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.

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