# If $v_1,…,v_r$ are the eigenvectors that correspond to distinct eigenvalues, then they are linearly independent.

Prove:

If $v_1,...,v_r$ are the 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.

Please give an example and tell me how this theorem works!

• Hints: $T v_j = \lambda _j v_j$ . Recall definition of linearly independent. Then you should be able to prove that. – needhelp Dec 16 '14 at 3:30

Remember that $\;\{v_1,...,v_r\}\;$ are lin. dependent iff there is $\;1\le i\le r\;$ s.t. $\;v_i\;$ lind. dep. on $\;v_1,...,v_{i-1}\;$, so let $\;i\;$ be the first such index for which this happens:

$$(1)\;\;v_i=\sum_{k=1}^{i-1}a_kv_k\implies\lambda_iv_i=\sum_{k=1}^{i-1}a_k\lambda_iv_k$$

$$(2)\;\;\lambda_iv_i=Av_i=\sum_{k=1}^{i-1}a_kAv_k=\sum_{k=1}^{i-1}a_k\lambda_kv_k$$

Now substract right sides of (1)-(2) and...end the argument.

• +1... what I was trying to do until I got caught up in jibberjabber. – Emily Dec 16 '14 at 3:41
• @Arkamis Hehe . I supposed you misunderstood the question when I read your answer. There are several proofs similar to the above one in different parts (independence of different automorphisms in Galois Theory and etc.), so sometimes is easy to go astray. – Timbuc Dec 16 '14 at 3:43
• I certainly got caught halfway through forgetting what was a hypothesis, and what we were trying to prove. – Emily Dec 16 '14 at 3:46

Here is a slightly different proof than usual for fields $\mathbb{R}$ or $\mathbb{C}$:

Suppose we order the eigenvalues so that $|\lambda_1| > |\lambda_2| > \cdots > |\lambda_r|$.

Suppose $v=\sum_k \alpha_k v_k = 0$, where the $v_k$ correspond to the $\lambda_k$. Then $({1 \over \lambda_1} A)^m v = \sum_k \alpha_k ({\lambda_k \over \lambda_1})^m v_k = 0$. Then $\lim_m ({1 \over \lambda_1} A)^m v = \alpha_1 v_1 = 0$ shows that $\alpha_1 = 0$.

Now repeat the process with $({1 \over \lambda_2} A)^m v$, etc.

I'm not sure what you mean by an example, but you could take $A=\operatorname{diag}(1,2,...,n)$, then the eigenvectors are $e_1,...,e_n$ which are obviously linearly independent.

• First, thanks for the editing of my answer. Second, the above needs a little care in case $\;\lambda_k=0\;$ for some $\;k\;$ . Third, you're taking limits in what so far is just a vector space. For that you'd need some topology, which could probably be way too far from the subject. – Timbuc Dec 16 '14 at 3:52
• @Timbuc: (i) You're welcome. (ii) Only $\lambda_r$ can be zero, in which case the above shows that $\alpha_1 = \cdots \alpha_{r-1} = 0$ which leaves $\alpha_r v_r = 0$ which gives $\alpha_r = 0$. (iii) The weakness of the above approach is that I am assuming that the field is either $\mathbb{R}$ or $\mathbb{C}$. Given that, however, one can choose any convenient norm. – copper.hat Dec 16 '14 at 4:11
• Another weakness is that the absolute values are not necessarily distinct. – copper.hat Mar 23 '18 at 5:41

Another way to look at it,

You want to show that $c_1v_1+c_2v_2+\dots+c_rv_r=0$ implies each $c_i$ is 0. So, you know for each eigenvector $v_i$, $\lambda_iv_i = Av_i$ s.t. $(A-\lambda_iI)v_i=0$. Multiply each side of $\sum^r_{i=1}c_iv_i=0$ by $(A-\lambda_1I)(A-\lambda_2I)\cdots(A-\lambda_{i-1}I)(A-\lambda_{i+1}I)\cdots(A-\lambda_rI)$ which will show $c_i$ is 0. Repeat the process for each $c_i$.