How can i proof the following:

Let $\mathbb L: V\rightarrow V $ be a linear mapping. Let $v_1,v_2,..,v_n$ non-zero eigenvectors with eigenvalues $c_1,c_2,..,c_n$ respectively, also let the eigenvalues be pairwise differents, show that $v_1,v_2,..,v_n$ are linearly independents.

I tried starting to write 0 as a linear combination of $v_1,v_2,..,v_n$ and then do the mapping, and use induction to proof but i could not finish, have another way to proof?

  • $\begingroup$ @Yassin: I think it should specify that $v_1,\dots,v_n$ are all non-$0.$ Otherwise, it needn't be true even in the $n=1$ cas3. $\endgroup$ Commented Jun 5, 2016 at 14:28
  • $\begingroup$ @CameronBuie You is right, i missed it, i will do the correction. Thank you. $\endgroup$ Commented Jun 5, 2016 at 14:32
  • $\begingroup$ @CameronBuie Oh my... $\endgroup$
    – Servaes
    Commented Jun 5, 2016 at 14:32
  • 2
    $\begingroup$ @CameronBuie eigenvectors are generally considered to be non-zero by definition. $\endgroup$
    – Arnaud D.
    Commented Jun 5, 2016 at 14:38
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    $\begingroup$ @YassinRany eigenvalues and eigenvectors are such that $\mathbb{L}v=cv$. If $v=0$ this is trivially true for any $c$, which is not interesting or helpful, hence the condition that $v\neq 0$. $\endgroup$
    – Arnaud D.
    Commented Jun 5, 2016 at 14:57

2 Answers 2


Your idea sounds good. I don't know where you got stuck, but here's something that should help you for a proof by induction :

You can prove that if $v_1,\dots,v_n$ are linearly dependent, then $v_1,\dots,v_{n-1}$ are linearly dependent. Indeed, suppose $$\alpha_1v_1+\dots+\alpha_n v_n=0\tag{A}\label{A}$$ with $\alpha_i\neq 0$. Applying $\mathbb{L}$, you get that $$\alpha_1c_1v_1+\dots+\alpha_n c_nv_n=0.\tag{B}\label{B}$$ Multiply \eqref{A} by $c_n$ and take the difference with \eqref{B}; you get $$\alpha_1(c_1-c_n)v_1+\dots+\alpha_{n-1} (c_{n-1}-c_n)v_{n-1}=0\tag{C}\label{C}$$ and all $c_i-c_n\neq 0$.

  • $\begingroup$ it worked, thank you. actually. My problem was in how to show that it is true to k+1 elements after assume by induction hypothesis that is true to k elements, but with your aforementioned reasoning it worked very well :D thank you very much. $\endgroup$ Commented Jun 5, 2016 at 14:40

Suppose that $\alpha_1v_1+...+\alpha_n v_n = 0$ (suppose all $\alpha_i$ are non-zero or else we have a smaller similar system). Apply $L$ a few times and get

$$\alpha_1v_1+...+\alpha_n v_n = 0$$ $$c_1\alpha_1v_1+...+c_n\alpha_n v_n = 0$$ $$...$$ $$c_1^{n-1}\alpha_1v_1+...+c_n^{n-1}\alpha_n v_n = 0$$

Now the determinant of the system is $$ \left|\begin{matrix}\alpha_1 & ... & \alpha_n \\ c_1\alpha_1 & ... & c_n\alpha_n \\ \vdots & \ddots & \vdots\\ c_1^{n-1}\alpha_1 & ... & c_n^{n-1}\alpha_n\end{matrix} \right| = \alpha_1...\alpha_n \prod_{i<j} (c_j-c_j) $$ (there's a Vandermonde determinant hidden there)

Since this determinant is non-zero, the Gauss reduction algorithm will result in a fully triangular system with non-zero elements on the diagonal. Thus the last equation will imply $v_n = 0$ which is a contradiction. Therefore we need to have all $\alpha_i$ equal to zero (recall that if some $\alpha_i$ are zero in the beginning we consider only the system determined by the non-zero coefficients)


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