A relation between nullity and the number of Jordan blocks of a finitely generated module over a PID

Let F be a field, V = F^n for some natural number n \geq 2, and T a linear transformation on V. Make V into an F[x]-module via T as usual. Let \lambda_i be the (distinct) eigenvalues of T (suppose these are all in F (if not, extend F)). By FTFGMPID, we have V = \bigoplus_i \bigoplus_j F[x]/(x-\lambda_i)^{a_{i,j}}. We will call each a_{i,j} the size of its corresponding Jordan block; note that this is precisely its dimension as an F-vector space. Letting i be fixed (so \lambda_i is a fixed eigenvalue), we say that V_i = \bigoplus_j F[x]/(x-\lambda_i)^{a_{i,j}} is the generalized eigenspace of V with eigenvalue \lambda_i.

Let k \geq 0, and fix an eigenvalue \lambda_i. Show that the nullity of T-\lambda_i I on (T-\lambda_i I)^kV_i is the same as the nullity of T-\lambda_iI on (T-\lambda_iI)^kV, and that these are equal to the number of Jordan blocks of V having eigenvalue \lambda_i and size greater than k.

(Recall that if \varphi is an F-linear transformation, then \mathsf{nullity}_F\ \varphi = \mathsf{dim}_F\ \mathsf{ker}\ \varphi.)

First an editorial note. The mapping T-\lambda_iI is really (a priori) a linear transformation on V. However, any transformation which is a polynomial in T fixes each Jordan block of V, so we can think of it as a transformation on V_i as well. With this interpretation, to show that the nullity of T-\lambda_iI on (T-\lambda_i I)^kV_i is the same as the nullity of T-\lambda_iI on (T-\lambda_iI)^kV, it is enough to show that the kernel of T-\lambda_iI is contained in (T-\lambda_iI)^kV_i. To see this, note that if j \neq i, then by Lemma 3 of this previous exercise, (x-\lambda_i) F[x]/(x-\lambda_j)^{a_{j,t}} = F[x]/(x-\lambda_j)^{a_{j,t}}, and thus the kernel of T-\lambda_iI on the Jordan blocks not in V_i is trivial.

Now using Lemmas 2 and 3 from this previous exercise, we have (x-\lambda_t)^kV = \bigoplus_i \bigoplus_j (x-\lambda_t)^k F[x]/(x-\lambda_i)^{a_{i,j}} \cong_{F[x]} \bigoplus_{j,a_{t,j} > k} F[x]/(x-\lambda_t)^{a_{t,j}-k} \oplus \bigoplus_{j,a_{t,j} \leq k} 0 \oplus \bigoplus_{i \neq t} V_i. Note that every F[x]-module isomorphism is also and F-vector space isomorphism.

Similarly, (x-\lambda_t)(x-\lambda_t)^kV \cong_{F[x]} \bigoplus_{j,a_{t,j} > k+1} F[x]/(x-\lambda_t)^{a_{t,j}-(k+1)} \oplus \bigoplus_{j,a_{t,j} \leq k+1} 0 \oplus \bigoplus_{i \neq t} V_i, and that the mapping realizing this isomorphism is also an F-vector space isomorphism.

By Theorem 7 on page 412 of D&F and by the First Isomorphism Theorem, we have \mathsf{dim}_F (x-\lambda_t)^kV = \mathsf{dim}_F\ \mathsf{im}\ (x-\lambda_t) + \mathsf{dim}_F\ \mathsf{ker}\ (x-\lambda_t). Note that the dimension of F[x]/(x-\lambda)^m over F is just m.

Thus we have the following, considering x-\lambda_t as an endomorphism of (x-\lambda_t)^kV.

\mathsf{nullity}_F (x-\lambda_t)  =  \mathsf{dim}_F\ \mathsf{ker}\ (x-\lambda_t)
 =  \mathsf{dim}_F (x-\lambda_t)^kV - \mathsf{dim}_F \mathsf{im}\ (x-\lambda_t)^{k+1}V
 =  \left[ \sum_{j,a_{t,j} > k} (a_{t,j} - k) + \mathsf{dim}_F\ \bigoplus_{i \neq t} V_i \right] - \left[ \sum_{j,a_{t,j} > k+1} (a_{t,j} - k - 1) + \mathsf{dim}_F\ \bigoplus_{i \neq t} V_i \right]
 =  \sum_{j,a_{t,j} > k} (a_{t,j} - k) - \sum_{j,a_{t,j} > k+1} (a_{t,j} - k - 1)
 =  \sum_{j,a_{t,j} = k+1} (a_{t,j} - k) + \sum_{j,a_{t,j} > k+1} (a_{t,j} - k - a_{t,j} + k + 1)
 =  \sum_{j,a_{t,j} = k+1} 1 + \sum_{j,a_{t,j} > k+1} 1
 =  \sum_{j,a_{t,j} > k} 1.

Note that this final sum is precisely the number of Jordan blocks with eigenvalue \lambda_t whose size is greater than k, as desired.

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