TITLE: Determining if a Matrix is Diagonalizable without computing Eigenvalues
QUESTION [6 upvotes]: Is there any simple way to determine if a matrix is diagonalizable without having to compute eigenvalues? 
I'm motivated by the idea that for $\mathbb{R}^n$, to determine if a matrix is diagonalizable via an orthogonal transformation, you just need to check if it's symmetric. Also, for $\mathbb{C}^n$, to determine if a matrix is diagonalizable via a unitary transformation, you just need to check if it's normal. So I'm just curious if one can drop the orthogonal/unitary requirements while still having an easy method to check if a matrix is diagonalizable.

REPLY [5 votes]: A matrix is diagonalizable over $\mathbb{R}$ if and only if all zeros of its minimal polynomial are simple. If the minimal polynomial is $f(t)$, its zeros are simple if and only if the gcd of $f(t)$ and its derivative $f'(t)$ is $1$.
So the problem is to compute the minimal polynomial. The obvious approach is
to compute the rational canonical form (also called the Frobenius normal form).
The key properties of this are that it can be computed over $\mathbb{R}$, and
that it does not require any information about the eigenvalues.