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.