The eigenspace of T, \lambda is the set of all eigenvectors of T corresponding to \lambda, plus the 0 vector. constituents T \in \mathcal{L}(V) \lambda \in \mathbb{F}, an eigenvalue of T requirements \begin{equation} E(\lambda, T) = \text{null}\ (T - \lambda I) \end{equation} i.e. all vectors such that (T- \lambda I) v = 0. where, E is an eigenspace of T. additional information sum of eigenspaces is a direct sum E(\lambda_{1}, T) + … + E(\lambda_{m}, T) is a direct sum. See eigenspaces are disjoint. dimension of sum of eigenspaces is smaller than or equal to the dimension of the whole space A correlate of the above is that:

Proof: Recall that:

because U_1 + \dots + U_{m} is a direct sum IFF \dim (U_1 + \dots + U_{m}) = \dim U_1 + \dots + \dim U_{m}. Now, the sum of subspaces is the smallest subspace, so \dim (E(\lambda_{1}, T) \oplus … \oplus E(\lambda_{m}, T) ) \leq \dim V. And hence:

as desired. \blacksquare