NUS-MATH530 Geometric Multiplicity

Let \lambda_{m} be an eigenvalue for T an operator on complex finite-dimensional V. Let m be the geometric multiplicity of \lambda_{m}. We desire that the algebraic multiplicity is at least m. Let \dim v = n.
We have that m is the geometric multiplicity of \lambda_{m}, meaning:

\begin{equation} \dim E(\lambda_{m}, T) = m \end{equation}

This means we can take m linearly independent eigenvectors from V. Extend this list now to a basis of V with v_1, …v_{m}, u_{1}, u_{n-m}.
Construct a matrix via this basis. By construction, the first m \times m of this matrix would appear diagonal (as each Tv = \lambda v). Furthermore, the diagonal of this sub-matrix would simply contain \lambda repeated m times.
Take now A = \mathcal{M}(T)-\lambda I.
Take the determinant of this matrix A now against the first column, yielding a characteristic polynomial with at least m factors with \lambda. Hence, the algebraic multiplicity of \lambda_{m} is at least m, as desired. \blacksquare