1
$\begingroup$

Prove that if $\{b_n\}$ are eigenvectors with distinct eigenvalues $\{\lambda_n\}$, then $\{b_n\}$ are linearly independent.

Note: Proofs for this are available on this site and elsewhere. My question is to verify my proof below, which, at least to me, is simpler.

Lemma: If $B$ is a finite set of linear independent vectors, there is at most one linear combination of $B$ equal to a given vector $v$.

Proof: Consider a finite set $\{b_n\}$ of linearly independent vectors. Assume $\sum_n c_nb_n = v$ and $\sum_n c'_nb_n = v$, with at least one $c_n \neq c'_n$. Then $\sum_n (c_n - c'_n)b_n = v - v = 0$, which is impossible since $\{b_n\}$ is linearly independent.

Main Proof: Let $\{b_n\}$ be linearly independent eigenvectors of matrix $A$ with distinct eigenvalues $\{\lambda_n\}$. Let $d$ be a distinct eigenvector of $A$ with eigenvalue $\delta \notin \{\lambda_n\}$. We claim that $d$ is linearly independent of $\{b_n\}$.

Assume $d$ is not linearly independent of $\{b_n\}$. Then there exists $\{c_n\}$ such that $\sum_n c_nb_n = d$. Consequently, $Ad = \sum_n \lambda_n c_n b_n$. But $Ad = \delta d = \sum_n \delta c_n b_n$. This would give two different linear combinations of $\{b_n\}$ that yield $Ad$, violating the lemma. Hence, $d$ must be linearly independent of $\{b_n\}$.

Questions: Is this proof correct, rigorous, and well written? How can it be improved?

As far as I can tell, this proof does not use induction. Is that correct?


Update

Thank you the responders. I believe I need to add the following step, which indeed uses induction. Can you please verify that with this step, the proof is correct?

Let $B_0$ be the empty set, and for all $n \in \mathbb N$, let $B_n = B_{n-1} \cup \{b_n\}$. If $B_n$ is a set (possibly empty) of linearly independent eigenvectors with distinct eigenvalues, $B_{n+1}$ must also be, as proven above. $B_0$ is such a set, and so therefore $B_n$ is such for all $n$. This completes the proof.

$\endgroup$
5
  • $\begingroup$ Are you sure this isn't a proof without induction? That sure looks like an inductive step. $\endgroup$ Mar 17 at 19:37
  • $\begingroup$ @eyeballfrog - Yes, I'd say they had done the inductive step, and left out the base case, and the words "the proof is by induction". I also think that the subtle distinction between "one vector is lin indep from a set of vectors" and "a lin indep set of vectors" makes it harder to see in this case. $\endgroup$ Mar 17 at 20:05
  • $\begingroup$ Thanks! Can you please review the fix. $\endgroup$ Mar 17 at 20:12
  • $\begingroup$ You are missing at least one step: from $\sum \lambda_n c_nb_n = \sum \delta c_nb_n$, you should conclude that $\lambda_n c_n=\delta c_n$ for all $n$. From the fact that $d$ is an eigenvector, and therefore not equal to zero, you know that at least on $c_i$ is nonzero. Then $\lambda_i c_i = \delta c_i$, and now you can cancel the $c_i$ to conclude that $\lambda_i=\delta$, contradicting the assumption that $\delta$ was distinct from all $\lambda_i$. $\endgroup$ Mar 17 at 20:31
  • 1
    $\begingroup$ There is a theorem (proven by induction) that says that if $b_1,\ldots,b_n$ is a list of vectors, then it is linearly dependent if and only if there exists a $j$, $1\leq j\leq n$, such that $b_1,\ldots,b_{j-1}$ is linearly independent, and $b_j\in\mathrm{span}(b_1,\ldots,b_{j-1})$ You can then start with a list $b_1,\ldots,b_n$ of eigenvectors of pairwise distinct eigenvalues; if they are linearly independent, you are done. If not, then $n\gt 1$, and there pick $b_j$ as above and do your argument (with the added step). But this uses induction hidden in the minimality of $j$. $\endgroup$ Mar 17 at 20:34

0

You must log in to answer this question.