Intuition: why distinct eigenvalues $\implies$ linearly independent eigen vectors? Suppose you have an $~n\times n~$ matrix with $~n~$ distinct (not repeated) eigenvalues. 
There is a theorem telling us that the eigen vectors corresponding to these eigenvalues must be linearly independent.  
I can basically follow the proof, but I am looking for an intuitive explanation of why this is the case. 
Can anyone offer some insight?
 A: I'm really sorry for the hand made graphs but I believe they can help
Think about the opposite case, can dependent eigen vectors have distinct eigen values?
In  $\mathbb{R}^2$, the only way for two vectors to be dependent is to be collinear

as both have the same span, it's easy to see why they can't have distinct eigen values.
Now in $\mathbb{R}^3$ the only way for $3$ vectors to be dependent is to be coplanars

Here it is a little bit trickier to visualize, you can't have a transformation that causes two vectors to stretch with distinct factors without distorting the plane. 
For example assume the three vectors $p_1,p_2,p_3$ to be eigen vectors for transformation $T$ can $p_1$ stretched with $\lambda_2=2$ , $p_3$ stretched with $\lambda_3=3$. 
It can't be the case since the resulted plane will not be linear:

For both cases $\mathbb{R}^2,~\mathbb{R}^3$ the only way to keep linearity with dependent eigen vectors would be to have non distinct eigen values. 
Hence, since the transformation is linear and the eigen values are distinct the eigen vectors cannot be dependent. 
You can go from here and generalize to: for $\mathbb{R}^n$ if there exist $n$ distinct eigen value then there must be $n$ linearly independent eigen vector.
A: A good way to think about eigenvalues/eigenvalues are that the eigenvectors are the vectors that are stretched or flipped(which you can think of "negative stretching"), and the eigenvalues are how much they get stretched by.  
So if we have some eigenvectors, say with eigenvalues 1,2,3.  Then we have eigenvectors $v_1,v_2,v_3$.  Let's say that $v_3$ is a linear combination of the other two.  The intuition is basically that if we can stretch the first two by their eigenvalues, then the action on the third with already be determined and "can't" be three.  So the only way for them to be able to stretch without effecting the others is for them to be linearly independent.  
This intuition is similar if we consider the vector space over $\mathbb{C}$ rather than $\mathbb{R}$, as complex multiplication corresponds to a combination of rotating and stretching.      
A: You may understand it better if you try a simple proof for low dimensional spaces
With one eigenvector, things are easy. If $f$ is linear and $f(\vec{v})=\lambda \vec{v}$ then $f(k\vec{v}) = k f(\vec{v})=k\lambda \vec{v} = \lambda(k\vec{v})$ (in short, all multiples of $\vec{v}$ are eigenvectors for the same eigenvalue, so they cannot be eigenvectors for a different one)
Things get trickier when we add more vectors: Let's say now $\vec{v},\vec{w}$ are linearly independent eigenvectors for the eigenvalues $\lambda, \mu$ respectively. Let's say we have a linear combination $a\vec{v}+b\vec{w}$ that is also an eigenvector for an eigenvalue that we'll call $\eta$.
Since $f$ is linaer, $f(a\vec{v}+b\vec{w})=af(\vec{v})+bf(\vec{w})=a\lambda \vec{v} + b\mu \vec{w}$
But, since $f(a\vec{v}+b\vec{w})$ is an eigenvector, $a\lambda \vec{v} + b\mu \vec{w} = \eta(a\vec{v}+b\vec{w})$ Now:


*

*If either $a$ or $b$ are zero, then we are back to the one-eigenvector case.

*If both $a$ and $b$ are non-zero, then $\eta=\lambda$ and $\eta=\mu$ are both true, so $\lambda=\mu$
Graphics for this proof have already been provided in kk96kk's answer
