Usually the independence of eigenvalues are shown for matrices, I did a proof without considering matrices.

Let $T:V\to V$ be a linear operator and let $X=${$v_1,v_2,...,v_m$} be eigenvectors each corresponding to a different eigenvalue then $X$ is linearly independent.

Proof: Suppose this is wrong and $X$ is linearly dependent. Then there are atleast two of which (if there were only one, that would make an eigenvector to be $0$) are nonzero $a_i \in F$ such that: $$ a_1v_1+a_2v_2+⋯+a_mv_m=0 \tag{*} $$

Let us pick only the vectors whose coefficients are nonzero and write:

$$α_1 u_1+α_2 u_2+⋯+α_k u_k=0$$ where for each $i$, $u_i=v_j$ for some $j$. (We don't pick the same vector twice here). Now all $\alpha$'s are different than $0$ in this context. Now let us multiply each side of this equation with $a_1^{-1}$ (since $a_1\neq 0$, $a_1^{-1}\in F$) to get:

$$ u_1+α_1^{-1}α_2u_2+⋯+α_1^{-1}α_ku_k=0 \tag{**} $$

Since each $u_i$ is an eigenvector we may write $Tu_i=\lambda_iu_i$ and we apply $T$ to the both sides of this equation to get:

$$\lambda_1u_1+\lambda_2α_1^{-1}α_2u_2+⋯+\lambda_kα_1^{-1}α_ku_k=0$$ and since $\lambda_1 \neq 0$ we can multiply each side with $\lambda_1^{-1}$ to get:

$$u_1+λ_1^{-1} λ_2 α_1^{-1} α_2 u_2+⋯+λ_1^{-1} λ_k.α^{-1} α_k u_k=0$$

From this equation and the $(**)$ equation we get:

$$ α_1^{-1} α_2 u_2+⋯+α_1^{-1} α_k u_k=λ_1^{-1} λ_2 α_1^{-1} α_2 u_2+⋯+λ_1^{-1} λ_k.α_1^{-1} α_k u_k$$

and this equation can be multiplied by $\alpha_1$ to get:

$$ α_2 u_2+⋯+α_k u_k=λ_1^{-1} λ_2 α_2 u_2+⋯+λ_1^{-1} λ_k.α_1^{-1} α_k u_k $$ which is $$ (1-λ_1^{-1} λ_2 ) α_2 u_2+⋯+(1-λ_1^{-1} λ_k ) α_k u_k=0 $$ since all the $a_i$ are nonzero, none of these coefficients can be $0$, had it been the case we would have for some $i\neq 1$,$1-λ_1^{-1} λ_i=0$ which is $λ_1=λ_i$ and this can't be since we picked $\lambda$'s to be distinct eigenvalues. Now if we rename each coefficient as $(1-λ_1^{-1} λ_2 ) α_2=β_2$, $(1-λ_1^{-1} λ_3 ) α_3=β_3$, $\ldots$, $(1-λ_1^{-1} λ_k ) α_k=β_k$ we get:

$$ β_2 u_2+β_3 u_3+⋯+β_k u_k=0 $$ where all $\beta_i$ are nonzero. This is precisely the same case within the equation $(*)$ (except we have $k-1$ vectors now) so we may apply the same procedure to this equation to get:

$$ (1-λ_2^{-1} λ_3 ) β_3 u_3+⋯+(1-λ_2^{-1} λ_k ) β_k u_k=0 $$ and for the same reasoning each coefficient is nonzero and we rename coefficients once more as $$γ_3=(1-λ_2^{-1} λ_3 ) β_3$$ and so on to get another equation but this time without $u_1$ and $u_2$. Since we have finite vectors, we keep doing this process, after $(k-1)$th iteration we get the equation:

$$ (1-λ_{k-1}^{-1} λ_k ) A_k u_k=0 $$

and since $A_k \neq 0$ (due to the process) and $u_k \neq 0$ we must have $$ 1-λ_{k-1}^{-1} λ_k=0 $$

which leads to the contradiction $λ_{k-1}=λ_{k}$.

What do you think about this proof? Is it valid?

  • $\begingroup$ Hi, Ata, could you make your question and your proof apart? We'd better give question clearly only by itself, and your proof could be an answer which the community will give comments on it. $\endgroup$
    – Mountain
    Jun 8 at 17:48
  • $\begingroup$ @Mountain The question is "is my proof valid". This is not an invitation for people to give their own proofs. $\endgroup$ Jun 9 at 7:04

3 Answers 3


It looks good to me, but I think the usual proof is simpler: if some linear combination $\sum_{i=1}^mc_iv_i$ is zero, then for each $k\in\{1,2,\ldots,m\}$, we have $$ 0 =\prod_{j\ne k}(T-\lambda_j\operatorname{id})\left(\sum_{i=1}^mc_iv_i\right) =\sum_{i=1}^mc_i\prod_{j\ne k}(T-\lambda_j\operatorname{id})v_i =c_k\prod_{j\ne k}(\lambda_k-\lambda_j)v_k. $$ Therefore $c_k=0$.


I don't see any errors in your proof, and this is basically how I would structure the proof. I would suggest two edits to make it shorter and more readable.

