Subspaces of Vector Space V in R4: U = {x ∈ R4 : x1 - 2x2 - 3x3 + x4 = 0}

[jameswill1am]

Homework Statement

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

Homework Equations

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

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.

 

[radou]

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

 

[jameswill1am]

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

 

[radou]

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

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)?

 

[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?

 

[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.

 

[vela]

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?

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

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

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

 

[jameswill1am]

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.

Yeah that makes more sense silly me. Thanks all!

 

[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 dependent? How do I make a basis out of that?

 

[Mark44]

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?