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 $y$?


2 Answers 2


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.

  • $\begingroup$ $R(U)$ being a $T-$ invariant subspace does not guarantee that $\lambda$ is a root of the characteristic polynomial of $T\upharpoonright_{R(U)}$, does it? $\endgroup$ Oct 31, 2019 at 15:44
  • 1
    $\begingroup$ 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 $V$. $\endgroup$
    – AGF
    Oct 31, 2019 at 18:36
  • $\begingroup$ I understand your point, thank you very much. $\endgroup$ Oct 31, 2019 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 $K_{\lambda} = N((T-\lambda I)^{m})\neq \{0\}\iff(\det(T-\lambda I))^{m} = 0 \iff\det(T-\lambda I) = 0$


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