# Vector spaces and subspaces

1. Jan 28, 2010

### jameswill1am

1. The problem statement, all variables and given/known data
For each of the following subsets U of the vector space V decide whether or not U is a
subspace of V . Give reasons for your answers. In each case when U is a subspace, find a
basis for U and state dim U

2. Relevant equations

$$V=P_{3} ; U=\left\{p\in\ P_{3}:p(0)-p'(0)=0\right\}$$

3. The attempt at a solution

so i set it out as (a_3p^3+...+a_1p^1+a_0)-(3a_3p^2+2a_2p+a_1)=0 then rearranged all that to get; a_3p^3+...+(a_1-2a_2)p^1+(a_0-a_1)=0

Now i'm not sure what to do next. I think its because i'm not comfortable with the idea of polynomials as regards to vector spaces. does the fact that a couple of them share coefficients as it were mean that they would be dependent and so reduce the dimension? i guess making it a one dimensional subspace? All help and explanations much appreciated.

p.s. another subset is $$x=(x_{1},x_{2},x_{3}): x^{2}_{3}=x^{2}_{1}+x^{2}_{2}$$
I want to say this is not a subspace because of the fact its not a linear equation? would that be correct? Doesn't feel like very rigorous reasoning so it feels wrong.

2. Jan 28, 2010

What is the condition (let's say, a "test" condition) for U to be a subspace of some vector space V?

3. Jan 28, 2010

### jameswill1am

that the elements of U are closed under addition and scalar multiplication operations and it contains the zero vector.

4. Jan 28, 2010

Correct. Any idea of how to test it? Take two vectors (polynomials) from U, let's say p and q. Is their linear combination αp + βq also in U? Is the zero vector in U (i.e. does it satisfy the condition on the coefficients implied by p(0) - p'(0) = 0)?

5. Jan 28, 2010

### jameswill1am

Ok right so the zero vector would be in U right? Just the one with all zero coefficients?

And the condition for p(0) - p'(0) = 0 is that $$p_{0}=p_{1}$$ so provided you chose polynomials P and Q that satisfied this then (ap+bq)-(ap+bq)' = ap+bq-ap'-bq' = ap-ap' + bq-bq' = a(p-p')+b(q-q') = a0 + b0 so that works to.

Am i on the right track so far?

So then given the condition that $$p_{0}=p_{1}$$ does that reduce the dimension from 3 to 2?

6. Jan 28, 2010

### some_dude

To conclude $$W \subset V$$ is a subspace of $$V$$, it suffices to verify that for arbitrary scalar $$\alpha$$ and arbitrary vectors $$x,y \in W$$, that $$(\alpha\cdot x + y)\in W$$ as well.

So, consider two vectors $$p, q \in U$$, and either (1) prove that for any scalar $$\alpha \in \mathbb{R}$$ that $$(\alpha\cdot p + q)$$ is also in $$U$$, or (2) show a case where it need not be.

7. Jan 28, 2010

### vela

Staff Emeritus
Yes, you've shown that the subset is closed under scalar multiplication and vector addition, so it's a subspace.

From 4 to 3, actually. P3 is spanned by {1, x, x2, x3}, so it's a four-dimensional space.

8. Jan 29, 2010

### jameswill1am

Yeah that makes more sense silly me. Thanks all!

9. Jan 29, 2010

### jameswill1am

Another question! V is a vector space in R4. And U={x=[x1,x2,x3,x4]: x1-2x2-3x3+x4=0}

so I'm happy this is a sunspace but I have to determine a basis and dimension.

Am I correct thinking that clearly there are non zero scalars such that x1+...+x4 equals zero, namely 1,-2,-3,1. So they are linearly dependant? How do I make a basis out of that?

10. Jan 29, 2010

### Staff: Mentor

Some corrections: V is a subspace of R4. R4 is a vector space in its own right. Actually there doesn't seem to be a need for V, since it isn't mentioned in the rest of the problem.

U is a subspace (as it turns out) of R4 such that if x is in U, then x1 - 2x2 - 3x3 +x4 = 0, where x1, x2, x3, and x4 are the coordinates of vector x. They are not vectors, so it makes no sense to talk about them being linearly dependent/independent.

This equation represents a "hyperplane" in R4. Have you seen any other problems where an equation such as the one above is the starting point for finding a basis for the subspace determined by that equation?