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Thread: 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 $\displaystyle \beta$ for V such that [T]_$\displaystyle \beta$ 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 $\displaystyle \{w_1,...,w_k\}$ for $\displaystyle W$, and then complete it to a basis $\displaystyle \{w_1,w_2,...,w_k,v_{k+1},...,v_n\}$ of $\displaystyle V$.

    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 $\displaystyle T(w_1) = a_1w_1+a_2w_2+...+a_kw_k+d_{k+1}v_{k+1}+...+d_nv_n ,$
    the first column of the matrix consists of the numbers $\displaystyle a_1,a_2,...,a_k,d_{k+1},...,d_n.$

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

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