The GramSchmidt process is a simple algorithm for producing an orthogonal or ort
ID: 3138382 • Letter: T
Question
The GramSchmidt process is a simple algorithm for producing an orthogonal or orthonormal basis for any nonzero subspace of R". Given any basis xi,x2,...,*p of a nonzero subspace W of R, one can find an orthogonal basis vi, V2..,V of W using Gram-Schmidt process described in theorem 11 section 6.4. a) Find an orthogonal basis of the column space of the following matrix (check before hand what you use is INDEED a basis). -1 6 6 3 -83 1 2 6 1 -4-3 b) Normalize the basis V in part a to find an orthonormal basis U of W. c) Find the orthogonal projection of vector y- onto Col(A). (Hint: the algorithm of finding orthogonal projection onto a spanset we learnt in class works ONLY for orthogonal/orthonormal basis so make sure you use one of the new bases you found in part a or b.) d) Find the distance from y to Col(A).Explanation / Answer
a). The RREF of A is
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It implies that the columns of A are linearly independent. Hence, {v1,v2,v3} = {(-1,3,1,1)T,(6,-8,-2,-4)T , (6,3,6,-3)T} is a basis for col(A).Now, let u1= v1 =(-1,3,1,1)T, u2 = v2- proju1(v2 ) = v2-[(v2.u1)/(u1.u1)]u1= v2- [(-6-24-2-4)/(1+9+1+1)]u1 = (6,-8,-2,-4)T +3(-1,3,1,1)T= (3,1,1,-1)T and u3 = v3- proju1(v3 )- proju2(v3 )= v3- [(v3.u1)/(u1.u1)]u1-[(v3.u2)/(u2.u2)]u2= v3- [(-6+9+6-3)/( 1+9+1+1)]u1-[(18+3+6+3)/(9+1+1+1)]u2= (6,3,6,-3)T –(1/2)(-1,3,1,1)T-(5/2) (3,1,1,-1)T= (-1,-1,3,-1) T. Then {u1,u2,u3} = {(-1,3,1,1)T,( 3,1,1,-1)T, (-1,-1,3,-1) T } is an orthogonal basis for col(A).
(b). Let e1=u1/||u1||=(-1,3,1,1)T /?(1+9+1+1)=(-1/?12,3/?12,1/?12,1/?12)T, e2=u2/||u2|| =( 3,1,1,-1)T / ?(9+1+1+1) = (3/?12,1/?12,1/?12,-1/?12)T and e3=u3/||u3|| =(-1,-1,3,-1) T/?(1+1+9+1)T =(-1/?12,-1/?12 , 3/?12,-1/?12) T . Then {e1,e2,e3} is an orthonormal basis for col(A).
Please post part ( c) again
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