TITLE: Diagonalized matrix - Is each column an eigenvector?
QUESTION [10 upvotes]: 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.

REPLY [6 votes]: 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.