(c) Find a basis for the column space Col(A) of A, then find the dimension dim (
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(c) Find a basis for the column space Col(A) of A, then find the dimension dim ( (d) Find a basis for the null space Null(A) of A, then find dim (Null(A)) 2. You are given the vectors a= (1,0,1,1) and b = (1,1,0,1) in R4 (a) What is the dimension of V span (a, b)? - Explain your reasoning! (b) What is the dimension of the orthogonal complement of C (a, b} in R1? (c) Find two lincarly independent (i.e. non-parallel) vectors in C (d) Is it possible to find linearly independent vectors in V? Justify your answer 3. Let V be the subspace of R1 spanned by u1,0,0), v(1,1,1,0), and w(,1,1,1) (a) Find an ordered orthonormal basis (ONB) B- (a, b, c) for V (b) Ifx (1,2,1,2), ind the orthogonal projection of x onto V.Explanation / Answer
2. (a). The vectors a = (1,0,1,1) and b =(1,1,0,1) are apparently linearly independent . Therefore, the dimension of V = span{a,b} is 2.
(b). Let X = (x,y,z,w)T be an arbitrary vector in C, where C = {a,b}. Then X.a = 0 or, x+z+w = 0 or, z = -x-w, and X.b = 0 or, x+y+w = 0 or, y = -x-w. Then X = (x,-x-w,-x-w,w)T = x(1,-1,-1,0)T +w(0,-1,-1,1)T. Thus, {(1,-1,-1,0)T,(0,-1,-1,1)T } is a basis for C so the dimension of C is 2.
(c ). (1,-1,-1,0)T and (0,-1,-1,1)T are 2 linearly independent vectors in C .
(d). The spanning set of V has 2 vectors namely a and b so that all the vectors in V are linear combinations of a and b. Therefore, it is not possible to find 4 linearly independent vectors in V.
3. (a).We have u=(1,1,0,0), v=(1,1,1,0) and w=(1,1,1,1). Let a=u/||u||=u/[(12+12+0+0)]=(1/2, 1/2, 0 , 0), b=v/||v||=v/[(12+12+12+0)]=(1/3, 1/3, 1/3, 0) and c=w/||w||=w/[(12+12+12+12)]=(1/2,1/2, ½,1/2). Then B = {a,b,c} is an ONB for V.
(b).We have x = (1,2,1,2) so that proj u(x)=[(x.u)/(u.u)]u=[(1+2+0+0)/(1+1+0+0)]u=3u/2 =(3/2,3/2,0,0), projv(x)=[(x.v)/(v.v)]v=[(1+2+1+0)/(1+1+1+0)]v=4v/3 =(4/3,4/3,4/3 ,0), and projw(x)=[(x.w)/(w.w)]w = [(1+2+1+2)/(1+1+1+1)]w =3w/2 =(3/2,3/2,3/2,3/2). Then projV (x) = proju(x)+ projv(x)+ projw(x) =(3/2,3/2,0,0)+ (4/3,4/3,4/3 ,0)+ (3/2,3/2,3/2,3/2) = (13/3,13/3,17/6,3/2).
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