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A question about the rank of a linear operator

  1. Apr 9, 2009 #1
    Let T is a linear transformation from a vector space V to V itself. The dimension of V, denoted by dim(V), is finite.

    If the rank of T, denoted by rk(T), is equivalent to the rank of TT, i.e., rk(T)=rk(TT)

    why is the intersection of image of T(denoted by im(T)) and the kernel of T(denoted by ker(T)) is the zero vector, i.e., im(T)[tex]\cap[/tex]ker(T)={0}?

    Thanks for any help.
  2. jcsd
  3. Apr 9, 2009 #2
    Think first of the special case where T is a projection. Then the image of T and the image of TT are clearly the same. In such a case, if the kernel of T intersected the image, then T wouldn't be a projection.
  4. Apr 9, 2009 #3
    You are right.

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