→ alan23273850: 我參考的解法有用到一個重要 lemma，就是 [x,y] = [ 08/05 15:30
→ alan23273850: ix,iy], 而這個東東沒人提示的話不容易自己想到。 08/05 15:30
引述《alan23273850 (God of Computer Science)》之銘言：
: 各位板友大家好，打給賀，太軋賀，小弟又有問題想上來請教各位：
: https://i.imgur.com/M3hr1uq.png
29.
Let V be a vector space over C.
Suppose [．,．] is a real inner product on V,
where V is regarded as a vector space over R,
such that [x,ix] = 0 for all x in V.
Let <．,．> be a complex-valued function defind by
<x,y> = [x,y] + i[x,iy] for x, y in V.
Prove that <．,．> is a complex inner product on V.
[證]
設 x = a + ib, y = c+id, 其中 a, b, c, d are real.
則 [x,y] 可寫成 [(a,b),(c,d)], 它(因為是實內積)滿足:
(1) 對稱쨺 [x,y] = [y,x] for x, y in V
(2) 雙線性:
[x+y,z] = [x,z] + [y,z] for x,y,z in V,
[x,y+z] = [x,y] + [x,z] for x,y,z in V,
[x,ky] = [kx,y] = k[x,y] for k in R, x,y in V
(3) 正確定性:
for all x in V, [x,x] ≧ 0,
[x,x] = 0 if and only if x = 0.
令 x = a+ib, a,b in R, 則 ix = -b+ia
[x,ix] = [(a,b),(-b,a)]
= [(a,0)+(0,b),(-b,0)+(0,a)]
= [(a,0),(-b,0)] + [(0,b),(-b,0)]
+ [(a,0),(0,a)] + [(0,b),(0,a)]
= [(a,0)+(-1)(b,0)] + [(0,b),(0,a)]
= -[(a,0],(b,0)] + [(0,a),(0,b)]
所以, 條件 [x,ix] = 0 for all x in V
意思是: for all (a,0),(0,a),(b,0),(0,b) in V,
[(a,0],(b,0)] = [(0,a),(0,b)]
故
[(a,b),(c,d)] = [(a,0),(c,0)] + [(0,b),(0,d)]
= [(0,a)+(0,c)] + [(b,0),(d,0)]
= [(b,a),(d,c)]
意謂此實內積 [．,．] 在同時對兩運算元做實部和虛部對調運算
時結果不變.
另外,
[(a,-b),(c,d)] = [(a,0],(c,0)] + [(0,-b),(0,d)]
= [(a,0],(c,0)] + (-1)[(0,b),(0,d)]
= [(a,0],(c,0)] + [(0,b),(0,-d)]
= [(a,b),(c,-d)]
對第一運算元取共軛等同於對第二運算元取共軛, 而同時對兩運算
元取共軛結果不變:
[(a,-b),(c,-d)] = [(a,b),(c,d)]
由於實內積定義中的 k 也是實數, 我們考慮把 k 換成 ik, k in R:
[x,(ik)y] = [x,k(iy)] = k[x,iy]
and
[(ik)x,y] = [k(ix),y] = k[ix,y]
= k[(-b,a),(c,d)]
= k[(a,-b),(d,c)]
= k[(a,b),(d,-c)]
= -k[(a,b),(-d,c)] = -k[x,iy]
In particular,
[x,iy] = -[ix,y]
現在把純量乘數 k 換成一般複數. 令 h,k in R,
[(h+ik)x,y] = [hx,y] + [ikx,y]
= h[x,y] + k[ix,y]
= h[x,y] - k[x,iy]
= [x,(h-ik)y]
目的是要證明 <x,y> 在複向量空間 V 定義了一個複內積, 現在我
們來驗證此運算滿足所需條件. 首先是共軛對稱:
<x,y> = [x,y] + i[x,iy]
= [y,x] + i[iy,x]
= [y,x] - i[y,ix]
= conjugate(<y,x>)
半雙線性: Let x,y,z in V, k in C,
<x+y,z> = [x+y,z] +i[x+y,iz]
= [x,z] + [y,z] + i([x,iz]+[y,iz])
= <x,z> + <y,z>
<x,y+z> = [x,y+z] + i[x,i(y+z)]
= [x,y] + [x,z] +i([x,iy]+[x,iz])
= <x,y> + <x,z>
令 k = a+ib, k* = a-ib, a,b in R
<kx,y> = [(a+ib)x,y] + i[(a+ib)x,y]
= [ax,y]+[ibx,y] + i[ax,iy]+i[ibx,iy]
= a([x,y]+i[x,iy]) - b([x,iy]+i[x,-y])
= a([x,y]+i[x,iy]) + ib(i[x,iy]+[x,y])
= k<x,y>
and
<kx,y> = [kx,y] + i[kx,iy]
= [x,k*y] + i[x,ik*y]
= <x,k*y>
正確定性: For any x in V,
<x,x> = [x,x] + i[x,ix] = [x,x]
is positive.
至此, 證明了, <．,．> is a complex inner product.
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