direct sum of generalized eigenspaces

Question

I'm studying linear algebra and learning about generalized eigenspaces, and i have 3 questions regarding a specific proof which i think i have to write down before i can ask the questions (I'm translating it to english, but it should hopefully be clear anyway).

First one piece of terminology: $EG_{\lambda}$ is defined as the generalized eigenspace corresponding to the eigenvalue $\lambda$, i.e.:$\bigcup\limits_{i=1}^{\infty} Ker(T-\lambda I)^i.$

Proposition: Let T be an operator on V with eigenvalues $\lambda_1,\lambda_2,....,\lambda_r$, then: $$V=GE_{\lambda_1}\oplus GE_{\lambda_2}\oplus....\oplus GE_{\lambda_r}.$$

Proof: It's a proof by induction over the dimension of $V$, with $n$ being the dimension.

Base case: If $n=1$, then it's obviously true.

Inductive argument: Let $n\geq 2$ and assume the proposition is true for all vectorspaces with dimension $\lt n$. T has an eigenvalue $\lambda_1$. Let $V_1=Ker(T-\lambda_1I)^n$ and $V_2=Ran(T-\lambda_1 I)^n$. We know that $V=V_1\oplus V_2$. If $\lambda_1$ is the only eigenvalue then $V_2=\{0\}$ and $V=GE_{\lambda_1}$ and we are done. if $\lambda_1$ isn't the only eigenvalue then, since $V_2$ is invariant, we can restrict T to $V_2$ with the eigenvalues $\lambda_2,\lambda_3,....,\lambda_r$. According to the inductive assumption, $V_2=GE_{\lambda_2}\oplus GE_{\lambda_3}\oplus....\oplus GE_{\lambda_r}$, and we get: $V=GE_{\lambda_1}\oplus V_2=GE_{\lambda_1}\oplus GE_{\lambda_2}\oplus....\oplus GE_{\lambda_r}$. And we are done!

First of all, it doesn't say what field V is over, but i think it's over a complex field (the course has been handling complex and real fields only), since we assume T has at least one eigenvalue, and that is only true if it's over a complex field (as far as i know).

Question 1: The statement: "If $\lambda_1$ is the only eigenvalue then $V_2=\{0\}$", is not clear to me. I've tried to prove it to my self and googling, but i can't figure it out. It is probably something trivial since it is not motivated in the proof, but i can't figure it out. Why would the fact that there is only one eigenvalue make this true is beyond me.

Question 2: In this statement: "we can restrict T to $V_2$ with the eigenvalues $\lambda_2,\lambda_3,....,\lambda_r$". Why does T have those eigenvalues when restricted to $V_2$?. It seems intuitively to be true, but how can we be sure that the eigenvectors corresponding to $\lambda_2,\lambda_3,....,\lambda_r$ isn't in $GE_{\lambda_1}$. For example, maybe there is some eigenvector v, corresponding to say $\lambda_4$, and some $k$ such that: $(T-\lambda_1 I)^kv=0$.

Question 3: How does the inductive argument work, i mean it's structure. In every case before where i've dealt with inductive proofs, it's been of the sort: "Prove a base case (usually $n=0$), and then assume it's true for $k=n-1$ and show that $P(k) \implies P(n)$". But in this proof it is stated explicitly "Assume it's true for all $k\lt n$" and prove that it's true for $n$. We don't know if $k=n-1$ or if it's $k=n-100$ (Unless we know that $dim(GE_{\lambda_1})=1$, then how do we know that?). I can do an almost identical proof (given that i know the answer to question 1 and 2) of the proposition if we let $n$ be the number of eigenvalues, and then it would be a "normal" proof by induction. But how does this version work?

edit: $V_2=Ran(T-\lambda_1 I)^n$, not $V_2=Ran(T-\lambda_2 I)^n$

Answer 1

Q0: Yes, we need to work over a field containing all of the eigenvalues of $T$. (Note that this is more general than saying "algebraically closed field," and gets you the same decomposition over $\mathbb{R}$ if $T$ has all real eigenvalues.)

Q1: If $\lambda_1$ is the only eigenvalue then $T - \lambda_1 I$ has only the zero eigenvalue, so is nilpotent (e.g. by the Cayley-Hamilton theorem), so $(T - \lambda_1 I)^n = 0$. In other words, in this case the generalized eigenspace of $\lambda_1$ is all of $V$.

Q2: If $v \in \text{Ran}(T - \lambda_1 I)^n)$ were to have an eigenvalue of $\lambda_1$ then we would have $(T - \lambda_1 I) v = 0$. But by definition we have $v = (T - \lambda_1 I)^n w$ for some $w$, so this gives $(T - \lambda_1 I)^{n+1} w = 0$, which gives $(T - \lambda_1 I)^n w = 0$ because the generalized eigenspace stabilizes at $n$ (e.g. by the Cayley-Hamilton theorem again). So $v = 0$.

