※ 引述《superconan (超級柯南)》之銘言：
: What are the eigenvalues and eigenvectors of a projection ? A reflection ?
: 目前我知道投影矩陣 P = A(A^TA)^(-1)A^T 且 P^2 = P
: Let t is an eigenvalue of P and v is an eigenvector of P, then Pv = tv. — (1)
: Pv = P^2v = P(Pv) = P(tv) = t(Pv) = t(tv) = t^2v. — (2)
: By (1), (2) => t^2 = t => t^2 - t = 0 => t(t-1) = 0 => t = 0, 1
: 其他的就不知道怎麼求了
: 麻煩高手解惑，謝謝～
A projection operator on a vector space V determines a direct sum
decomposition of V into ker P and Im P. Since P^2=P, we see that
P(Pv)=Pv and P[(I-P)v]=0.
Therefore vectors in Im P are all eigenvectors of P corresponding
to eigenvalue 1 while the vectors in ker P are eigenvectors of P
corresponding to eigenvalue 0.
This also shows that P is diagonizable. (t^2-t is the minimal polynomial of P)
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