why this operator $T$ is always diagonalizable? Let $V = \mathbb{R}^3$ and $B=(v_1,v_2,v_3)$ ordered basis for $V$
Let $T:V \to V$ linear operator  and given the representation matrix with respect to the basis $B$ $$[T]_B^B = {\left[\begin{array}{ccc} 3 & 0 & 8 \\ 0 & 0 & -1 \\ 8 & -1 & 5 \end{array}\right]}.$$
Why is it true that $T$ is always diagonalizable?
I do not understand how I can conclude anything about eigenvector? this the only way I think on approaching this kind of question
 A: I will try to answer this question in a little bit more general setting.
One of the common definitions of diagonalizability is that a linear operator $A\colon V\to V$ is diagonalizable if and only if there exists a basis of $V$ consisting of eigenvectors of $A$. The most general class of linear operators satisfying this property are the so called normal operators. We say that a linear operator $A$ is a normal operator if and only if it commutes with its Hermitian adjoint: $AA^*=A^*A$.
Since a linear operator on $V$ can be expressed in a matrix form we also have a notion of a normal matrix, i.e. the matrix $A$ is normal if and only if $AA^*=A^*A$, where $A^*$ is the conjugate transpose of $A$. Note that for the real case $A^*=A^T$, so a real matrix $A$ is normal if and only if $AA^T=A^TA$. This property is invariant under the change of basis since
$$AA^T=A^TA\Longleftrightarrow P^{-1}AA^TP=P^{-1}A^TAP$$
for any invertible matrix $P$.
In this particular problem it's easy to note that $T$ is represented as a symmetric matrix in the given basis $\mathcal{B}$, i.e. $([T]_{\mathcal{B}})^T=[T]_{\mathcal{B}}$, so $T$ is normal, since $TT^T=T^2=T^TT$.