Q3: It's a proof by strong induction, which is equivalent to ordinary induction. You prove the base case $P(1)$ and then the inductive step is to show that if $P(k)$ is true for all $k < n$ then $P(n)$ is true.

Edit: Without Cayley-Hamilton we can prove directly the necessary result, which is the following.

Proposition: Let $T : V \to V$ be a linear map on an $n$-dimensional vector space. Then for all $N \ge n$ we have $\text{im}(T^N) = \text{im}(T^n)$ and $\text{ker}(T^N) = \text{ker}(T^n)$.

Proof. The sequence of subspaces $I_k = \text{im}(T^k)$ are decreasing in the sense that $I_0 \supseteq I_1 \supseteq I_2 \dots$, and in particular their dimensions are decreasing. Since $\dim I_0 \le n$ it follows that for some $k \le n$ we must have $\dim I_k = \dim I_{k+1}$, which gives $I_k = I_{k+1}$. This means that $T(\text{im}(T^k)) = \text{im}(T^k)$, hence $T$ is invertible when restricted to $\text{im}(T^k)$. From that point on we must have $I_k = I_{k+1} = I_{k+2} = \dots$ so the sequence of subspaces stabilizes, and $k \le n$ so in particular $I_n = I_{n+1} = \dots $.

Similarly for the kernels define the sequence of subspaces $J_k = \text{ker}(T^k)$ which is increasing in the sense that $J_0 \subseteq J_1 \subseteq \dots$, and in particular their dimensions are increasing. Since $\dim J_k \le n$ for all $n$ it follows that for some $k \le n$ we must have $\dim J_k = \dim J_{k+1}$, which gives $J_k = J_{k+1}$. This means that $T^{-1}(\text{ker}(T^k)) = \text{ker}(T^k)$, meaning $T$ is invertible when restricted to $\text{ker}(T^k)$. From that point on we must have $J_k = J_{k+1} = J_{k+2} = \dots$ as above. $\Box$

Corollary: If $T : V \to V$ has no nonzero eigenvalues then $T^n = 0$ (and in particular $T$ is nilpotent).

Proof. If $T$ has no nonzero eigenvalues then it cannot be invertible when restricted to any proper subspace of $V$ (otherwise we would be able to produce an eigenvector with nonzero eigenvalue in that subspace), so when running the argument above for the subspaces $I_k = \text{im}(T^k)$ we get that once they stabilize they must be equal to zero, and in particular $I_n = 0$. $\Box$

Corollary: The generalized eigenspace $E_{\lambda} = \bigcup_{k \ge 1} \text{ker}((T - \lambda I)^k)$ is equal to $\text{ker}(T - \lambda I)^n$.

We can actually do better than this: it turns out that $E_{\lambda} = \text{ker}(T - \lambda)^m$ where $m$ is the multiplicity of $\lambda$ as a root of the minimal (not characteristic) polynomial of $T$, and this bound is sharp: $E_{\lambda} \neq \text{ker}(T - \lambda)^{m-1}$. See, for example, this answer.

Answer 2

We might be over-complicating the matter.

If $Ran(T - \lambda I)^n = \{ 0 \}$, then $(T - \lambda I)^n = 0$ (i.e. the zero matrix), and $\forall v\in V: (T - \lambda I)^nv = 0$.

The equivalency of all three statements about $(T - \lambda I)^n$ is truthful at the level of definition.

Tell me if you think the following is a short and snappy rewrite of the proposed proof, that makes its rigor indisputably clear:

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Claim: The generalized eigenspaces of $T\in\text{Hom}(V,V)$ span $V$.

Proof:

Given $T\in \text{Hom}(V,V)$, the fundamental theorem of algebra guarantees at least one eigenvalue $\lambda\in \mathbb C$. Choose $n$ sufficiently high that Ran$(T-\lambda I)^n = \text{Ran} (T- \lambda I)^{n+1}$.

Case 1: Ran$(T-\lambda I)^n = W \neq \{0\}$

• Then $(T-\lambda I)$ is invertible on $W$
• By inductive hypothesis (induction on dim$V$), the generalized eigenspaces of $(T - \lambda I)$, (corresponding to some set of eigenvalues, $\{\lambda_k\}$) span all of $W$.
• The OP already proved that $V = \text{Ran}(T-\lambda I)^n \oplus \text{Ker}(T-\lambda I)^n$. We are done (NOTE: the generalized $\lambda_k$-eigenspace of $(T-\lambda I)$, is the generalized $(\lambda_k+\lambda)$-eigenspace of $T$).

Case 2: Ran$(T-\lambda I)^n = \{0\}$

• This preceding statement is equivalent to $\forall v: (T-\lambda I)^nv = 0$ (definition of the Range of an operator).
• Then the $\lambda$ generalised eigenspace (i.e. ker$(T- \lambda I)^n$) is simply $V$ itself. We are done.