I am reading up on some functional analysis to prepare for an upcoming subject. I came across the following theorem and proof.

Let $X$ be a linear space and $T:X\to X$ be a linear operator. Then the eigenvectors corresponding to different eigenvalues are linearly independent. Then the proof proceeds as:

Assume that $x_1,\ldots,x_n$ are eigenvectors corresponding to $\lambda_1,\ldots,\lambda_n$ distinct eigenvalues. Then suppose that $x_m$ is the first eigenvector such that $$x_m = \mu_1x_1+\ldots +\mu_{m-1}x_{m-1}.$$ Then a contradiction follows and I have no problems with this part of the proof. My problem/confusion is with the first part. What if we have a subset of eigenvectors that are not linearly independent that has cardinality greater than $\mathbb R$. Then how are we supposed to order those and talk about a 'first' eigenvector.

Maybe I am missing something trivial, I feel like my understanding of these concepts is shaky which does not help. Maybe someone can explain what I am missing or confirm my confusion in that this is indeed a problem. Perhaps there is an alternative proof that deals with this situation. Thanks!


1 Answer 1


A family of vectors is linearly dependent iff there exists a nontrivial linear combination of them equal to zero. Note that a linear combination is an expression of the from $$ \sum_{i\in I}\mu_i x_i$$ such that all but finitely many of the $\mu_i$ are zero!

So to recap, let $I$ be an index set, for each $i$ let $x_i\in X$ be a (nonzero!) eigenvector of eigenvalue $\lambda_i$ and assume that $i\ne j$ implies $\lambda_i\ne \lambda_j$. If we asssume that the family of vectors $\{x_i\}_{i\in I}$ is linearly dependent, then there exists a function $c\colon I\to \mathbb R$, $i\mapsto c_i$ such that $\operatorname{supp}(c):=\{\,i\in I\mid c_i\ne 0\,\}$ is finite and nonempty and $\sum_{i\in I}c_ix_i=0$. Since we assume that $\operatorname{supp}(c)$ is finite, there exists among all such $c$ one that minimizes the natural number $|\operatorname{supp}(c)|$. Clearly, we have $|\operatorname{supp}(c)|\ge 2$ as all $x_i$ are nonzero. Then for such a minimal $c$, one picks $i_0\in I$ and notes that also $T(0)-\lambda_{i_0}\cdot 0=\sum_{i\in I}\underbrace{(\lambda_i-\lambda_{i_0}) c_i}_{d_i}x_i=0$, and that $\operatorname{supp}(d)=\operatorname{supp}(c)\setminus\{i_0\}$. If $i_0\in\operatorname{supp}(c)$, this contradicts the minimality of $c$. Hence $\{x_i\}_{i\in I}$ is linearly independent.

  • $\begingroup$ Thanks for your answer, I get what you are saying but I am afraid I am not sure how it helps. The contradiction in the proof comes from the fact that $x_m$ is the first eigenvector that can be written as a linear combination of the first $m-1$ eigenvectors. But how can we talk about such ordering if we have an index set of cardinality greater than $\mathbb R$? In other words, how can we possibly say this is the first eigenvector such that... $\endgroup$
    – Slugger
    Apr 13, 2015 at 17:12
  • $\begingroup$ @Slugger the point that Hagen is making is that we can always reduce the indices under consideration to a finite set by throwing out the $x_i$ with zero coefficients. $\endgroup$ Apr 13, 2015 at 17:20
  • $\begingroup$ @Slugger Good point. No order is needed as can be seen from my additional remarks. $\endgroup$ Apr 13, 2015 at 17:21
  • $\begingroup$ Ah perfect I see now! Thanks for clearing up the confusion. I did not realize that linear combinations should satisfy the finite support property you describe. $\endgroup$
    – Slugger
    Apr 13, 2015 at 17:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .