Here is the formal statement:

Let $\lambda_1, \lambda_2, \lambda_3$ be distinct eigenvalues of $n\times n$ matrix $A$. Let $S=\{v_1, v_2, v_3\}$, where $Av_i = \lambda_i v_i$ for $1\leq i\leq 3$. Prove $S$ is linearly independent.

Many resources online state the general proof or the proof for two eigenvectors. What is the proof for specifically 3? I tried to derive the 3 eigenvector proof from the 2 eigenvector proofs, but failed.


marked as duplicate by Strants, Kevin Long, Arnaud D., Vladhagen, Xam Mar 23 '18 at 21:17

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  • $\begingroup$ Please show us what you have tried so far. $\endgroup$ – Vishaal Sudarsan Mar 23 '18 at 5:38

Here's one idea that comes to mind, although I don't promise there isn't a slicker way to do it. Suppose $c_1v_1 + c_2v_2 + c_3v_3 = 0.$ Applying $A$ gives $$\lambda_1c_1v_1 + \lambda_2c_2v_2 + \lambda_3c_3v_3 = 0.$$ On the other hand, multiplying the original equation by $\lambda_1$ gives $$\lambda_1c_1v_1 + \lambda_1c_2v_2 + \lambda_1c_3v_3 = 0.$$ Comparing the two displayed equations, we get $$(\lambda_2-\lambda_1)c_2v_2 + (\lambda_3-\lambda_1)c_3v_3 = 0.$$ Since you say you can prove any two eigenvectors corresponding to distinct eigenvalues are linearly independent, you now know that $(\lambda_2-\lambda_1)c_2 = (\lambda_3-\lambda_1)c_3 = 0.$ But since all the $\lambda_i$ were distinct, this means $c_2 = c_3 = 0.$ Thus the original equation says $c_1v_1 = 0,$ but since eigenvectors are by definition non-zero, we see that $c_1 = 0,$ completing the proof.

Notice the inductive nature of the proof -- the same idea will work for $n$ eigenvectors.

  • $\begingroup$ Awesome! This is where I was going, but I didn't think to use the previously proved result regarding two eigenvectors being LI. I now see why the general (n vectors) proof is inductive. I find this more helpful than just reading the general proof. $\endgroup$ – user3724404 Mar 23 '18 at 15:32
  • $\begingroup$ @user3724404 This proof can also be used to show that two eigenvectors are LI, so the other proof isn't even really needed $\endgroup$ – Jared Goguen Mar 23 '18 at 17:00

Here's the $n$-eigenvector proof:

We assume

$A\vec v_i = \lambda_i \vec v_i, \; 1 \le i \le n, \tag 1$


$\lambda_i \ne \lambda_j, \; 1 \le i, j \le n; \tag 2$

assume there is a linear dependence between the eigenvectors:

$\displaystyle \sum_1^n a_i \vec v_i = 0, \; \exists [a_i \ne 0, 1 \le i \le n]; \tag 3$

since relations such as (3) are assumed to exist, there is (at least) one having a minimum number of non-zero coefficients $a_i$; we assume (3) is such; we note the number of non-zero $a_i$ must be $\ge 2$, otherwise (3) is of the form

$a_j \vec v_j = 0, \tag 4$

which implies $a_j = 0$, forbidden by hypothesis. Then

$A(\displaystyle \sum_1^n a_i \vec v_i) = 0, \tag 5$


$\displaystyle \sum_1^n a_i \lambda_i \vec v_i = 0; \tag 6$

we may assume without loss of generality that $a_1 \ne 0$; if we multiply (3) by $\lambda_1$ we have

$\displaystyle \sum_1^n a_i \lambda_1 \vec v_i = 0; \tag 7$

we subtract (7) from (6):

$\displaystyle \sum_2^n a_i (\lambda_i - \lambda_1) \vec v_i = 0; \tag 8$

since for all $i$

$\lambda_i - \lambda_1 \ne 0, \tag 9$

(8) is a linear relation between eigenvectors with fewer non-zero coefficients than (3); this contradiction shows (3) is impossible and hence the eigenvectors are linearly independent.

  • 1
    $\begingroup$ +1 for a nice answer, but aren't the differences from my own basically cosmetic? I'm just curious what you thought was missing that made you want to respond also. $\endgroup$ – RCT Mar 23 '18 at 6:32
  • $\begingroup$ @RCT: more or less. I just wanted to write up the $n$ dimensional version! And +1 to you too! Cheers! $\endgroup$ – Robert Lewis Mar 23 '18 at 6:34
  • 1
    $\begingroup$ @RobertLewis Fair enough! I thought it was apparent how the induction would proceed, but your version looks prettier in any case. $\endgroup$ – RCT Mar 23 '18 at 6:36

Suppose $v=\sum_k \alpha_k v_k = 0$.

$(A-\lambda_1 I)v = \sum_{k>1} \alpha_k (\lambda_k-\lambda_1)v_k = 0$.

$(A-\lambda_2 I)(A-\lambda_1 I)v = \sum_{k>2} \alpha_k (\lambda_k-\lambda_2) (\lambda_k-\lambda_1)v_k = 0$. $$\vdots$$ $\prod_{i=1}^{n-1} (A-\lambda_i I) v = \alpha_n \prod_{i=1}^{n-1}(\lambda_n-\lambda_i) v_n = 0$.

Hence $\alpha_n = 0$. Then the previous equation gives $\alpha_{n-1} = 0$, etc, etc.


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