


    Consequently, this means dim(Col(A)) and dim(col(B)) is 3 right?
     

    Attached Files:

  2. jcsd
  3. Oct 30, 2012 #2

    Mark44

    Staff: Mentor

    Yes.
    Right.
    Yes.

    Let's look at this situation from the perspective of transformations. Your matrices are 3 X 3, meaning that they represent transformations from R3 to R3 (or possibly a lower dimension subspace of R3). Since both matrices reduce to I3, both matrices map a nonzero vector in R3 to some other nonzero vector. The only vector mapped to <0, 0, 0> is the zero vector itself. Another way to say this is N(A) = N(B) = {0}.

    If it had turned out that the nullspace for one of the matrices was one-dimensional, then some vectors in R3 are getting mapped to 0. For this matrix, the range would be a two-dimensional subspace of R3. In other words, a plane embedded in space. An example of this is T(<x, y, z>) = <x, y, 0>. This transformation projects a vector v down to the x-y plane.
     
    Last edited: Oct 30, 2012
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