Eigenvectors yog TSIS cim, rau ntau yam. Hloov cov cim, thiab ib qho eigenvector tseem yog eigenvector rau tib eigenvalue. Qhov tseeb, muab faib los ntawm qhov tsis tu ncua, thiab ib qho eigenvector tseem yog qhov ntawd. Cov cuab yeej sib txawv tuaj yeem xaiv qhov sib txawv normalizations.
Koj paub li cas yog eigenvalues txawv?
"Distinct" tus lej tsuas yog txhais tau tias sib txawv. Yog a thiab b yog eigen qhov tseem ceeb ntawm tus neeg teb xov tooj T thiab tom qab ntawd lawv yog"qhov txawv" eigenvalues. Yog hais tias lawv tshwm sim los ua 0 thiab 1, ces, vim lawv txawv, lawv yog "sib txawv ".
Koj puas muaj eigenvectors txawv?
Yog tias lub matrix muaj ntau dua ib qho eigenvector qhov cuam tshuam eigenvalues tuaj yeem sib txawv rau qhov sib txawv eigenvectors. … Geometrically, qhov kev txiav txim ntawm lub matrix ntawm ib qho ntawm nws cov eigenvectors ua rau lub vector stretch (los yog shrink) thiab / los yog rov qab kev taw qhia.
Ib yam eigenvalues puas muaj qhov sib txawv eigenvectors?
It muaj tsuas yog ib qho eigenvalue, uas yog 1. Txawm li cas los xij ob qho tib si e1=(1, 0) thiab e2=(0, 1) yog eigenvectors ntawm no matrix. Yog tias b=0, muaj 2 qhov sib txawv eigenvectors rau tib eigenvalue a. Yog tias b≠0, ces tsuas muaj ib qho eigenvector rau eigenvalue a.
Puas yog eigenvector decomposition tshwj xeeb?
◮ Decomposition is not unique when ob eigenvalues tib yam. ◮ Los ntawm convention, txiav txim nkag ntawm Λ nyob rau hauv nqe lus txiav txim. Tom qab ntawd, eigendecomposition yog qhov tshwj xeeb yog tias txhua qhov eigenvalues yog qhov tshwj xeeb.
