My question is regarding this post : compact and normal operator is diagonalizable, there is a detail in the argument that I don't understand why is $u(K^{\perp}) \subseteq K^{\perp}$ ?
2026-04-02 03:43:48.1775101428
compact and normal operator is diagonlisable
51 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in FUNCTIONAL-ANALYSIS
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