# A question on definition of generalized eigenspaces/ characteristic polynomial

I am studying for quals and have a couple of questions on generalized eigenspaces/characteristic polynomials. Suppose $T : V \to V$ is a linear map of a finite dimensional vector space over $k = \overline{k}$. Then Theorem 8.23 of Axler states:

Theorem 8.23 (Axler): Write $\lambda_1,\ldots,\lambda_m$ for the eigenvalues of $T$. There is a decomposition $$V \cong \ker(T-\lambda_1I)^{\dim V} \oplus \ldots \oplus \ker(T -\lambda_mI)^{\dim V}.$$

Question 1: Is it true that $\ker(T-\lambda_iI)^{\dim V} = \ker(T -\lambda_iI)^{d_i},$ where $d_i$ is the multiplicity of the linear factor $t - \lambda$ in the characteristic polynomial $\chi_T(t)$? I have a feeling this is true.

Question 2: If question 1 is true, can we see it from the structure theorem of modules over a PID?

## 2 Answers

$\ker\left ((T - \lambda I)^{\dim V}\right )$ is the generalized eigenspace of $\lambda$. $\alpha_\lambda$ is the smallest exponent such that $(T - \lambda I)^{\alpha_\lambda}$ is zero on that space, so $\alpha_\lambda \le \dim V$. But the definition of $\alpha_\lambda$ says $\ker\left ((T - \lambda I)^{\dim V}\right ) \subseteq \ker\left ((T - \lambda I)^{\alpha_\lambda}\right )$, while $\alpha_\lambda \le \dim V$ implies $\ker\left ((T - \lambda I)^{\alpha_\lambda}\right ) \subseteq \ker\left ((T - \lambda I)^{\dim V}\right )$, so the two are equal, and $\ker\left ((T - \lambda I)^{\beta_\lambda}\right )$ is caught between them.

Axler's paper Down with Determinants answers your first question fairly well, I think. Lemma 3.1 shows that $\ker\left ((T - \lambda I)^{\dim V}\right )$ is the generalized eigenspace of the eigenvalue $\lambda$. Proposition 3.4 is the result quoted above. In theorem 4.1, he defines the "geometric multiplicity" (he doesn't call it that, but I've seen that terminology used elsewhere) $\alpha_\lambda$ to be the smallest value such that $(T - \lambda I)^{\alpha_\lambda} = 0$ on the eigenspace of $\lambda$. And in the proof of theorem 5.2, he notes with a brief argument, that $\alpha_\lambda \le \beta_\lambda$, the (algebraic) multiplicity of $\lambda$ that you have labelled $d$ in your question. So $\ker\left ((T - \lambda I)^{\dim V}\right ) = \ker\left ((T - \lambda I)^{\beta_\lambda}\right ) = \ker\left ((T - \lambda I)^{\alpha_\lambda}\right )$ can be seen with only a little more effort.

As for question 2, without looking up the structure theorem you refer to, I suspect it is similar to theorem 4.1 in this paper, and likely could be used in the same way.

• In general $\alpha_\lambda$ need not be equal to $\beta_\lambda$, so I don't see immediately how you deduced the equality of kernels above. – Ben Lim Sep 6 '15 at 1:09