Rau qhov kev xav niaj hnub no ntawm kev ua haujlwm, nws "nco" nws cov codomain, thiab peb xav kom lub npe ntawm nws cov inverse ua tag nrho cov codomain, yog li ib qho kev txhaj tshuaj tsuas yog invertible yog tias nws tseem yog bijective.
Puas txhaj tshuaj imply inverse?
Yog tias koj txoj haujlwm f: X → Y yog txhaj tshuaj tab sis tsis tas yuav tsum tau surjective, koj tuaj yeem hais tias nws muaj inverse muaj nuj nqi txhais ntawm daim duab f(X), tab sis tsis nyob rau Tag nrho cov Y. Los ntawm kev muab cov txiaj ntsig tsis txaus ntseeg ntawm Y∖f(X), koj tau txais sab laug inverse rau koj txoj haujlwm.
Koj paub li cas yog matrix txhaj?
Cia A ua matrix thiab cia Ared yog kab txo daim ntawv A. Yog Ared muaj tus thawj 1 hauv txhua kab, ces A yog txhaj. Yog Ared muaj ib kem uas tsis muaj tus thawj 1 hauv nws, ces A tsis txhaj.
Puas yog square matrix txhaj?
Note that a square matrix A is injective (los yog surjective) iff nws yog ob qho tib si txhaj tshuaj thiab surjective, i.e., iff nws yog bijective. Bijective matrices tseem hu ua invertible matrices, vim hais tias lawv yog tus yam ntxwv los ntawm lub neej ntawm ib tug cim square matrix B (tus inverse ntawm A, denoted los ntawm A−1) xws li AB=BA=I.
YPuas txhaj yog thiab tsuas yog nws muaj sab laug sab laug?
Claim: f yog txhaj tshuaj yog tias nws muaj qhov thim rov qab. Pov thawj: Peb yuav tsum (⇒) ua pov thawj tias yog f yog txhaj tshuaj ces nws muaj sab laug inverse, thiab (⇐) tias yog f muaj sab laug inverse, ces nws yogtxhaj tshuaj. (⇒) Piv txwv tias f yog txhaj. Peb xav tsim ib txoj haujlwm g: B→A xws li g ∘ f=idA.
