[微方] 證明題

[TampaBayRays (坦帕灣光芒)]

   Ax=b when det A=0

(a)Suppose that A is a real-valued n*n matrix.Show that (Ax,y)=(x,A的轉置y)for

   any vectors x and y.

(b)If A is not necessary real, show that (Ax,y)=(x,A*y)for any vectors x and y.

   (*不是乘,是A*乘y)

(c)If A is Hermitian, show that (Ax,y)=(x,Ay)for any vectors x and y.


順便借問一題線代

  ┌-1 4 2 ┐                    ┌ 1-2-1 ┐
A=│-1 3 1 │ eigenvalue=1→ A-I=│ 0 0 0 │
  └-1 2 2 ┘                    └ 0 0 0 ┘

  ┌ 0 3 1 ┐                    ┌ 1 0 0 ┐
B=│-1 3 1 │ eigenvalue=1→ B-I=│ 0 1 0 │
  └ 0 1 1 ┘                    └ 0 0 0 ┘

A的rank=1 dim EA(1)=2

B的rank=2 dim EB(1)=1

請問dim要怎麼看??

謝謝

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◆ From: 114.43.154.238
推 justinj     :dim就算不知道..看你寫的也猜的出來吧..dim=n-rank
→ TampaBayRays:嗯 謝謝 我也是這樣猜的 想確定一下
推 justinj     :但我查資料(太久沒碰了)rank(A)=2=dim(A)=2...
→ justinj     :rank叫秩,dim叫維度...那EA(1)是指A-I嗎..如果是那
→ justinj     :dim應該只剩1.....
推 justinj     :抱歉...算錯...rank(A)=3=dim(A)....
→ TampaBayRays:對 EA(1)是指A-I