引述《alan23273850 (God of Computer Science)》之铭言： : 各位板友大家好，打给贺，太轧贺，小弟又有问题想上来请教各位： : 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. [证] 设 x = a + ib, y = c+id, 其中 a, b, c, d are real. 则 [x,y] 可写成 [(a,b),(c,d)], 它(因为是实内积)满足: (1) 对称쨺 [x,y] = [y,x] for x, y in V (2) 双线性: [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) 正确定性: 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, 则 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, 条件 [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)] 意谓此实内积 [．,．] 在同时对两运算元做实部和虚部对调运算 时结果不变. 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)] 对第一运算元取共轭等同於对第二运算元取共轭, 而同时对两运算 元取共轭结果不变: [(a,-b),(c,-d)] = [(a,b),(c,d)] 由於实内积定义中的 k 也是实数, 我们考虑把 k 换成 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] 现在把纯量乘数 k 换成一般复数. 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] 目的是要证明 <x,y> 在复向量空间 V 定义了一个复内积, 现在我 们来验证此运算满足所需条件. 首先是共轭对称: <x,y> = [x,y] + i[x,iy] = [y,x] + i[iy,x] = [y,x] - i[y,ix] = conjugate(<y,x>) 半双线性: 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> 正确定性: For any x in V, <x,x> = [x,x] + i[x,ix] = [x,x] is positive. 至此, 证明了, <．,．> is a complex inner product. --

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