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λ is a root of p(x), the minimal polynomial of T (linear operator on complex V). Then λ is an eigenvalue of T. How to prove that the degree of (x-λ) in p(x) equals the size of the largest λ-Jordan block (i.e. a Jordan block with λ on its diagonal)?

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  • $\begingroup$ Simply remember that each Jordan Block of size $k$ is a nilpotent matrix of degree $k$. Hence the result $\endgroup$ – b00n heT Feb 1 '14 at 21:28
  • $\begingroup$ You're welcome $\ddot\smile$ $\endgroup$ – b00n heT Feb 1 '14 at 21:42
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Denote the null space of $p(T)$ as $N_p$. It's a fact that $N_p$ decomposes as the direct sum of nullspaces for its factors $$N_{(T-\lambda_1I)^{d_1}} \oplus \cdots \oplus N_{(T-\lambda_nI)^{d_n}}$$ So each $d_i$ is the smallest number $d$ for which $(T - \lambda_i I)^d$ is zero on the generalized eigenspace for $\lambda_i$.

Consider the generalized eigenspace for $\lambda_i$. Over this space, $T - \lambda_i I$ is a block matrix consisting of nilpotent matrices. The largest such nilpotent matrix is the size of the largest Jordan block. The size of this nilpotent matrix is the smallest $d_i$ for which $(T - \lambda_i I)^{d_i} = 0$. So it's also the degree $d_i$ in the minimal polynomial.

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