When is algebraic multiplicity = geometric multiplicity?

[Boorglar]

In my last Linear Algebra class we saw Eigenvalues and Diagonalizations. It turns out that an n x n matrix is diagonalizable if its eigenbasis has n linearly independent vectors.

If the characteristic equation for the matrix is $(λ - λ_1)^{e_1}(λ - λ_2)^{e_2}...(λ - λ_k)^{e_k} = 0$ then 1) eigenspaces corresponding to different eigenvalues are linearly independent and 2) the dimension of the eigenspace corresponding to $λ_i ≤ e_i$. That is, the algebraic multiplicity of the eigenvalue is greater than or equal to the geometric multiplicity.

But is there a general criterion for telling when algebraic and geometric multiplicities are equal? Given a concrete matrix, I can use Gaussian elimination to solve for the eigenvectors, but is there a more general way of dealing with this problem simply by looking at the form of the matrix?

EDIT: If I remember, there was something about real symmetric matrices but I forgot some of the theorems.

 

[micromass]

The algebraic multiplicities equal the geometric multiplicities exactly when the matrix is diagonalizable. So you're searching for a criterion to decide whether a matrix is diagonalizable or not. To my knowledge, just looking at the matrix usually does not suffice. So such a criterion does not exist.

The theorem you mention is the very important spectral theorem. This is one theorem where it's easy to see that a matrix is diagonalizable. Indeed, is a matrix is real symmetric, then it is diagonalizable. So in that case, we can always infer that the algebraic multiplicities equal the geometric multiplicities. But to my knowledge, there is no easy generalization of the spectral theorem that works for any diagonalizable matrix.

 

[Boorglar]

Oh well that's too bad. Still, is there an algorithm (which may involve operations on the matrix) more efficient than Gaussian Elimination to determine if a matrix is diagonalizable or not?