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. \(\blacksquare\)