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Dimention proof question

  1. Mar 7, 2009 #1
    there is a matrix A of M3X3 (K) over field K
    so rank A=r , marked U as A of M3X3 (K) so AX=0

    prove that dim U=3(3-r))
    ??


    i was told that each column gives me 3-r we have 3 columns so
    it 3(3-r)

    i cant understand how we get 3-r out of each column??

    there is W={x exists in R^3|Ax=0}
    dim W=3-r
    1-1 function hase Ker (t)=0
    Im (T)=WxWxW

    i understand that W is the kernel of T
    then they say

    "dim W=3-r"
    so '"r" is the Image of the matrix
    but why its from one column
    ??
     
    Last edited: Mar 7, 2009
  2. jcsd
  3. Mar 7, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This is very hard to understand! I assume you mean that A is a 3 by 3 matrix having rank r and U is the subset of 3 by 3 matrices, X, such that AX= 0. The fact that A has rank 3 means that its image, A(K3), has dimension r. That also means that only r of its columns, thought of as vectors in K3, are independent and that kernel of A, the set of vectors, v, in K3 such that Av= 0 has dimension 3-r. If AX= 0, then X must have image a subset of the kernel of A: X must have rank less than or equal to 3-r.
     
  4. Mar 7, 2009 #3
    how did they prove that
    dim U=3(3-r)
    ??
     
  5. Mar 7, 2009 #4
    whats A(k^3)
    ??
     
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