If A constituted a sub-space we would have for all t in R there exists a u in R such that:

2(t, 5-t)=(u, 5-u),

then we have from the first coordinate 2t=u, and from the second coordinate 10-5t=5-u, or 5-5t=u, but both of these cannot be true for general t, so A does not constitute a subspace of R^2.

Now consider any two elements of B; U=(u, -u) and V=(v, -v), and any a and b in R, then:

aU+bV=(au+bv, -au-bv)

which is also in B, so B is a subspace of R^2.

Now V, T, Z and P are all in R^3, so at most 3 of them are linearly independent (they can all be written as a linear combination of (1,0,0), (0,1,0), and (0,0,1) and so the space spanned by them can depend on at most three parameters).Moreover, consider 4 vectors V=(3; 2; 7) T=(1; -5; 1) Z=(0; -3, 1) P=(-1;3; 2) a basis for the linear space spanned by these vectors is formed by 2, 3 or 4 vectors, or none of the preceding?thanks a lot

Now U=T+P=(0,-2,3), and as Z=(0,-3,1) is not a multiple of this together they span a 2-D subspace of R^3, and no linear combination of them could be equal to T (as it has a non-zero 1st component) Therefore collectively the vectors span R^3 which is requires a basis of 3 vectors.

RonL