Proof that the Eigenvectors of three distinct Eigenvalues are linearly dependent without using Induction

Question

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.

Answer 1

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 $b=0$.

Answer 2

**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 $v_1\ne 0$.