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The below result is from Linear Algebra Done Right, 4th edition, Sheldon Axler:

5.11 linearly independent eigenvectors

Suppose $T \in \mathcal{L}(V)$. Then every list of eigenvectors of $T$ corresponding to distinct eigenvalues of $T$ is linearly independent.

Firstly, cool result. Secondly, I'm having trouble understanding one part of Axler's proof. He does a proof by contradiction. Since he assumes the result is false, he states:

... there exists a smallest positive integer $m$ such that there exists a linearly dependent list $v_1, \ldots, v_m$ of eigenvectors of $T$ corresponding to distinct eigenvalues $\lambda_1, \ldots, \lambda_m$ of T (note that $m \geq 2$ because an eigenvector is, by definition, nonzero).

My question is why must there specifically exist a smallest positive integer $m$? Is that referring to the note in the parenthesis, i.e., $m$ cannot be smaller than $2$? This part is integral to the proof, so I must know.

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3 Answers 3

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Every nonempty set of positive integers contains a smallest element.

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  • $\begingroup$ Is it true for every topology on $\mathbb Z$, such as the Zariski topology or a non-Hausdorff one? The only way I can think to improve the answer (if that isn't the case) is to specify that every nonempty set of positive integers contains a smallest element under the natural $<$, being generated by $1 < 2 < 3 < \cdots$. $\endgroup$
    – Snared
    Apr 25 at 2:39
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    $\begingroup$ In this context it is obvious which version of "smallest" is being referenced, just thinking about things getting taken out of context. The answer is perfectly fine.. Thank you. $\endgroup$
    – Snared
    Apr 25 at 2:41
  • $\begingroup$ Hi Mr. Axler. Thank you for answering! I understand now; you are choosing the list of eigenvectors whose length is the smallest. Thank you once again, and a fantastic textbook you have written; truly a work of art and one for the student in mind. $\endgroup$
    – Paul Ash
    Apr 25 at 2:46
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    $\begingroup$ @Snared It's worth noting that this answer has nothing to do with topology, it depends only on the ordering. And there are certainly orderings of the positive integers that allow for no smallest element - for example, the reverse of the standard ordering. $\endgroup$ Apr 25 at 21:21
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Negating the original statement yields "there exists a linearly dependent list of eigenvectors corresponding to distinct eigenvalues." There are multiple such lists: maybe one list has 5 eigenvectors, another list has 3 eigenvectors, etc. He is choosing one that is at least as short as any other list. For example, maybe all such lists have $\ge 3$ eigenvectors; then he is choosing one particular list with exactly $3$ eigenvectors.

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The proof can be done without contradiction.

Theorem. Let $T\in\mathcal{L}(V)$ and let $v_1,\dots,v_m$ be a list of eigenvectors corresponding to distinct eigenvalues $\lambda_1,\dots,\lambda_m$ of $T$. Then the list $v_1,\dots,v_m$ is linearly independent.

Proof. The statement is obvious if $m=0$. Suppose, as induction hypothesis, that the statement holds for lists of $m$ eigenvectors and consider a list of eigenvectors $v_1,\dots,v_m,v_{m+1}$, corresponding to distinct eigenvalues $\lambda_1,\dots,\lambda_m,\lambda_{m+1}$. Suppose $$ \sum_{k=1}^{m+1} c_kv_k=0 $$ for scalars $c_1,\dots,c_k,c_{k+1}$. Then we also have \begin{gather} 0=\lambda_{m+1}\sum_{k=1}^{m+1}c_kv_k=\sum_{k=1}^{m+1}\lambda_{m+1}c_kv_k \\[6px] 0=T\biggl(\,\sum_{k=1}^{m+1} c_kv_k\biggr)=\sum_{k=1}^{m+1}\lambda_kc_kv_k \end{gather} By subtracting we get $$ \sum_{k=1}^{m}(\lambda_k-\lambda_{m+1})c_kv_k=0 $$ and by the induction hypothesis we conclude that $$ (\lambda_1-\lambda_{m+1})c_1=0,\dots,(\lambda_m-\lambda_{m+1})c_m=0 $$ As the eigenvalues are distinct, we deduce that $$ c_1=0,\dots,c_m=0 $$ Hence also $c_{m+1}v_{m+1}=0$ and, since $v_{m+1}\ne0$ (it's an eigenvector), we can conclude that also $c_{m+1}=0$. QED

Comment. Axler's proof is essentially the same, but uses the minimum principle instead of induction. What style to use is a matter of taste. I'd prefer this direct proof, because it shows what's being done: you want to remove one of the eigenvectors (and the last one is the best candidate in order to get a smooth proof) and use $T$ for the purpose, together with the crucial assumption that the eigenvalues are distinct.

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  • $\begingroup$ Excellent! Thank you for providing this alternative proof. I agree that it gives more insight. $\endgroup$
    – Paul Ash
    Apr 26 at 15:32
  • $\begingroup$ @PaulAsh Would you believe I got a downvote for this? $\endgroup$
    – egreg
    Apr 26 at 16:09
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    $\begingroup$ That’s unfortunate… it should be required that for any downvote, an explanation should be given as to why it was done. $\endgroup$
    – Paul Ash
    Apr 26 at 17:29
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    $\begingroup$ The downvoter has being contradictory. $\endgroup$ Apr 26 at 19:33
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    $\begingroup$ @PaulAsh Thanks for spotting the typos. $\endgroup$
    – egreg
    Apr 27 at 8:33

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