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?
Best Answer
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.