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Math Help - Show that there is a basis for beta

  1. #1
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    Show that there is a basis for beta

    Let V be an n-dimensional vector space, and let T:V->V be a linear transformation. Suppose that W is a T-invariant subspace of V having dimension k. Show that there is a basis \beta for V such that [T]_ \beta has the form : (this is supposed to be a matrix im still working on my latex)

    (A B
    O C) where A is a kxk matrix and O is the (n-k)xk zero matrix.
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  2. #2
    Member HappyJoe's Avatar
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    You can start by choosing a basis \{w_1,...,w_k\} for W, and then complete it to a basis \{w_1,w_2,...,w_k,v_{k+1},...,v_n\} of V.

    The first column of the matrix of T with respect to this basis is the coordinates of T(w_1) expressed in this basis: Writing T(w_1) = a_1w_1+a_2w_2+...+a_kw_k+d_{k+1}v_{k+1}+...+d_nv_n  ,
    the first column of the matrix consists of the numbers a_1,a_2,...,a_k,d_{k+1},...,d_n.

    But W is an invariant subspace of T, so since w_1 is a member of W, then T(w_1) will also be a member of W. But this means that T(w_1) can be expressed in the basis \{w_1,w_2,...,w_k\} of W, so that all the scalars d_{k+1},\ldots,d_n are 0. In conclusion, the last n-k entries in the first column of the matrix are 0.

    Repeat this procedure for all the basis vectors w_i to see that for each of the first k columns, it is the case that the last n-k entries are 0.
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