How to prove that eigenvectors from different eigenvalues are linearly independent How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?

Consider the matrix $A$ with two distinct  eigen values $\lambda_1$ and $\lambda_2$. First note that the eigen vectors cannot be same , i.e., If $Ax = \lambda_1x$ and $Ax = \lambda_2x$ for some non-zero vector $x$. Well , then it means $(\lambda_1- \lambda_2)x=\bf{0}$. Since $\lambda_1$ and $\lambda_2$ are scalars , this can only happen iff $x = \bf{0}$ (which is trivial) or when $\lambda_1 =\lambda_2$ .
Thus now we can safely assume that given two eigenvalue , eigenvector pair say $(\lambda_1, {\bf x_1})$ and $(\lambda_2 , {\bf x_2})$ there cannot exist another  pair $(\lambda_3 , {\bf x_3})$ such that ${\bf x_3} = k{\bf x_1}$ or ${\bf x_3} = k{\bf x_2}$ for any scalar $k$. Now let ${\bf x_3} = k_1{\bf x_1}+k_2{\bf x_2}$ for some scalars $k_1,k_2$
Now,
$$ A{\bf x_3}=\lambda_3{\bf x_3} \\ $$
$$ {\bf x_3} = k_1{\bf x_1} + k_2{\bf x_2} \:\:\: \dots(1)\\  $$
$$ \Rightarrow A{\bf x_3}=\lambda_3k_1{\bf x_1} + \lambda_3k_2{\bf x_2}\\$$
but, $\:\: {\bf x_1}=\frac{1}{\lambda_1}Ax_1$ and ${\bf x_2}=\frac{1}{\lambda_2}Ax_2$. Substituting in above equation we get,
$$A{\bf x_3}=\frac{\lambda_3k_1}{\lambda_1}A{\bf x_1} + \frac{\lambda_3k_2}{\lambda_2}A{\bf x_2} \\$$
$$\Rightarrow {\bf x_3}=\frac{\lambda_3k_1}{\lambda_1}{\bf x_1} + \frac{\lambda_3k_2}{\lambda_2}{\bf x_2} \:\:\: \dots (2)$$
From equation $(1)$ and $(2)$ if we compare coefficients we get $\lambda_3 = \lambda_1$ and  $\lambda_3 = \lambda_2$ , which implies  $\lambda_1 = \lambda_2 = \lambda_3$ but according to our assumption they were all distinct!!! (Contradiction)
NOTE: This argument generalizes in the exact same fashion for any number of eigenvectors. Also, it is clear that if $ {\bf x_3}$ cannot be a linear combination of two vectors then it cannot be a linear combination of any $n >2$ vectors (Try to prove!!!)
