Please help in details . Thanks a lot. Appreciate:) Consider the subspace of R^5
ID: 3036535 • Letter: P
Question
Please help in details . Thanks a lot. Appreciate:) Consider the subspace of R^5 defined as w = Span(v_1 v_2.v_3, v_4,v_5), where v_1 = (2, 0, 2, 0, 8), v_2 = (1, 1, 3, 0, 7), v_3 = (0, 1, 2, 1, 5), v_4 = (1, 1, 3, 1, 9) v_5 = (1, 2, 5, 2, 14). Find dim(W) and a basis for W. Find dim(W^ ) and a basis for W^ Find a vector w element R^5 such that w element W. Find a matrix A such that W is described as the solution space of Ax = 0. Find a matrix B such that W^ is described as the solution space of Bx = 0.Explanation / Answer
Let A =
2
1
0
1
1
0
1
1
1
2
2
3
2
3
5
0
0
1
1
2
8
7
5
9
14
The RREF of A is
1
0
0
½
½
0
1
0
0
0
0
0
1
1
2
0
0
0
0
0
0
0
0
0
0
1.Apparently, only v1,v2 and v3 are linearly independent and v4, v5 are linear combinations of v1,v2 and v3. Hence dim(W) = 3. A basis for W is { (2,0,2,0,8)T, (1,1,3,0,7)T, ( 1,1,3,1,9)T} or, {(1,0,0,0,0)T, (0,1,0,0,0)T, (0,0,1,0,0)T}.
2. Let x = (p,q,r,s,t)T be an arbitrary element of W. Then, (p,q,r,s,t)T.( 1,0,0,0,0)T = 0 so that p = 0. Also,
(p,q,r,s,t)T.( 0,1,0,0,0)T = 0 so that q = 0 and (p,q,r,s,t)T.( 0,0,1,0,0)T = 0 so that r = 0 . Hence x = (0,0,0,s,t)T = s(0,0,0,1,0)T +t(0,0,0,0,1)T. Therefore, dim(W )=2 ans a basis for W is{(0,0,0,1,0)T, (0,0,0,0,1)T}.
3. The vector (0,0,0,1,0)T W.
4. A =
2
0
2
0
8
1
1
3
0
7
1
1
3
1
9
5. B=
0
0
0
1
0
0
0
0
0
1
2
1
0
1
1
0
1
1
1
2
2
3
2
3
5
0
0
1
1
2
8
7
5
9
14
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