推 morning3569 : 謝謝 ~ 07/06 00:12
※ 引述《morning3569 (我甘心做一條水草)》之銘言
: 1. Suppose that λ,μ are distinct eigenvalues of A with corresponding ei
gensp
: aces E_λ,E_μ. Show that E_λ∩E_μ = { 0 }
let v in both eigenspaces
then Av=lv and Av=uv implies 0=(l-u)v
which means v = 0
: 2. Let T ：V --> W be a linear transformation and let {w1,w2,......,w
k} be
: a L.I. subset of R(T). Show that if S = {v1,v2,....,vk} is chosen so that
T(v
: i) = wi for i = 1,2,....,k, then S is linearly independent.
assume S not linearly indep.
then exists nonzero (c1, ..., ck)
such thay c1v1+...+ckvk = 0
thus c1w1+...+ckwk
= c1T(v1)+...+ckT(vk)
= T(c1v1+...+ckvk) = T(0) = 0
i.e. wi is not linearly indep.
: 第一題是證明兩個特徵向量是線性獨立嗎？？
: 第二題我沒什麼頭緒，我寫出 R(T) = span {{T(v1),T(v2),....,T(vk)}} 我就不知
道怎
: 麼寫了
: 另外想問
: W_1 = {(a1,a2,a3,a4,a5)∈R5|a1-a2-2a4 = 0}
: W_2 = {(a1,a2,a3,a4,a5)∈R5|a2=a3 and a1+a5 = 0}
: dim(W_1+W_2) = 4 嗎？
it's clear that dim(W_1) = 4
now pick v=(0,0,0,1,0)
then v in W_2, thus v in W_1 + W_2
but v not in W_1, thus
4 = dim(W_1) < dim(W_1+W_2) <= dim R^5 = 5
i.e. dim(W_1+W_2) = 5
: ※ 編輯: morning3569 (111.252.0.93), 07/05/2018 22:48:41
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