# Basis of generalized eigenspace as a disjoint union of cycles

I am studying the Jordan canonical form from the book Linear Algebra by Friedberg, Insel, and Spence (4th Edition). I have a question regarding the following theorem:

Theorem 7.7. Let $$T$$ be a linear operator on a finite-dimensional vector space $$V$$, and let $$\lambda$$ be an eigenvalue of $$T$$. Then the generalized eigenspace $$K_{\lambda}$$ has an ordered basis consisting of a union of disjoint cycles of generalized eigenvectors corresponding to $$\lambda$$.

I provide part of the proof for context.

Proof: The proof is by induction on $$n = \text{dim}(K_{\lambda})$$. The result is clear for $$n = 1$$. Suppose the result is true whenver $$\text{dim}(K_{\lambda}) < n$$ and suppose $$\text{dim}(K_{\lambda}) = n$$. Let $$U = (T-\lambda I) \upharpoonright_{K_{\lambda}}$$. Then $$R(U)$$ is a subspace of $$K_{\lambda}$$ of lesser dimension. Now we consider $$T\upharpoonright_{R(U)}$$. Then $$R(U)$$ is the generalized eigenspace corresponding to $$\lambda$$ of $$T\upharpoonright_{R(U)}$$ and therefore, we may use the inductive hypothesis.

Question: In order to utilize the inductive hypothesis one needs to show that $$\lambda$$ is an eigenvalue of $$T\upharpoonright_{R(U)}$$ (as stated in the Theorem) i.e. $$\exists x\in R(U)$$ such that $$(T-\lambda I)x = 0$$ or in other words $$\exists y\in K_{\lambda}$$ such that $$(T-\lambda I)^2y = 0$$. How does one show the existence of such a $$y$$?

Maybe this is the confusing bit, but there is just the trivial case where $$R(U) = 0$$.

Otherwise, if $$R(U) \neq 0$$ then for any nonzero $$x \in R(U)$$ there exists $$y \in K_\lambda$$ of rank $$m > 1$$ such that $$x = U(y)$$. So, $$x$$ is a generalized eigenvector of $$T \restriction R(U)$$ of rank $$m-1$$. Since $$R(U)$$ definitely contains the generalized eigenspace of $$T \restriction R(U)$$, we can conclude $$R(U)$$ is the generalized eigenspace corresponding to $$\lambda$$ of $$T \restriction R(U)$$.

Maybe another thing to note is that $$T \restriction R(U)$$ is a legitimate operator on $$R(U)$$ (aka $$R(U)$$ is an invariant subspace of $$T$$) since $$T$$ and $$T - \lambda I$$ commute.

• $R(U)$ being a $T-$ invariant subspace does not guarantee that $\lambda$ is a root of the characteristic polynomial of $T\upharpoonright_{R(U)}$, does it? – Karthik Kannan Oct 31 at 15:44
• When we show $R(U) \neq 0$ is the generalized eigenspace of $T \restriction R(U)$ for $\lambda$ this shows $\lambda$ is an eigenvalue for $T \restriction R(U)$ (just look at a generalized eigenvector of rank $1$), so a root of the characteristic polynomial. What $R(U)$ being a $T$-invariant subspace says is that $T \restriction R(U)$ is actually an operator on $R(U)$, which is important for the induction since we start with an operator $T$ on some space $V$. – AGF Oct 31 at 18:36
• I understand your point, thank you very much. – Karthik Kannan Oct 31 at 19:05

In the trivial case $$R(U) = \{0\}$$, we have $$(T-\lambda I)x = 0$$ for all $$x\in K_{\lambda}$$ so $$K_{\lambda} = E_{\lambda}$$, the eigenspace corresponding to $$\lambda$$. Then $$K_{\lambda}$$ has an ordered basis consisting of a union of disjoint cycles of eigenvectors (i.e. cycles of length $$1$$) and the Theorem is proved. Else, $$R(U)$$ is the non-trivial generalized eigenspace corresponding to $$\lambda$$ of $$T\upharpoonright_{R(U)}$$ and therefore $$\lambda$$ must be an eigenvalue of $$T\upharpoonright_{R(U)}$$, and we may proceed with the inductive hypothesis.

Note: I have used the fact that $$K_{\lambda}$$ is non-trivial iff $$E_{\lambda}$$ is non-trivial since if $$m\geq 2$$ is the multiplicity of $$\lambda$$ then $$K_{\lambda} = N((T-\lambda I)^{m})\neq \{0\}\iff(\det(T-\lambda I))^{m} = 0 \iff\det(T-\lambda I) = 0$$