[Math] Dimension of generalized eigenvector space

Question

From Wikipedia

For a matrix A, there may not always exist a full set of linearly independent eigenvectors that form a complete basis – a matrix may not be diagonalizable. This happens when the algebraic multiplicity of at least one eigenvalue $λ$ is greater than its geometric multiplicity (the nullity of the matrix $(A-\lambda I)$, or the dimension of its nullspace).

In such cases, a generalized eigenvector of $A$ is a nonzero vector $v$, which is associated with $λ$ having algebraic multiplicity $k ≥1$, satisfying
$$
(A-\lambda I)^k\mathbf{v} = \mathbf{0}.
$$
The set of all generalized eigenvectors for a given $λ$, together with the zero vector, form the generalized eigenspace for $λ$.

So it seems that generalized eigenvectors are introduced to solve the problem that the algebraic and geometric multiplicities of an eigenvalues may not be identical.

Then for each eigenvalue, how is its algebraic multiplicity identical to the dimension of its generalized eigenspace?

Thanks!

Answer 1

This question has been answered in comments:

Over an algebraically closed field, such as the complex numbers, we can express "how" the algebraic multiplicity equals the dimension of a corresponding generalized eigenspace by making change of basis that puts the matrix in Jordan canonical form. See this previous question for a few more details. – hardmath Nov 24 '12 at 18:05