## 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.