Let W be the set of ll ectors of the form shown on the right, where b and c are
ID: 3116377 • Letter: L
Question
Let W be the set of ll ectors of the form shown on the right, where b and c are artitrary Find vectors u and v Why does this show that W is a subspace of R37 such that W Spantu, v). 6b+7c 4c Using the given vector space, write vectors u and v such that W-Span(u, v) (Use a comma to separate answers as needed.) Choose the correct theorem that indicates why these vectors show that W is a subspace of R3 O A. The column space of an m xn matrix A is a subspace of R OB. I, are in a vector spece V. then Sis a subspace of V O C. The null space of an mxn matrix is a subspace of R". Equivalently, the set of all solutions to a system Ax o of m homogeneous linear equations in n unknowns is a subspace of R° OD·Anindexed set v vo) of two or more vectors in a ve tor space . with v, #0 is a subspace of vrand only if some v isinSpan vr--1M) Click to select your answerls).Explanation / Answer
(a). We have (6b+7c,-b,4c)T = b(6,-1,0)T+c(7,0,4)T. Now, if u = (6,-1,0)T and v = (7,0,4)T,then W = span{u,v}. Thus,(u,v) = {(6,-1,0)T ,(7,0,4)T}. Also, W R3. The RREF of the matrix A = [u,v] =
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This implies that u and v are linearly independent.Further, let X = au+bv,Y = cu+dv (where a,b,c,d R) and let k be an arbitrary scalar. Then X+Y = au+bv + cu+dv = (a+c)u+(b+d)v W so that W is closed under vector addition. Also, kX = k(au+bv) = kau+kbv W so that W is closed under scalar multiplication. Also, the vector (0,0,0)T W ( as (6b+7c,-b,4c)T = (0,0,0)T when b=c=0). Hence W is a vector space, and , therefore, a subspace of R3. The option A is a correct answer as W = col(A) = span{e1,e2} R3 = span{e1,e2,e3}.
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