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I found this lemma on page 120 of Micheal Artin's "Algebra" (Second Edition) and didn't understand the proof (the guy omits most of them anyway). Could anyone supply one.

Take $v$ to be a generalized eigenvector of a linear operator $T$, with eigenvalue $\lambda$ and exponent $d$. The linear combination $c_ju_j+...+c_{d-1}u_{d-1}$ for $0<j<d-1$, and $u_i=(T-\lambda I)^iv$ is generalized eigenvector with exponent $d-1$ and eigenvalue $\lambda$.

I'm fairly new to the field of algebra, so if you could try to be as simple as possible I would appreciate it.

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This is not on page 120 of my copy of Artin's "Algebra". Which chapter/section is this in? – Isaac Solomon Jan 6 '13 at 23:44
Sorry, this is in the second edition. Chapter 4, Section 7, Lemma 4.7.2. – Pax Kivimae Jan 6 '13 at 23:45
up vote 1 down vote accepted

So, we have $(T-\lambda I)^d v = 0$, per definition.

First, note that for $0 < i < d$, we get that the vector $u_i = (T - \lambda I)^i v$ is a generalized eigenvector with exponent $d-1$, since:

$$(T- \lambda I)^{d-1} u_i = (T - \lambda I)^{d-1} (T- \lambda I)^i v = (T - \lambda I)^{d + i - 1} v = (T - \lambda I)^{i-1} \underbrace{(T - \lambda I)^d v}_{=0} = 0$$

and it follows that $c_i u_i$ for any scalar $c_i$ is a generalized eigenvector with exponent $d-1$.

So, for any $0 < j < d - 1$ we get that the linear combination $c_j u_j + \dots + c_{d-1}u_{d-1}$ is a generalized eigenvector with exponent $d-1$:

$$\begin{align}&(T-\lambda I)^{d-1} (c_j u_j + \dots + c_{d-1}u_{d-1}) = \\ &= c_j (T-\lambda I)^{d-1}u_j + \dots + c_{d-1}(T-\lambda I)^{d-1} u_{d-1} = \\ &= c_j \cdot 0 + \dots + c_{d-1} \cdot 0 = 0 \end{align}.$$

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Thank you. That is very helpful. – Pax Kivimae Jan 6 '13 at 23:51
You are welcome. :) – Calle Jan 6 '13 at 23:52

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