# Thread: Justify basis

1. ## Justify basis

Suppose that {v1, v2, v3, v4} is a basis for a subspace V of $R^n$. 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

2. Originally Posted by 450081592
Suppose that {v1, v2, v3, v4} is a basis for a subspace V of $R^n$. 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
Ok good.
I think that to prove that S1 a span of V, if you can show that v1, v2, v3 and v4 can be written as a linear combination of the vectors forming S1, it means it spans the same vector space, S1 in this case.

3. hi
$S_5$ is obviously not a basis of $V$, since,
$(v_1 + v_2 + v_3 + v_4)=(v_1+v_2)+(v_3+v_4)$.

4. $S_1$ is also not a basis since,
$v_4+v_1=(v_4+v_3)+(v_1+v_2)-(v_2+v_3)$.

5. We can note from the given base that $dim V = 4$ and therefore:

$S_5$ is not a base since $|S_5| = 5 > 4 = |\{v_1,v_2,v_3,v_3\}|$

Same for $S_4$: $|S_4| = 3 < 4$

For $S_3$ - what happens when you take the sum of its elements?

$S_1$ is similar to $S_3$.

For $S_2$, follow arbolis' advice.

6. Originally Posted by 450081592
Suppose that {v1, v2, v3, v4} is a basis for a subspace V of $R^n$. 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.

7. Thanks a lot for the examples, such a great help !!!!!