Is it only the symetric matrix that have an ortonormal base with its eigenvectors? There is a question on an old exam that asks: 
"Does it exist a base in $\mathbb{R^3}$, that consists of orthogonal eigenvectors of T. Explain your answer"
$T= \begin{bmatrix}-11&9&6\\-8&6&2\\ -6&6&7\end{bmatrix}$
Ive found all the eigenvectors and values but Im thinking because its not symetric, then there exist no such base. Is that correct, and why if so?
 A: You're right. Because $T$ is not symmetric, it does not have an orthonormal base of eigenvectors. This is guaranteed by the spectral theorem for symmetric matrices, so you can reach this conclusion simply by noting that $T$ is not symmetric.
If you correctly compute the eigenvalues, you should find that $T$ has real eigenvalues $3,-2,1$. However, no pair of associated eigenvectors is orthogonal.
A: 
Ive found all the eigenvectors and values but Im thinking because its not symetric, then there exist no such base. Is that correct, and why if so?

It is a base. To prove it you just need to prove that the vectors of the base are linear independent. That is done calculating the rank of the Eigenvector matrix. If it has full rank then it's a base in this case for $\mathbb{R^3}$
$rank(T) = 3$(full rank) and $Det(T)\neq 0$ so $T$ is also invertible therefore the matrix $A$ is diagonizable.
For proving the orthogonality of the basis,take the dot product of the vectors and it should be zero for any two different vectors.
