Let $A \in \mathcal{M}_n(\mathbb{F}),$ where $\mathbb{F}$ is a field, and suppose that $m_A(t)$ is the minimal polynomial of $A$. Let $w$ an eigenvector of $A$. Say $Aw = \lambda w.$ If $k$ is the degree of $m_A(t),$ then what is possible to say about $\dim \{v \mid Av = \lambda v \}?$
Obviously $k \le n.$ I know that if $k = n,$ then $\dim \{v \mid Av = \lambda v \} = 1,$ but for $k < n,$ I have no idea how can I proceed.
Best Answer
The dimension of the eigenspace $E_\lambda$ is known as the geometric multiplicity of the eigenvalue $\lambda$. It cannot exceed the algebraic multiplicity of this eigenvalue, i.e. its multiplicity as a root of the characteristic polynomial $\chi_A(t)$.
A criterion for diagonalisability asserts that $A$ is diagonalisable if and only if the geometric and algebraic multiplicities of each eigenvalue of $A$ are equal. It is also that the minimal polynomial $m_A(t)$ has only simple roots.