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I am asking this instead of perusing this answer because the accepted answer uses induction... which seldom help in understanding. At least for me, inductive proofs do a poor job in answering the "why" question behind the theorem but settles for proving that it is true...reason or not.

If $v_1,...,v_r$ are eigenvectors that correspond to distinct eigenvalues $\lambda_1, ...,\lambda_r$ of an $n \times n$ matrix $A$, then the set $\{v_1,...,v_r\}$ is linearly independent.

I attempted a proof, but it didn't work, as Carl Christian pointed out.

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    $\begingroup$ Your deletion of the eigenvalues is not permitted. Your matrix has dimension $3$, while the vector consisting of the $v_i$ has dimension $9$, so your product is not defined. Your determinant is zero when at least one of $a$, $b$ and $c$ are zero. $\endgroup$ – Carl Christian Jul 23 '19 at 21:51
  • $\begingroup$ The two systems of equations are not equivalent. You’ve basically divided the first equation by $\lambda_2-\lambda_1$ to eliminate that from the second term, but that doesn’t eliminate $\lambda3-\lambda_1$ from the third term: how do you know that they’re equal? $\endgroup$ – amd Jul 23 '19 at 22:00
  • $\begingroup$ You can get the main intuition for the case of two vectors. $\endgroup$ – Michael Jul 23 '19 at 22:00
  • $\begingroup$ There is a polynomial $P$ such that $P(\lambda_2)=\ldots=P(\lambda_r)=0$, while $P(\lambda_1)=1$. Then, if $2 \leq i \leq r$, $P(A)v_i=0$, yet $P(A)v_1=v_1$. Thus $v_1 \notin \ker\,P(A) \supset \text{span}\{v_2, \ldots,v_r\}.$ $\endgroup$ – Mindlack Jul 23 '19 at 22:03
  • $\begingroup$ While the easiest way is indeed by induction, you can start out with the equation $a_1x_1+a_2x_2+a_3x_3=0$ (where $x_1,x_2,x_3$ are eigenvectors of $A$ with distinct eigenvalues $\lambda_1\neq 0, \lambda_2\neq 0, \lambda_3$). Then obtain two more equations by multiplying by $A$ and $A^2$. $\endgroup$ – Michael Jul 23 '19 at 22:06
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That proof is wrong because you cannot pass from $0+b(\lambda_2-\lambda_1)v_2 + c(\lambda_3-\lambda_1)v_3=0$ to $0+bv_2 + cv_3=0$. You could if $\lambda_2-\lambda_1=\lambda_3-\lambda_1$, but you are not assuming that.

If $av_1+bv_2+cv_3=0$, then, as you know,\begin{align}0&=\lambda_1(av_1+bv_2+cv_3)-A.(av_1+bv_2+cv_3)\\&=b(\lambda_1-\lambda_2)v_2+c(\lambda_1-\lambda_3)v_3.\end{align}But then\begin{align}0&=\lambda_2\bigl(b(\lambda_1-\lambda_2)v_2+c(\lambda_1-\lambda_3)v_3\bigr)-A.\bigl(b(\lambda_1-\lambda_2)v_2+c(\lambda_1-\lambda_3)v_3\bigr)\\&=c(\lambda_2-\lambda_3)(\lambda_1-\lambda_3)v_3.\end{align}Therefore $c=0$. So, $av_1+bv_2+cv_3$ is simply $av_1+bv_2$ and you can start all over again in order to prove that $b=0$.

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  • $\begingroup$ Your answer was so clear I had no problem understanding the hint and developing it into a complete answer. gofile.io/?c=L3Cwlh Thank you! $\endgroup$ – kakashi10192020 Jul 24 '19 at 0:16
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Jul 24 '19 at 6:04
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**hint to finish*

Assume $$av_1+bv_2=0$$ then

$$f(av_1+bv_2)=a\lambda_1v_1+b\lambda_2v_2=0$$ and

$$a(\lambda_1-\lambda_2)v_1=0$$

thus $a=0$ since $\lambda_1\ne \lambda_2$ and $v_1\ne 0$.

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