Showing whether a matrix is diagonalizable over $C$? I have the following matrix \begin{bmatrix}
    1 & 2 & 3  \\
    2 & 4 & 5  \\
    3 & 5 & 6   
\end{bmatrix}
and I want to determine whether it is diagonalizable over the field of complex numbers. 
If we calculate the characteristic polynomial, we get $p(x) = -x^3+11x^2+4x-1$. Unforunately, this has no rational roots, and I don't know of any way to determine the complex roots of this polynomial as is, since it is a cubic. 
My other thought was that, this is a symmetric matrix. Are there any substantial results about whether or not a symmetric matrix is diagonalizable over the field of complex numbers?
 A: A real symmetric matrix is diagonalizable over the reals, which trivially implies that it is diagonalizable over the complex numbers.
In general, for complex matrices, the corresponding result is that a Hermitian matrix is diagonalizable (all the eigenvalues will also be real, which is a nice bonus). "Hermitian" means it's equal to its own conjugate transpose.
A: Since the given matrix is symmetric hence it is diagonalizable.
Also because it has three distinct  latent roots $\lambda= 11.3448, -0.515729, 0.170915$
coming from $|A-\lambda I|=0$. These latent roots are the eigenvalues of $A$ and the diagonal element of the diagonalized matrix.
A: Even if you can’t find the roots explicitly, knowing that the roots are distinct would be enough to conclude that the matrix is diagonalizable (if a! eigenvalue have algebraic multiplicity 1 then it also must have geometric multiplicity 1).
For a cubic it’s easy to tell that it has distinct roots: any irrational roots have to come with at least one conjugate, so a repeated irrational root can only occur with degree 4 or higher.
But more generally, we can always tell whether a polynomial $f(x)$ has repeated roots, just by taking the GCD of $f$ and the derivative $f’$.  The set of repeated roots is equal to the roots of the GCD (in particular, if there are no repeated roots then the polynomial GCD will be $1$).
Of course, the symmetry of this matrix is a much quicker way to see this is diagonalizable.  I just wanted to point out that your approach could still work.
A: Two facts you need to know about the symmetric matrix in general:

Any symmetric matrix
         1) has only real eigenvalues;
         2) is always diagonalizable;


The proof is:
1) Let λ ∈ C be an eigenvalue of the symmetric matrix A. Then Av = λv, $v ̸\neq 0$, and
$v^∗Av = λv^∗v$, $v^∗ = \bar{v}^T.$$
But since A is symmetric
$λv^∗v = v^∗Av = (v^∗Av)^∗ = \bar{λ}v^∗v.$
Therefore, λ must be equal to $\bar{λ}$ !
Also, you don't need to find complex root of this characteristic polynomial because it doesn't have one (all roots are real).
2) If the symmetric matrix A is not diagonalizable then it must have generalized eigenvalues of order 2 or higher. That is, for some repeated eigenvalue $λ_i$ there exists $v ̸= 0$ such that
$(A − λ_iI)2v = 0$,   $\;\;(A − λ_iI)v \neq 0$
But note that $0 = v^∗(A − λ_iI)2v = v^∗(A − λ_iI)(A − λ_iI) ̸\neq0$,
which is a contradiction. Therefore, as there exist no generalized eigenvectors of order 2 or higher, A must be diagonalizable.
Therefore by this result, the matrix in the question is diagonalizable.
