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 $y$?
Best Answer
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.