Proof of Linear independence of generalized eigenvectors without applying generalized eigenspace decomposition.

Question

Let $V$ be a finite-dimensional complex vector space. Let $T\in \mathcal L(V)$ be an endomorphism.

A vector $v\in V \setminus\{0\}$ is called a generalized eigenvector to an eigenvalue $\lambda\in\mathbb C$ of $T$ iff there exists $k> 0$ such that $$ (\lambda I - T)^kv=0. $$ From the generalized eigenspace decomposition it follows that generalized eigenvectors to different eigenvalues are linearly independent.

---

My Question is: Is there an elementary proof for this result? Maybe along the lines of the proof for linear independence of (ordinary) eigenvectors: Let $v_1,v_2$ be eigenvectors to eigenvalues $\lambda_1\ne \lambda_2$. Then $$ a_1 v_1 + a_2v_2=0 $$ implies (1: apply $T$, 2: multiply equation by $\lambda_2$, subtract) $$ a_1 (\lambda_1-\lambda_2)v_1=0 $$ hence $a_1=0$, and $a_2=0$.

Answer 1

Let $n>0$, let $v_1\dots v_n$ be non-zero generalized eigenvectors to distinct eigenvalues $\lambda_1\dots \lambda_n$. Then the vectors $v_1\dots v_n$ are linearly independent.

First I prove this lemma:

Lemma: If $\lambda_1\ne\lambda_2$ then $\ker (\lambda_1 I - A)^{k_1}\cap \ker(\lambda_2 I - A)^{k_2}=\{0\}$ for all $k_1\ge1$ and $k_2\ge1$.

Proof: Let $k_1\ge 1$, $k_2=1$, $v\in \ker (\lambda_1 I - A)^{k_1}\cap\ker (\lambda_2 I - A)$. Then $Av = \lambda_2v$, $(\lambda_1 I - A)^{k_1}v = (\lambda_1 - \lambda_2)^{k_1}v$, implying $v=0$.

Let $k_1>1$, $k_2>1$. Let $v\in \ker (\lambda_1 I - A)^{k_1}\cap \ker(\lambda_2 I - A)^{k_2}$. Then $(\lambda_2 I - A)^{k_2-1}v\in \ker (\lambda_1 I - A)^{k_1}\cap \ker (\lambda_2 I - A)$, which implies by the above considerations $(\lambda_2 I - A)^{k_2-1}v=0$, hence $v\in \ker (\lambda_1 I - A)^{k_1}\cap \ker(\lambda_2 I - A)^{k_2-1}$. By induction it follows $v=0$. [End of Proof]

Proof of the claim above: By induction with respect to $n$. The claim is obviously true for $n=1$.

Assume that the claim holds for some $n\ge1$. Let now $n+1$ vectors etc as above be given. Let $k_1\dots k_{n+1}$ be positive numbers such that $$ (\lambda_i I - A)^{k_i}v_i=0 \quad i=1\dots n+1. $$ Let $a_1\dots a_{n+1}$ be scalars such that $$ \sum_{i=1}^{n+1} a_i v_i =0. $$ Applying $(\lambda_{n+1}I-A)^{k_{n+1}}$ to this equation yields $$ \sum_{i=1}^{n} a_i (\lambda_{n+1}I-A)^{k_{n+1}}v_i =0. $$ By the Lemma above it follows $(\lambda_{n+1}I-A)^{k_{n+1}}v_i \ne0$. Since $(\lambda_{n+1}I-A)^{k_{n+1}}v_i \in \ker(\lambda_iI-A)^{k_i}$, it follows by the induction assumption that the vectors $(\lambda_{n+1}I-A)^{k_{n+1}}v_1\dots (\lambda_{n+1}I-A)^{k_{n+1}}v_n$ are linearly independent. Hence $a_1=\dots a_n=0$. This also implies $a_{n+1}=0$. Hence the claim is proven. [End of Proof]