Diagonalized matrix - Is each column an eigenvector? Everything below is correct. But I got a question at the bottom (yellow).
Given is matrix $A=\begin{pmatrix}
11 &  0 &  -6\\ 
 0 &  5 &  -6\\ 
-6 & -6 &  -2
\end{pmatrix}$ which is diagonalizable.
The eigenvalues are
$$\lambda_1= -7\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \lambda_2= 7\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } \lambda_3=14$$
The eigenspaces are
$E_{A}(\lambda_1) = \left\{ \begin{pmatrix}
\frac{z}{3}\\ 
\frac{z}{2}\\ 
z
\end{pmatrix} \mid z \in \mathbb{R}\right\}, E_{A}(\lambda_2) = \left\{ \begin{pmatrix}
\frac{3}{2}z\\ 
-3z\\ 
z
\end{pmatrix} \mid z \in \mathbb{R}\right\}$
$E_{A}(\lambda_3) = \left\{ \begin{pmatrix}
-2z\\ 
-\frac{2}{3}z\\ 
z
\end{pmatrix} \mid z \in \mathbb{R}\right\}$

Let's say I was supposed to to determine an orthonormal basis
  $(v_1,v_2,v_3) \in \mathbb{R}^{3}$ of the eigenvectors of $A$, can I
  just write the matrix in its diagonalized form and continue there
  determining an orthonormal basis? It should be possible iff each
  column is an eigenvector of the eigenspace, right? Is it?
So matrix A diagonalized is $A_D= \begin{pmatrix}
-7 & 0 & 0\\   0 & 7 & 0\\   0 & 0 & 14 \end{pmatrix}$
Now can I say that $v_1= \begin{pmatrix}
-7\\  0\\  0 \end{pmatrix}, v_2= \begin{pmatrix} 0\\  7\\  0 \end{pmatrix}, v_3= \begin{pmatrix} 0\\  0\\  14 \end{pmatrix}$ and
  determine the orthonormal basis with these? Sorry for making this question so long, I didn't know how to keep it short. I hope you can help me anyway.

 A: So, you want to find a basis of the eigenvectors but the $v_i$'s you listed correspond to the diagonalized matrix and thus correspond to the eigenvalues.
How do we find a basis for each eigenspace? Well consider $E_A(\lambda_1)$. Every vector in this space is a multiple of $(1/3, 1/2, 1)^T$ (this is easy to check), and thus $(1/3, 1/2, 1)^T$ is a basis for the eigenspace $E_A(\lambda_1)$. To make this normalized, we want the length of the vector to be exactly $1$. We can do this by calculating the norm of the vector and dividing each element by the norm. So for this basis, the length is $\sqrt{(1/3)^2 + (1/2)^2 + 1^2}$. Dividing every element in the vector by $\sqrt{(1/3)^2 + (1/2)^2 + 1^2}$ gives us the normalized vector $(2/7, 3/7, 6/7)^T$ (after some algebra and reducing fractions).
We can continue doing this for each eigenspace. We should get three normalized vectors.
A: Note that $A$ is a symmetric matrix, so it can be diagonalized by an orthogonal matrix ( see the properties of  symmetric matrices here). 
The matrix that diagonalize $A$ is not $A_D$ but the matrix $M$ in $A=MA_DM^{-1}$, that is the matrix that has as columns the eigenvectors of the corresponding eigenvalues in the diagonal of $A_D$.
In fact  the eigenspaces that you have found are orthogonal as you can easily verify. This means that for any value of $z \ne 0$ that you can chose, you have three orthogonal vectors and, if you want, you can normalize these vectors to obtain an orthonormal basis.
A: The matrix $A_D$ contains the eigenvalues along the diagonal entries; the columns are not the eigenvectors. You already have the eigenspaces, which are just the spans of the eigenvectors. 
For example, just take $z = 1$. It is usually best pick a basis of orthonormal eigenvectors, so just normalize each eigenvector once you have your basis.
A: The answer is no. If you want the eigenvectors, take them from the eigenspaces; they're the spans of each of the eigenvectors.
A: But the columns of $A_D$ are NOT eigenvectors of $A$. They rather describe the action of $A$ on those eigenvectors. The eigenvectors are essentially listed in those eigenspaces that you already found. For example, from $E_A(\lambda_1)$ we can see that an eigenvector corresponding to $\lambda_1=-7$ is
$$\mathbf{v}_1=\begin{pmatrix}1/3\\1/2\\1\end{pmatrix}$$
or any vector parallel to it, i.e. any nonzero multiple of it. In fact, if you need an orthonormal basis, you'll have to renormalize this vector to get its multiple whose length (norm) is $1$. Similarly for the other two.
Good news, though: you don't need to worry about the vectors being orthogonal to each other, because eigenvectors corresponding to different eigenvalues are automatically orthogonal.
