Suppose that {v1, v2, v3, v4} is a basis for a subspace V of . Which, if any, of the following subsets are bases for V?
1) S1 = {v1 + v2, v2 + v3, v3 + v4, v4 + v1}
2) S2 = {v1, v1 + v2, v1 + v2 + v3, v1 + v2 + v3 + v4}
3) S3 = {v1 - v2, v2 - v3, v3 - v4, v4 - v1}
4) S4 = {v1 + v2, v2+ v3, v3 + v4}
5) S5 = {v1, v1 + v2, v2 + v3, v3 + v4, v1 + v2 + v3 + v4}
Justify your answers?
Can someone just show me how to do one of them, then I wil be able to do rest of them?
I know a basis needs to be a span of V and linearly independent, I am not sure how to prove it is a span of V
You do not need to show that they span V. Since V has a basis consisting of four vectors, it is of dimension 4 and so any set of 4 independent vectors in V is a basis for V.
S3 has only three members and so cannot be a basis for V.
S5 has five members and so cannot be a basis for V.
The others have four vectors and so you only need to show that they are independent.
To show that Si= {u1, u2, u3, u4} is independent, look at au1+ bu2+ cu3+ du4= 0. Rewrite them as a linear combination of v1, v2, v3, v4 and use the fact that v1, v2, v3, v4 are independent.
For example, in S2, that would be a(v1)+ b(v1 + v2)+ c(v1 + v2 + v3)+ d(v1 + v2 + v3 + v4)= 0. That reduces to (a+ b+ c+ d)v1+ (b+ c+ d)v2+ (c+ d)v3+ dv4= 0. Since v1, v2, v3, and v4 are independent, we must have a+ b+ c+ d= 0, b+ c+ d= 0, c+ d= 0, and d= 0. Putting d= 0 into the third equation, c+ 0= 0 so c= 0. Putting d= 0 and c= 0 into the second equation, b+ 0+ 0= 0 so b= 0. Putting d= 0, c= 0, and b= 0 into the first equation, a+ 0+ 0+ 0= 0 so a= 0. We must have all coefficients zero so this set of four vectors is independent and forms a basis.