如何用施密特法把向量组 a1=(1,1,1), a2=(1,2,3), a3=(1,4,9)正交...答:解:b1=a1=(1,1,1)b2=a2-(a2,b1)/(b1,b1)b1 = (1,2,3)-(6/3)(1,1,1)=(-1,0,1)b3=a3-(a3,b2)/(b2,b2)b2-(a3,b1)/(b1,b1)b1 = (1,4,9)-(8/2)(-1,0,1)-(14/3)(1,1,1)= (1/3,-2/3,1/3).满意请采纳^_^ ...
斯密特正交变化过程是如何推算出来的,思路是什么?答:把一组线性无关的向量变成一单位正交向量组的方法在一些书和文献中称为施密特(Schimidt)正交化过程.把a1,a2,...ar规范正交化,取b1=a1 b2=a2-[b1,a2]b1/[b1,b1]...br=ar-[b1,ar]b1/[b1,b1]-[b2,ar]b2/[b2,b2]-...-[br-1,ar]br-1/[br-1,br-1]容易验证b1,...br两两正交...
试用施密特法把向量组a1=(1,1,1)^T,a2=(1,2,3)^T,a3=(1,4,9)^T正交...答:正交化套公式就行了 b1=a1 b2 = a2 - (b1,a2)/(b1,b1)b1 = (1,2,3)^T - 6/3 (1,1,1)^T = (-1,0,1)^T b3 类似, 你练习一下吧
求施密特正交单位化一道题答:楼主请出题。施密特正交化和单位化的公式在图片里。下面是具体解答过程:先正交化,令b1=a1=(1,1,0,0),b2 = a2 - (a2,b1)b1/(b1,b1) = ( 0.5,-0.5,1,0);b3 = a3 - (a3,b1)b1/(b1,b1) - (a3,b2)b2/(b2,b2) = (-1,0,0,1) + (0.5,0.5,0,0) + (0.5,-...