Suppose m is an eigenvalue of the n×n square matrix **M**. It is easy to see that the set of all vectors v for which **M**v = mv (the set of the eigenvectors of v, along with 0) is a linear space. Its dimension d is called the (geometric) multiplicity of the eigenvalue m. Since m is an eigenvalue, d≥1.

Having a nontrivial (v≠0) solution **M**v = mv means that the characteristic polynomial of the matrix, p_{M}(t)=det(tI-M), has a root at t=m. So we can write p_{M}(t)=(t-m)^{k}⋅q(t) for some polynomial q(t) with q(m)≠0. The degree k is called the *algebraic* multiplicity of m.

It's not too hard to show that d≤m. Since the sum of all the m's is n, the sum of all the d's is *at most* n. We conclude that over an algebraically closed field, **M** is a diagonalizable matrix iff it has n perpendicular eigenvectors, iff the algebraic and geometric multiplicities of each eigenvalue are the same.

**Example:** Consider the matrix

[[1 1]
[0 1]].

The characteristic polynomial is (t-1)

^{2}. So there is only one eigenvalue (1), and it has algebraic multiplicity 2. The geometric multiplicity is either 1 or 2. But if it were 2, then the matrix would be the

unit matrix. Thus, it must be 1.