# Linear independence of generalized eigenvectors

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 $a_2=0$.

• i do not see what exactly are you looking for..
– user87543
Jun 26, 2014 at 8:25
• I am looking for a proof of linear independence of generalized eigenvectors without applying generalized eigenspace decomposition. Ideally the proof should us arguments of the 'level' as the proof of linear independence of eigenvectors.
– daw
Jun 26, 2014 at 8:49
• @daw Re my answer: Sorry, I was assuming uniqueness of the eigenvectors. I'll delete the answer. Jun 26, 2014 at 9:10

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 $$\sum_{i=1}^{n} a_i v_i =0.$$ 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]

• a more direct way to see this is to notice that if $v_1, v_2$ are linearly dependent, then $v_1=\mu v_2$ for some $\mu$, and thus $v_2$ is also generalized eigenvector of $\lambda_1$, which by your lemma is impossible
– glS
Jul 26, 2018 at 12:13
• @glS I think your direct argument doesn’t work if we’re considering whether v3 might be a linear combination of v1 and v2 with different eigenvalues. Jan 29, 2019 at 20:45