(1) You use phrases like "apply the same procedure", "for the same reasoning", "we keep doing this process". These are all clues that you are writing a proof by induction. If you explicitly write it as an inductive proof, it will be easier for an experienced reader to check.

(2) The way you wrote the proof, the case $\lambda_1 = 0$ needs to be handled separately, but if you are a little careful you can dodge this. Specifically, you multiply one equation by $\lambda_1^{-1}$ before subtracting it from another, but you could multiply the second equation by $\lambda_1$ instead. It is often possible to avoid making $0$ a special case by being careful about this.

Here is how I'd write it. See this answer for a very similar presentation.

Theorem: Let $T: V \to V$ be a linear operator, and let $v_1$, $v_2$, ..., $v_m$ be nonzero eigenvectors corresponding to distinct eigenvalues. Then $v_1$, $v_2$, ..., $v_k$ are linearly independent.

Proof: We induct on $k$; the base case $k=1$ holds because $v_1 \neq 0$.

So, suppose the result holds for $k-1$ and suppose, for the sake of contradiction, that $$c_1 v_1 + c_2 v_2 + \cdots + c_k v_k = 0 \qquad (\ast)$$ with the $c_i$ not all zero. Multiplying by $T$, we have $$c_1 \lambda_1 v_1 + c_2 \lambda_2 v_2 + \cdots + c_k \lambda_k v_k = 0$$ and multiplying by $\lambda_1$ we have $$c_1 \lambda_1 v_1 + c_2 \lambda_1 v_2 + \cdots + c_k \lambda_1 v_k = 0.$$ So, subtracting these, $$c_2(\lambda_2 - \lambda_1) v_2 + \cdots + c_k(\lambda_k-\lambda_1) v_k = 0.$$

By induction, $v_2$, ..., $v_k$ are linearly independent, so we deduce that $c_2(\lambda_1-\lambda_2) = \cdots = c_k (\lambda_1 - \lambda_k) = 0$. Since $\lambda_1 \neq \lambda_2$, ..., $\lambda_k$, this means that $c_2 = \cdots = c_k=0$. But then $(\ast)$ means that $c_1 v_1=0$, and we have assumed $v_1 \neq 0$, so $c_1 = \cdots = c_k=0$ and this isn't a true linear dependence. $\square$

Both my notes are small bits of polishing though. If the question is just "is my original proof right", then my answer is "yes".

  • $\begingroup$ Thank you for your advices, I will most certainly be following them in the future. I was quite lucky here that those coefficients were eigenvalues so that they weren't 0. $\endgroup$
    – Ata
    Jun 8 at 19:08
  • 1
    $\begingroup$ @Ata Eigenvectors can't be 0. Eigenvalues can. $\endgroup$ Jun 9 at 7:07
  • $\begingroup$ Two notes: (1) @FedericoPoloni I have seen books differ on this, but I think the most usual conventions are that a "$\lambda$-eigenvector" is any vector $\vec{v}$, including $\vec{0}$, such that $T \vec{v} = \lambda \vec{v}$, but an "eigenvalue" is a scalar $\lambda$ for which there is a nonzero $\lambda$-eigenvector. That's why I wrote the phrase "nonzero eigenvector" several times in my proof. $\endgroup$ Jun 9 at 13:33
  • $\begingroup$ (2) It does look like the "since $\lambda_1 \neq 0$" step in the original proof is unjustified. I thought the argument was "there are at least two eigenvalues, so reorder so that the one we call $\lambda_1$ isn't $0$." But I don't actually see that in the text. $\endgroup$ Jun 9 at 13:33
  • $\begingroup$ Re, I think that that is not the most usual convention. Rather, I (and Wikipedia) think that usually an eigenvector is non-$\vec0$ by definition, whereas an eigenspace includes $\vec0$. But of course that doesn't say what convention any particular person is used to—and referring to non-$\vec0$ eigenvectors is at worst a redundancy; according to no-one's definition is it incorrect. $\endgroup$
    – LSpice
    Jun 9 at 16:32

I think the below argument is cleaner.

Let $\lambda_1, \lambda_2$ be distinct non-zero eigenvalues with corresponding eigenvectors $v_1,v_2$ (it is immediately clear a zero eigenvector must be independent). By definition, $Tv_1=\lambda_1v_1$ and $Tv_2=\lambda_2v_2$.

Assume the eigenvectors are linearly dependent, so $a_1v_1+a_2v_2=0$ for some non-zero $a_i$. Then $T(a_1v_1+a_2v_2)=T(0) \implies a_1T(v_1)+a_2T(v_2)=0 \implies a_1\lambda_1v_1+a_2\lambda_2v_2=0$. By our assumption, this can be re-written $\lambda_1(a_1v_1)-\lambda_2(a_1v_1)=(\lambda_1-\lambda_2)(a_1)v_1=0$. But this is impossible since $v_1 \neq 0 \in V$, and since both $a_1 \neq0$ and $\lambda_1 \neq \lambda_2$.

  • 5
    $\begingroup$ This is correct but this applies only if there are two vectors in the given set. $\endgroup$
    – Ata
    Jun 8 at 18:08

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