For any 1<=k<=n, find a Jordan form for linear map T:Pn(F)->Pn(F): T(p(z)) = p^k(z), kth derivative of p(z)

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  • 1 decade ago
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    I assume that Pn(F) is the vector space of univariate polynomials with coefficients in F of degree at most n. If F has characteristic zero, a basis of F is {e1, ..., e(n+1)}, where ej = n!/(j−1)! z^{j−1} for all j. Then e1 = n! so d/dz(e1) = 0 and for each j=2, ..., n+1, d/dz(ej) = (j−1) n!/(j−1)! z^{j−2} = e(j−1). Therefore, d/dz has one (n+1) by (n+1) Jordan block with eigenvalue 0. If, for some 2 ≤ k ≤ n+1, we take a kth power of d/dz, this will break up into k eigenvalue 0 subblocks; each of the subblocks will have basis {ej | j congruent to i modulo k}, where i is in 0, ..., k−1. For example if k=3, d^3/dz^3 has three Jordan blocks, one with basis {e(n+1), e(n−2), e(n−5), ...}, one with basis {en, e(n−3), e(n−6), ...}, and one with basis {e(n−1), e(n−4), e(n−7), ...}. For k ≥ n+1, d^k/dz^k will be identically zero, so it will have n+1 eigenvalue 0 Jordan blocks.

    If F has characteristic p we have d/dz(z^p) = d/dz(z^{2p}) = ... = 0, so d/dz will have a number of Jordan blocks of size p, and also a smaller, remnant, block if n is not congruent to −1 modulo p. For 2 ≤ k ≤ p, d^k/dz^k will have Jordan blocks given by decomposing these blocks as above. For k ≥ p, d^k/dz^k will be identically zero, so it will have n+1 Jordan blocks.

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