# Degree of (x-λ) in minimal polynomial

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)?

• Simply remember that each Jordan Block of size $k$ is a nilpotent matrix of degree $k$. Hence the result – b00n heT Feb 1 '14 at 21:28
• You're welcome $\ddot\smile$ – b00n heT Feb 1 '14 at 21:42

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.