Let $e_1,e_2,e_3$ be three eigenvalues and let $x_1,x_2,x_3$ be three eigenvectors corresponding to them.

Suppose $x_1,x_2,x_3$ are linearly independent, then

$\alpha_1x_1 + \alpha_2x_2 + \alpha_3x_3 = 0 \Rightarrow \alpha_1=\alpha_2=\alpha_3=0$

So we have

$\alpha_1x_1 + \alpha_2x_2 + \alpha_3x_3 = 0 $ --- (1)

Multiplying both sides by matrix $A$, we get

$\alpha_1(Ax_1) + \alpha_2(Ax_2) + \alpha_3(Ax_3) = 0 $

$\Rightarrow \alpha_1(e_1x_1) + \alpha_2(e_2x_2) + \alpha_3(e_3x_3) = 0 $ --- (2)

Multiplying equation (1) by $e_1$, we get

$\alpha_1(e_1x_1) + \alpha_2(e_1x_2) + \alpha_3(e_1x_3) = 0 $ --- (3)

Now subtracting equation (3) from (2), we get

$\alpha_2(e_2-e_1)x_2 + \alpha_3(e_3-e_1)x_3 = 0$

$\Rightarrow \alpha_2\beta_{21}x_2 + \alpha_3\beta_{31}x_3 = 0$ --- (4)

where $\beta_{21}=(e_2-e_1), \beta_{31}=(e_3-e_1)$

So we have eliminated $x_1$ in equation (4).

We can repeat same procedure i.e. multiplying equation (4) by $A$ to get

$\alpha_2\beta_{21}(Ax_2) + \alpha_3\beta_{31}(Ax_3) = 0$

$\Rightarrow \alpha_2\beta_{21}(e_2x_2) + \alpha_3\beta_{31}(e_3x_3) = 0$ -- (5)

and multiplying equation (4) by $e_2$

$\alpha_2\beta_{21}(e_2x_2) + \alpha_3\beta_{31}(e_2x_3) = 0$ -- (6)

and subtracting equation (6) from (5) to eliminate $x_2$,

$\alpha_3\beta_{31}(e_3-e_2)x_3 = 0$

Now Since all eigenvalues are distinct, $\beta_{31}, (e_3-e_2) \neq 0$, so $\alpha_3 = 0$.

Similarly, we can show that $\alpha_1,\alpha_2=0$, hence $x_1,x_2,x_3$ are linearly independent.

In fact, we can show this for any dimension, thus having $n$ distinct eigenvalues implies $n$ eigenvectors are linearly independent.

Note : We can't have more than 3 linearly independent eigenvectors because $4^{th}$ eigenvector must correspond to any one of eigenvalue $e_1,e_2,e_3$, and thus it becomes linearly dependent on any one of $x_1,x_2,x_3$.

.