谁能给解释下“施密特正交化过程”的原理

谁能给解释下“施密特正交化过程”的原理
[v(1)]^T*{v(2) - [v(1)]^Tv(2)v(1)}
这个式子中^ T 还有这个式子看不太懂?
附：您的回答：先看2个（列）向量的正交化.
设2向量V(1),V(2)线性无关.
v(1) = V(1)/|V(1)|,v(2) = V(2)/|V(2)|是对应的2个单位向量.
则,
[v(1)]^T*{v(2) - [v(1)]^Tv(2)v(1)}
= [v(1)]^Tv(2) - [v(1)]^T*[v(1)]^Tv(2)*v(1)
= [v(1)]^Tv(2) - [v(1)]^Tv(2)*[v(1)]^T*v(1)【注意到,[v(1)]^Tv(2)是1个数.因此,可以提到乘积的前面来.而v(1)是单位向量,因此[v(1)]^Tv(1)=|v(1)|^2 = 1^2 = 1.】
= [v(1)]^Tv(2) - [v(1)]^Tv(2)*1
= 0
所以,
向量v(1)与v(2) - [v(1)]^Tv(2)*v(1)相互垂直.
u(1)=v(1),u(2) = {v(2) - [v(1)]^Tv(2)*v(1)}/|v(2) - [v(1)]^Tv(2)*v(1)|就是相互垂直的2个单位向量.
假设v(1),v(2),...,v(k)是k个相互垂直的单位向量,v(k+1)与v(1),v(2),...,v(k)线性无关.
则对于 i = 1,2,...,n.都有,
[v(i)]^T{v(k+1) - [v(1)]^Tv(k+1)v(1) - [v(2)]^Tv(k+1)v(2) - ...- [v(i)]^Tv(k+1)v(i) - ...- [v(k)]^Tv(k+1)v(k)}
= [v(i)]^Tv(k+1) - [v(i)]^T*[v(1)]^Tv(k+1)v(1) - [v(i)]^T*[v(2)]^Tv(k+1)v(2) - ...- [v(i)]^T*[v(i)]^Tv(k+1)v(i) - ...- [v(i)]^T*[v(k)]^Tv(k+1)v(k)
= [v(i)]^Tv(k+1) - [v(1)]^Tv(k+1)*[v(i)]^T*v(1) - [v(2)]^Tv(k+1)*[v(i)]^T*v(2) - ...- [v(i)]^Tv(k+1)[v(i)]^T*v(i) - ...- [v(k)]^Tv(k+1)[v(i)]^T*v(k)
= [v(i)]^Tv(k+1) - [v(1)]^Tv(k+1)*0 - [v(2)]^Tv(k+1)*0 - ...- [v(i)]^Tv(k+1)*1 - ...- [v(k)]^Tv(k+1)*0
= [v(i)]^Tv(k+1) - [v(i)]^Tv(k+1)
= 0.
因此,
向量{v(k+1) - [v(1)]^Tv(k+1)v(1) - [v(2)]^Tv(k+1)v(2) - ...- [v(i)]^Tv(k+1)v(i) - ...- [v(k)]^Tv(k+1)v(k)}和v(1),v(2),...,v(k)都相互垂直.
u(1)=v(1),u(2)=v(2),...,u(k)=v(k),u(k+1) = {v(k+1) - [v(1)]^Tv(k+1)v(1) - [v(2)]^Tv(k+1)v(2) - ...- [v(i)]^Tv(k+1)v(i) - ...- [v(k)]^Tv(k+1)v(k)}/|{v(k+1) - [v(1)]^Tv(k+1)v(1) - [v(2)]^Tv(k+1)v(2) - ...- [v(i)]^Tv(k+1)v(i) - ...- [v(k)]^Tv(k+1)v(k)}| 是相互垂直的k+1个单位向量