推 cakeboy:R(A)basis可以寫成(1,0) (0,1) 02/02 14:30
→ sm008150204:是沒錯 因為span整個 R^2 謝你你的補充 02/02 14:37
提供另一種想法 從幾何上來看
Let L be a linear transformation of the reflection about ax+by=0
It is easy to see that Lv= 1*v for all v on the line ax+by=0
Lv=-1*v for all v on the line bx-ay=0
which say that 1 is an eigenvalue and [b,-a] is an eigenvector
-1 is an eigenvalue and [a, b] is an eigenvector
Note that [b,-a] and [a, b] span R^2
thus P^(-1) A P = D so A = P D P^(-1)
1 [b a] [1 0][b -a]
A = ------- (自己展開一下)
a^2+b^2 [-a b] [0 -1][a b]
R(A) = span{[b,-a] , [a, b]}
N(A) = {[0,0]}
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