Assume that x_1, x_2, ..., x_k are k vectors in R^n. The Gram-Schmidt determinan
ID: 3037610 • Letter: A
Question
Assume that x_1, x_2, ..., x_k are k vectors in R^n. The Gram-Schmidt determinant of x_1, x_2, ..., x_k is the determinant of the k times k matrix G(x_1, x_2, ..., x_k) = G:= [a_i, j], where a_i, j = x_i middot x_j (i.e. the (i, j)-entry of G is the dot product between the vectors x_i and x_j). (a) Prove that G(x_1, x_2, ..., x_k) is always a symmetric matrix, for any vectors x_1, x_2, ..., x_k. (b) Assume that x_1 = [2 2 2 2 0 0], x_2 = [0 0 -1 -1 -1 -1], and x_3 = [1 1 0 0 1 1]. Find the Gram-Schmidt determinant of x_1, x_2, x_3. (c) Assume that v_1, v_2, v_3 are orthogonal vectors as defined in Theorem 11, page 357. Find the Gram- determinant of v_1, v_2, v_3. (d) Verify that v_1 middot v_1 = det G(x_1) v_2 middot v_2 = det G(x_1, x_2)/det G(x_1), v_3 middot v_3 = det G(x_1, x_2, x_3)/det G(x_1, x_2).Explanation / Answer
a. Given G =(aij) = (xi.xj) = (xj.xi) {as dot product of two vectors are commutative}
= (aji)
=> G is a symmetric matrix
b. x 1= (2,2,2,2,0,0) x2= ( 0,0,-1,-1,-1,-1) and x3= (1,1,0,0,1,1)
using dot product concept '
a11=x1.x1= 4+4+4+4=16 , a12= 0+0-2-2+0+0= - 4 , a13= 2+2+0+0+0+0=4
a21=a12= - 4 , a22= 1+1+1+1=4 , a23= 0+0+0+0-1-1= - 2
a31=a13=4 , a32=a23= - 2 and a33= 4
G = [ 16 - 4 4
- 4 4 - 2
4 -2 4]
det G = 128
For part c reference is given as page 357 . only if we know what is given in that page we can do part c