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。
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