# Confusion about the description of generalized eigenspace

When learning about generalized eigenspace, there are two statements from two different textbooks which I was learning, seems contradict to the other one.

In Chapter 8 of the book Linear Algebra Done Right, 3rd edition by Sheldon Alxer, a theorem(Theorem 8.11 , Description of generalized eigenspaces) was stated like this

Suppose $$\mathcal{T}\in \mathcal{L}(V)$$ and $$\lambda\in\mathbb{F}$$. Then $$G(\lambda, \mathcal{T})=\text{Null}(\mathcal{T}-\lambda \mathcal{I})^{\text{dim}V}$$

Here $$G(\lambda, \mathcal{T})$$ means the generalized eigenspace of $$\mathcal{T}$$ corresponding to eigenvalue $$\lambda$$, and "Null" stands for the kernal space.

While in another textbook(Linear Algebra, Special for mathematic major by Shangzhi Li), there is also a similar theorem. Here's what it said (translated from Chinese, which can be a little bit unprecise)

Suppose an operator $$\mathcal{T}$$ defined on a n-dimensional linear space $$V$$ has $$t$$ different eigenvalues $$\lambda_1,...,\lambda_t$$, and the characteristic polynomial has the form$$P_\mathcal{T}(\lambda)=(\lambda-\lambda_1)^{n_1}...(\lambda-\lambda_t)^{n_t}.$$Then for each eigenvalule $$\lambda_i(1\leq i\leq t)$$, a subspace $$\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{n_i}$$ was formed by a zero vector and all of the generalized eigenvectors with respect to $$\lambda_i$$, which has $$n_i$$ dimensions.

Now I'm getting quite unsure about the description of generalized eigenspace.

Comparing these two theorems. The first one states that the generalized eigenspace $$G(\lambda, \mathcal{T})$$ can be described as $$\text{Null}(\mathcal{T}-\lambda\mathcal{I})^{\text{dim}V}$$, while the second one suggests that the description of generalized eigenspace with respect to $$\lambda_i$$ seems ought to have the form $$\text{Null}(\mathcal{T}-\lambda_i\mathcal{I})^{n_i}$$, where $$n_i$$ is known to be the algebraic multiplicity of $$\lambda_i$$, appeared from the characteristic polynomial. However, it is clear that $$\text{Null}(\mathcal{T}-\lambda_i\mathcal{I})^{n_i}\neq \text{Null}(\mathcal{T}-\lambda_i\mathcal{I})^{\text{dim}V}$$ for some specified eigenvalue $$\lambda_i$$, which seems like a contradiction.

More interestingly, there is also an exercise problem in today's aftercalss-assignment which gives me an even more confusing conclusion.

Suppose $$\mathcal{T}\in\mathcal{L}(V)$$ has a minimal polynomial $$D_{\mathcal{T}}(\lambda)=(\lambda-\lambda_1)^{k_1}...(\lambda-\lambda_t)^{k_t}$$ Prove that $$G(\lambda_i, \mathcal{T})=\text{Null}(\mathcal{T}-\lambda\mathcal{I})^{k_i}$$

From now on I'm getting totally dizzy... is it possible that these three statements are not all correct? Or if I missed some important things? I do think it is not possible that $$\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{\text{dim}V}=\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{n_i}=\text{Null}(\mathcal{T}-\lambda_i \mathcal{I})^{k_i}$$

Can anyone help me with that? Thanks a lot!

• "However, it is clear ..." is wrong. The sequence $\ker (T - \lambda)^k$ stabilises, i.e. there is a $k_0$ such that $\ker (T - \lambda)^k = \ker (T - \lambda)^{k+1}$ for $k \geqslant k_0$. And that $k_0$ is not larger than the algebraic multiplicity of the eigenvalue $\lambda$. – Daniel Fischer Sep 30 '20 at 11:31
• @Daniel Fischer Could you please show me why such $k_0$ exists? And why it is not larger than the algebraic multiplicity of the eigenvalue? Thanks – Scanners Sep 30 '20 at 11:55
• Let $d_k = \dim \bigl(\ker (t - \lambda)^k\bigr)$. Then $0 = d_0 \leqslant d_1 \leqslant d_2 \leqslant \ldots$, and on the other hand $d_k \leqslant \dim V$ for all $k$. A bounded monotonic sequence of integers is eventually constant. That $k_0$ is the exponent of $(X - \lambda)$ in the minimal polynomial of $T$ is more or less immediate from the definitions. If you know the Jordan normal form, you can also see it in that. – Daniel Fischer Sep 30 '20 at 12:00
• @ Daniel Fischer♦ Okay I think I've got the key point, however it is still hard for me to understand why $k_0$ is the exponent of $X-\lambda$, can you elaborate on that a little bit? P.s. I do not think Axler's book includes very much about Jordan normal form. Can you recommend some textbooks which expain that section well? – Scanners Sep 30 '20 at 12:23
• @Scanners Axler's book does say a bit about why the dimension of the generalized eigenspace is the exponent of $(x - \lambda)$ (except that he "defines things backwards" in a sense). See in particular theorem 8.10 (p. 169 second edition) – Ben Grossmann Sep 30 '20 at 12:49

As Daniel says in the comments, all of these definitions are equivalent; that is,

$$\text{ker}(T - \lambda)^{\dim V} = \text{ker}(T - \lambda)^{n_i} = \text{ker}(T - \lambda)^{k_i}.$$

This can be seen using Jordan normal form but that's overkill. Let $$m(t) = \prod (t - \lambda_i)^{k_i}$$ be the minimal polynomial of $$T$$ and let $$v \in \text{ker}(T - \lambda_j)^{\dim V}$$ for some fixed $$j$$. Then we have

$$m(T) v = \left( \prod_{i \neq j} (T - \lambda_i)^{k_i} \right) (T - \lambda_j)^{k_j} v.$$

We want to show that $$v_j = (T - \lambda_j)^{k_j} v = 0$$. The identity above gives that $$v_j$$ lies in the kernel of $$\prod_{i \neq j} (T - \lambda_i)^{k_i}$$. This is contained in (and in fact is exactly) the sum of the generalized eigenspaces of each $$\lambda_i, i \neq j$$, and a basic fact about generalized eigenspaces of different eigenvalues is that they are linearly independent. On the other hand, $$v_j$$ lies in the generalized eigenspace of $$\lambda_j$$. Hence $$v_j = 0$$ as desired.

More explicitly, we can prove the following.

Lemma: Suppose $$p, q \in \mathbb{C}[t]$$ are two polynomials such that $$p(T) v = q(T) v = 0$$. Then $$g = \gcd(p, q)$$ also satisfies $$g(T) v = 0$$.

Proof. Abstractly the point is that $$\{ p \in \mathbb{C}[t] : p(T) v = 0 \}$$ is an ideal of $$\mathbb{C}[t]$$ and hence principal. Concretely, we can apply Bezout's lemma to find polynomials $$a, b$$ such that $$ap + bq = g$$, which gives $$a(T) p(T) v + b(T) p(T) v = 0 = g(T) v$$. $$\Box$$

So $$v_j = (T - \lambda_j)^{k_j} v$$ satisfies $$(T - \lambda_j)^{\dim V - k_j} v_j = 0$$, but it also satisfies $$\left( \prod_{i \neq j} (T - \lambda_i)^k \right) v_j = 0$$, and these two polynomials have no roots in common so their $$\gcd$$ is equal to $$1$$. Hence $$v_j = 0$$.