Suggest edit — eigenspace

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title: "eigenspace"
source: https://www.jemoka.com/posts/kbheigenspace/
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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 \(E(\lambda_{1}, T) + &hellip; + E(\lambda_{m}, T)\) 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) + &hellip; + \dim E(\lambda_{m}, T) \leq \dim V \end{equation}
Proof:
Recall that:
\begin{equation} \dim E(\lambda_{1}, T) + &hellip; + \dim E(\lambda_{m}, T) = \dim (E(\lambda_{1}, T) \oplus &hellip; \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 \(\dim (E(\lambda_{1}, T) \oplus &hellip; \oplus E(\lambda_{m}, T) ) \leq \dim V\).
And hence:
\begin{equation} \dim E(\lambda_{1}, T) + &hellip; + \dim E(\lambda_{m}, T) \leq \dim V \end{equation}
as desired. \(\blacksquare\)