Is the span of Eigenvectors equal to the span of the column space of the matrix? 
*

*I wanted to know if the span of eigenvectors is related to the span of the matrix itself. (Ex: equal to the column span of the matrix?)

*Is there a way to represent the columns of a matrix in terms of its eigenvectors and eigenvalues?

*Lastly, how the second point is related to finding the powers of the matrix

 A: For $1$: yes, there is a relation, but it isn't always a nice one. Sometimes, the matrix is defective: it doesn't have enough eigenvectors to span the space. However, you can always express a matrix in terms of generalised eigenvectors via something called Jordan Normal/Canonical Form. The nice relation is when the matrix is diagonalisable - see below - if it is instead defective, it has a non-diagonal Jordan form.
For $2)$: it depends. If your matrix is diagonalisable, then yes. What does that mean? It means that it has a diagonal representation, right, but importantly for this question that really means: if you change your basis into your eigenvectors, that there are enough linearly independent eigenvectors to span your space.
Examples:

$$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$This is Andreas' example. This matrix is not diagonalisable. It has "two" eigenvalues ($1$, with multiplicity $2$), but the only eigenvectors are multiples of $(1,0)^T$. $2D$ space cannot be spanned by a single eigenvector, so the best you can do is leave this in its Jordan Normal Form - which it is already in!


$$\begin{pmatrix}2&3\\3&2\end{pmatrix}$$This is diagonalisable. Its eigenvalues are $5,-1$, with representative eigenvectors $(1,1)^T,(1,-1)^T$. Notice that these are $2$ linearly independent eigenvectors, which is sufficient to span $2D$ space. Let's first build this matrix of eigenvectors: $$\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$ And I also want to find its inverse, so that we can go from the normal basis to this one: $$\begin{pmatrix}1&1\\1&-1\end{pmatrix}^{-1}=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{pmatrix}$$Now I need to think about the diagonal form: when acting on the eigenvectors, by definition this matrix will just scale them by the eigenvalues. Therefore I can write a diagonal matrix with $5,-1$ on the respective entries. Altogether now: $$\begin{pmatrix}2&3\\3&2\end{pmatrix}=\underset{P}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}\underset{D}{\begin{pmatrix}5&0\\0&-1\end{pmatrix}}\underset{P^{-1}}{\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{pmatrix}}$$This works because every vector in $\Bbb R^2$ can be expressed in terms of the eigenvectors of this matrix, because there were two different ones, and the matrix on the right converts a vector in terms of $\bf i, j$ into one in terms of the eigenvectors; the diagonal part scales the eigenvectors as it normally would, and the last matrix converts everything back into vectors in terms of $\bf i,j$.

What about powers?
For $3)$, consider this: when the matrix is diagonalisable, I can express it in terms of $PDP^{-1}$, where $D$ is the diagonal matrix and $P$ is the change of basis to the eigenbasis (linearly independent eigenvectors). Remember I can only do this sometimes. But suppose now is such a time, and I can write $A=PDP^{-1}$. Let's think about integer powers:
$$A^2=(PDP^{-1})^2=PD\color{red}{P^{-1}P}DP^{-1}=PD\color{red}{I}DP^{-1}=PDDP^{-1}=PD^2P^{-1}\\A^3=A^2A=PD^2P^{-1}PDP^{-1}=PD^2\color{red}{P^{-1}P}DP^{-1}=PD^2DP^{-1}=PD^3P^{-1}$$
A straightforward exercise in induction reveals that, for any natural number $n$:
$$A^n=PD^nP^{-1}$$
Why do we care? Well, $D^n$ is really easy to calculate. You just raise the diagonal entries to the relevant power. Then it's a lot easier:
Example:

$$\begin{align}\begin{pmatrix}2&3\\3&2\end{pmatrix}^n&=PD^nP^{-1}=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}5&0\\0&-1\end{pmatrix}^n\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{pmatrix}\\&=\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}5^n&0\\0&(-1)^n\end{pmatrix}\begin{pmatrix}\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{pmatrix}\\&=\cdots\\&=\frac{1}{2}\begin{pmatrix}(-1)^n+5^n&(-1)^{n+1}+5^n\\(-1)^{n+1}+5^n&(-1)^n+5^n\end{pmatrix}\end{align}$$

And a difficult problem becomes easier. Actually, since that matrix was symmetric, it wouldn't have been too hard to inductively find its powers, but this technique is very powerful and fairly general - it fails when the matrix is defective, because it won't have the nice diagonal form.
