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Math Help - Vector space / linear transformation question

  1. #1
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    Vector space / linear transformation question

    I've been going over this question for a while, and can't seem to make head or tail of it. Any help would be appreciated!

    Let V be a vector space s.t. dim(V) = n. T: V -> V is a linear transformation.

    Define {K_i} = \ker {T^i},{\rm{ }}i \ge 0, where {T^0} is the identity mapping. I need to show that {K_i} \subseteq {K_{i + 1}} (I've done that bit); and then deduce that \exists m \ge 0:{K_m} = {K_{m + 1}}, and so {K_m} = {K_{m + j}} for all j \ge 1, and so finally V = {K_m} \oplus {T^m}\left( V \right).

    In particular, I don't understand why the kernal can't just keep growing - can you not just keep including one more element in successive linear transformations each time? If your vector space has an infinite number of elements in it (e.g. the real numbers), then why does there have to come a point at which the kernel doesn't grow any bigger?
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  2. #2
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    Quote Originally Posted by adamcurran11 View Post
    I've been going over this question for a while, and can't seem to make head or tail of it. Any help would be appreciated!

    Let V be a vector space s.t. dim(V) = n. T: V -> V is a linear transformation.

    Define {K_i} = \ker {T^i},{\rm{ }}i \ge 0, where {T^0} is the identity mapping. I need to show that {K_i} \subseteq {K_{i + 1}} (I've done that bit); and then deduce that \exists m \ge 0:{K_m} = {K_{m + 1}}, and so {K_m} = {K_{m + j}} for all j \ge 1, and so finally V = {K_m} \oplus {T^m}\left( V \right).

    In particular, I don't understand why the kernal can't just keep growing - can you not just keep including one more element in successive linear transformations each time? If your vector space has an infinite number of elements in it (e.g. the real numbers), then why does there have to come a point at which the kernel doesn't grow any bigger?

    Since kernels of lin. trans. are subspaces, "to keep growing" mean they increase their dimension (because if you add a vector to a lin. space which already is a linear combination of some basis of this space, the vector's ALREADY in the space and then there's no growth). Since the whole lin. space's dimension's finite the growing can't continue forever, and thus there exists an integer m as in the question.

    Tonio
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