Let W be the set of al vectors of the form shown on the right, where a, b, and e
ID: 3186304 • Letter: L
Question
Let W be the set of al vectors of the form shown on the right, where a, b, and e represent arbitrary real numbers. Find a set S of vectors that spans W or give an example or an explanation to show that W is not a vector space 6a+ 5 a+b c c-4a Select the correct choice below and if necessary, fill in the answer box to complete your choice O A The set Wis a vector space and a spanning sot is O B. The set W is not a vector space because W doos not contain the zero vector (Use a comma to separate vectors as needed ) C. The set w n not a vector space because w-not dosed under vector addison O D. The set W is not a vector space because W is not closed under scalar mutipkcationExplanation / Answer
We have W=span{(6a+5b,0,a+b+c,c-4a)T: a,b,c are arbitrary real numbers}. Further,(6a+5b,0,a+b+c,c-4a)T =a(6,0,1,-4)T +b(5,0,1,0)T+c (0,0,1,1)T}.Thus, an arbitrary vector in W is a linear combination of the vectors (6,0,1,-4)T,(5,0,1,0)T, (0,0,1,1)T. Hence S = {(6,0,1,-4)T,(5,0,1,0)T, (0,0,1,1)T } is a spanning set for W.
To show that W is a vector space, let X = (6a+5b,0,a+b+c,c-4a)T and Y =(6d+5e,0,d+e+f,f-4d)T be 2 arbitrary vectors in W and let k be an arbitrary real scalar. Then X+Y = (6a+5b,0,a+b+c,c-4a)T + (6d+5e,0,d+e+f,f-4d)T = (6(a+d)+5(b+e) ,0,(a+d)+(b+e)+(c+f), (c+f)-4(a+d))T. It implies that X+Y is in W so that W is closed under vector addition. Also, kX = k(6a+5b,0,a+b+c,c-4a)T = (6ka+5kb,0,ak+kb+kc,kc-4ka)T . It implies that kX is in W so that W is closed under scalar multiplication. Further, when each of a,b,c is 0, X = (0,0,0,0)T. Hence the zero vector also is in W. Therefore, W is a vector space.
Option A: The set W is a vector space and the set S is {(6,0,1,-4)T,(5,0,1,0)T, (0,0,1,1)T }.
The coordinate vector of the polynomial 1-9t2 -t3= 1+0*t-9t2 -t3 with respect to the basis B is (1,0,-9,-1)T. The coordinate vector of the polynomial t+6t3= 0*1+1*t+0*t2+6*t3 with respect to the basis B is (0,1,0,6)T.
The coordinate vector of the polynomial 1+t-9t2 = 1*1+1*t -9*t2 +0*t3 with respect to the basis B is (1,1,-9,0)T.
Let A be the matrix with the above coordinate vectors as columns. Then A =
1
0
1
0
1
1
-9
0
-9
1
6
0
To determine whether the given polynomials are linearly independent, we have to reduce A to its RREF which is
1
0
0
0
1
0
0
0
1
0
0
0
It implies that none of the 3 polynomials is either a scalar multiple of any other polynomial in the set or is a linear combination of the other 2 polynomials in the set. Hence the given polynomials are linearly independent.Option D is the correct answer.
1
0
1
0
1
1
-9
0
-9
1
6
0
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