Invariant Subspace

A subspace U⊆VU\subseteq V is called T^\hat{T}-invariant if operator T^\hat{T} keeps everything in U inside U. If it is T-invariant, then this must always be true:

T^(U)={T^∣ψ⟩∣∣ψ⟩∈U}⊆U\begin{gather*} \hat{T}(U)=\{\hat{T}\left\lvert \psi \right\rangle|\left\lvert \psi \right\rangle\in U\}\subseteq U \end{gather*}

Simplest T invariant subspace is (with dimensions dd)

U=span{∣u⟩}T^∣y⟩=λ∣u⟩\begin{gather*} U=span\{\left\lvert u \right\rangle\}\\ \hat{T}\left\lvert y \right\rangle=\lambda \left\lvert u \right\rangle \end{gather*}

When λ\lambda is Eigenvalue and ∣u⟩\left\lvert u \right\rangle is eigenvector and ∣u⟩≠0\left\lvert u \right\rangle\neq 0 and you can define these Vector by checking

(T^−λI)∣u⟩=00=det(T^−λI)=(λ1−λ)(λ2−λ)..(λd−λ)\begin{gather*} (\hat{T}-\lambda I)\left\lvert u \right\rangle=0\\ 0=det(\hat{T}-\lambda I)\\ =(\lambda_1-\lambda)(\lambda_2-\lambda)..(\lambda_d-\lambda) \end{gather*}

In general, this determinant has degree λi\lambda_i are zeroes of character polynomial λi∈C\lambda_i\in C can be repeated