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@jemoka / Jemoka Knowledge Base / raw/concept/kbheigenspace.md
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--- title: "eigenspace" source: https://www.jemoka.com/posts/kbheigenspace/ --- 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: \begin{equation} \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T) \leq \dim V \end{equation} Proof: Recall that: \begin{equation} \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T) = \dim (E(\lambda_{1}, T) \oplus … \oplus E(\lambda_{m}, T) ) \end{equation} 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: \begin{equation} \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T) \leq \dim V \end{equation} as desired. \(\blacksquare\)