In parts ak determine whether the statement is true or false
Solution
(a) By the definition of a subspace, the statement is True.
(b) Since every set is a subset of itself, hence every vector space is a subspace of itself. The statement is true.
(c) The statement is false. The subset also needs to be closed under ‘multiplication’ and ‘addition’. e.g. {0, 1} is not a subspace of R while span{0, 1} is a subspace of R.
(d) The statement is false. Every subspace of R will have 3-vectors while R2, having 2-vectors is not a subset of R3 and hence not a subspace of R3.
(e) The statement is false as every subspace of Rn must have the 0 vector.
(f) By the definition of span, the statement is true.
(g) The statement is true. If both U,W are subspaces of V, then 0 U and 0 W, so that 0 UW. Further, if u,w U W, then u, w U and u,w W. Since U, W are subspaces, they are closed under addition which means that u + w U and u + w W. This implies that u + w UW. Also, if u UW , then u is in U and in W, so that cu U and cu W for any scalar c. Hence cu UW. Thus UW is a subspace of V.
(h) The statement is false. Let U = span{(1,0)} and W = span{(0,1)} . Then (1,0) U and (0,1) W, but u +w = (1,1) either U or W so that u+w U U W.
(i) The statement is false. The sets U = {(1,1)T, (0,1)T} and W = {(1,0)T,(0,1)T} both span R2, but U W.
(j) The statement is true. The set of nxn upper triangular matrices , being closed under vector addition and scalar multiplication, and also having the 0 matrix, is a vector space.
(k) The statement is false. (x-1)3 = x3-3x2+3x-1, (x-1)2= x-2x+1 and x-1. The RREf of the coefficients, i.e. the RREF of the matrix A =
1
0
0
-3
1
0
3
-2
1
-1
1
-1
is B =
1
0
0
0
1
0
0
0
1
0
0
0
so that 1 span{ (x-1)3, (x-1)2, (x-1)}. Hence, the set { (x-1)3, (x-1)2, (x-1)} does not span P3.
| 1 | 0 | 0 |
| -3 | 1 | 0 |
| 3 | -2 | 1 |
| -1 | 1 | -1 |

