Creative ways to show that a given matrix is diagonalizable? I know the standard method of calculate the characteristic polynomial, then get the eigenvalues, and look for the dimension of the null space associated to each eigenvalue, then see if their algebraic multiplicity coincides with their geometric multiplicity but I wonder if there are some creative ways instead of the standard. 
Do you know any non standard method? 
Thanks for your kind attention.
 A: The standard way to show that $A$ is diagonalisable should be to show that it is annihilated by a polynomial that splits into distinct monic factors of degree$~1$. So you could take the product of $(X-\lambda)$ for all distinct eigenvalues$~\lambda$ (for instance found as roots of the characteristic polynomial) and check that it annihilates$~A$. Or you could check that the minimal polynomial of $A$ splits without repeated factors. But sometimes you get an annihilating polynomial from your hypothesis, for instance if it is given that $A^k=I$ for some integer$~k$, where $X^k-1$ splits into distinct factors if the coefficient field is$~\Bbb C$.
I don't claim this is particularly creative. It is just a standard characterisation of diagonalisability.
A: This list contains a few basic tools. 
1) Use the spectral theorem. ( check if it is normal )
2) If all the eigen values are distinct, then it is diagonalizable.
3) Use the characteristic polynomial as you have mentioned.
4) Check if the matrix belongs to a class of matrices which are already known to be diagonalizable . (Involutions,projections,Normal,..)
5) Check if the matrix belongs to a class of matrices which are already known to be non diagonalizable . ( Nilpotent, some rotation matrices, .. )
