推 kusorz:eigenvalues 0 1 1 怎麼看出來的? 03/05 22:41
※ 引述《std810471 (家程)》之銘言：
: Let T : R3 → R3 be the projection of the space R3 on the plane x + 2y + 3z = 0 . Find the
: eigenvalues of T and find a basis of the corresponding eigenspaces.
令 W = {x,y,z 屬於 R | x + 2y + 3z = 0}
= span {(-2,1,0),(-3,0,1)}
可得 Wper = span {(1,2,3)}
則 T 之eigenvalues 為 0,1,1
且 N(T-0I) = span {(1,2,3)}
N(T-1I) = span {(-2,1,0),(-3,0,1)}
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※ 編輯: a53285315 來自: 111.255.68.204 (03/03 21:01)