What is the proof that x, x^2, x^3 form a basis of V?

[TranscendArcu]

Homework Statement

http://img856.imageshack.us/img856/5586/screenshot20120121at328.png

The Attempt at a Solution

I propose the vectors x,x^2,x^3 form a basis of V. To test for linear independence, let 0 = a_1 x + a_2 x^2 + a_3 x^3, where a \in R. A polynomial is 0 iff all of its coefficients are zero. Thus, linear independence is proved. That is, a_1 = a_2 = a_3 = 0

To prove that x,x^2,x^3 spans V, let p(x) = b_1 x + b_2 x^2 + b_3 x^3 \in V, where b \in R. We need numbers c_1,c_2,c_3 such that b_1 x + b_2 x^2 + b_3 x^3 = c_1 x + c_2 x^2 + c_3 x^3. This implies b_1 = c_1, b_2 = c_2, b_3 = c_3. Thus, p(x) = c_1 x + c_2 x^2 + c_3 x^3 and clearly p(0) = c_1 (0) + c_2 (0) + c_3 (0) = 0. Therefore, x,x^2,x^3 span V.

Am I doing this right?

 

[TranscendArcu]

This is the second part of the question:

http://img819.imageshack.us/img819/9069/screenshot20120121at350.png

I began with \int \frac{d}{dx} p(x) dx = 0x + c_1, where c_1 \in R. Thus, I reasoned that p(x) must be of the form: c_1 + 0(x) + 0(x^2) + 0(x^3). A polynomial is zero iff all of its coefficients are zero. c_1 must necessarily equal zero in order for the zero vector to result. Thus, linear independence is proven. To show that c_1 spans S, let p(x) = b_1 + 0(x) + 0(x^2) + 0(x^3) \in S, where b_1 \in R. We need p(x) = b_1 = c_1. This implies b_1 = c_1. So p(x) = c_1 and clearly \frac{d}{dx} p(x) =\frac{d}{dx} c_1 = 0. Thus, c_1 spans S.

Man, I don't think I'm doing these right...

 

[vela]

Homework Statement

http://img856.imageshack.us/img856/5586/screenshot20120121at328.png

The Attempt at a Solution

I propose the vectors x,x^2,x^3 form a basis of V. To test for linear independence, let 0 = a_1 x + a_2 x^2 + a_3 x^3, where a \in R. A polynomial is 0 iff all of its coefficients are zero. Thus, linear independence is proved. That is, a_1 = a_2 = a_3 = 0

To prove that x,x^2,x^3 spans V, let p(x) = b_1 x + b_2 x^2 + b_3 x^3 \in V, where b \in R. We need numbers c_1,c_2,c_3 such that b_1 x + b_2 x^2 + b_3 x^3 = c_1 x + c_2 x^2 + c_3 x^3. This implies b_1 = c_1, b_2 = c_2, b_3 = c_3. Thus, p(x) = c_1 x + c_2 x^2 + c_3 x^3 and clearly p(0) = c_1 (0) + c_2 (0) + c_3 (0) = 0. Therefore, x,x^2,x^3 span V.

Am I doing this right?

In the second part, you can't assume p(x) has the form $b_1 x + b_2 x^2 + b_3 x^3$ right off, otherwise you're assuming what you're trying to prove. When you say $p(x) \in V$, you know that p is a polynomial and that p(0)=0. From what you know about polynomials, you should be able deduce that p(x) has the form required, from which it follows that p(x) is in the span of {x, x², x³}.

 

[TranscendArcu]

Okay. Can I assume that p(x) has the form b_0 +b_1 x + b_2 x^2 + b_3 x^3? But, clearly p(0) cannot equal zero unless b_0 = 0, which immediately gets us to the required form.

Is that better?

 

[vela]

Yes, that's fine. Because you know V is a subspace of P₃ and presumably you know that {1, x, x², x³} is a basis for P₃, you can write p(x) in that form.

 

[TranscendArcu]

This is the last problem. I need only find a basis.

http://img861.imageshack.us/img861/7183/screenshot20120121at517.png

I propose that the vectors 1,x^2 form a basis for W. To prove linear independence, we write, 0 = a_0 + a_2 x^2 where a \in R. A polynomial is zero iff all coefficients equal zero. To prove that 1,x^2 span W let p(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3, but b_1 = b_3 = 0 otherwise condition fails. We need numbers c_0, c_2 such that b_0 +b_2 x^2 = c_0 +c_2 x^2. This implies that b_0 = c_0 and b_2 = c_2. Thus p(x) = c_0 + c_2 x^2 in which case \forall x p(x) = p(-x), as desired. Thus 1,x^2 form a basis for W.

 

[vela]

This is the last problem. I need only find a basis.

http://img861.imageshack.us/img861/7183/screenshot20120121at517.png

I propose that the vectors 1,x^2 form a basis for W. To prove linear independence, we write, 0 = a_0 + a_2 x^2 where a \in R. A polynomial is zero iff all coefficients equal zero. To prove that 1,x^2 span W let p(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3, but b_1 = b_3 = 0 otherwise condition fails.

You should go into more detail about why you must have b₁=b₃=0.

We need numbers c_0, c_2 such that b_0 +b_2 x^2 = c_0 +c_2 x^2. This implies that b_0 = c_0 and b_2 = c_2. Thus p(x) = c_0 + c_2 x^2 in which case \forall x p(x) = p(-x), as desired. Thus 1,x^2 form a basis for W.

This part is unnecessary. After you show that b₁=b₃=0, you know that p(x)=b₀+b₂x². Since it's a linear combination of 1 and x², it's in the span of {1, x²}.

 

[TranscendArcu]

Should I say something like:

If p(x) = p(-x) then it follows that all of the n-degree components of p(x) should be equal to their respective n-degree components of p(-x). However, clearly this can't be so since x ≠ -x (in the first degree) and (x)³ ≠ (-x)³ (in the third degree). Thus, in order to preserve the equality of the condition, it is necessary to zero these terms.

?

 

[vela]

This is the second part of the question:

http://img819.imageshack.us/img819/9069/screenshot20120121at350.png

I began with \int \frac{d}{dx} p(x) dx = 0x + c_1, where c_1 \in R. Thus, I reasoned that p(x) must be of the form: c_1 + 0(x) + 0(x^2) + 0(x^3). A polynomial is zero iff all of its coefficients are zero. c_1 must necessarily equal zero in order for the zero vector to result. Thus, linear independence is proven. To show that c_1 spans S, let p(x) = b_1 + 0(x) + 0(x^2) + 0(x^3) \in S, where b_1 \in R. We need p(x) = b_1 = c_1. This implies b_1 = c_1. So p(x) = c_1 and clearly \frac{d}{dx} p(x) =\frac{d}{dx} c_1 = 0. Thus, c_1 spans S.

You made a slight error. You showed that p(x)∈S has the form p(x)=c₁. The basis vector is 1, not c₁, i.e. p(x) = c₁1. You want to show the vector 1 spans S, not c₁ as you said.

Since you've already shown that p(x)∈S implies p(x)=c₁=c₁1, you're done because you've shown that p(x) can be written as a linear combination of the set of vectors {1}.

 

[vela]

Should I say something like:

If p(x) = p(-x) then it follows that all of the n-degree components of p(x) should be equal to their respective n-degree components of p(-x). However, clearly this can't be so since x ≠ -x (in the first degree) and (x)³ = (-x)³ (in the third degree). Thus, in order to preserve the equality of the condition, it is necessary to zero these terms.

?

It would be more straightforward to let $p(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3$ and calculate g(x)=p(x)-p(-x). Then show that g(x)=0 implies b₁=b₃=0.