# Determining if a Matrix is Diagonalizable without computing Eigenvalues

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.

• I remember trying to figure this out a while ago. In the general case the best I was able to find was that the geometric multiplicity and algebraic multiplicity of the eigenvalues must be the same. Of course you have to compute the eigenvalues for this. Unless you have some additional structure on the matrix, like symmetry, I don't think we can do better Mar 29, 2013 at 0:12
• I disagree with mtiano. A matrix is normal if, and only if it is unitarily diagonalizable. One would think that dropping unitarily would lead to something treatable. One the other hand if there was such a (known) criteria it would surely be well known. I think I agree with mtiano after all. Mar 29, 2013 at 0:14
• Are you looking for a numerical technique or a theoretical equivalence? Mar 29, 2013 at 0:17
• A numerical technique Mar 29, 2013 at 0:18
• What are you meaning when you say numerical technique? I think guest might be saying he wants an algorithm, not actual numerical procedure. Mar 29, 2013 at 0:22

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.