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

Let $F$ be a field, $V$ and $n$-dimensional $F$-vector space, and $T$ a linear transformation on $V$. Suppose the eigenvalues of $T$ are contained in $F$, and let $\lambda$ be one such eigenvalue. Let $k$ be a natural number, and let $r_k = \mathsf{dim}_F\ (T-\lambda I)^k V$. Prove that $r_{k-1} - 2r_k + r_{k+1}$ is the number of Jordan blocks of $T$ with eigenvalue $\lambda$ and having size $k$.

We will use the notation established in this previous exercise. To wit, $\lambda_i$ are the eigenvalues of $T$, and we have $V = \bigoplus_i \bigoplus_j F[x]/(x-\lambda_i)^{a_{i,j}}$, where $a_{i,j}$ is the ‘size’ (that is, dimension over $F$) of its corresponding Jordan block. For brevity we say $V_i = \bigoplus_j F[x]/(x-\lambda_i)^{a_{i,j}}$.

We showed in that previous exercise that $\mathsf{dim}_F\ (T-\lambda_t I)^kV = \sum_{j,a_{t,j} > k} (a_{t,j} - k) + \mathsf{dim}_F\ \bigoplus_{i,i\neq t} V_i$. Thus we have the following.

 $r_{k-1} - 2r_k + r_{k+1}$ = $\sum_{j,a_{t,j} > k-1} (a_{t,j} - k + 1) - 2 \sum_{j,a_{t,j} > k} (a_{t,j} - k) + \sum_{j,a_{t,j} > k+1} (a_{t,j} - k - 1)$ = $\sum_{j,a_{t,j} > k+1} (a_{t,j} - k+1 - 2(a_{t,j}-k) + a_{t,j}-k-1) + \sum_{j,a_{t,j} = k+1} (a_{t,j} - k + 1 - 2(a_{t,j}-k) + \sum_{j,a_{t,j} = k} (a_{t,j} - k + 1)$ = $\sum_{j,a_{t,j} = k} 1$

This final sum is precisely the number of Jordan blocks with eigenvalue $\lambda_t$ and having size $k$, as desired.