Basis and Elements of W

[Displayer102]

I have been stuck on this problem for awhile and have no sense of it, I have gotten stuff written down but sadly don't have any confidence in my answer. Any help would be absolutely appreciated thank you!!

This is what I have now.
$$
\begin{array}{l}
\mathrm{W}^{\perp}=\{p \in P 2 |\langle p, x+1\rangle=0\} \\
\langle p x+1\rangle=p(-1) x+1(-1)+p(0) x+1(0)+p(1) x+1(1)=p(-1) \\
(-1+1)+p(0)(0+1)+p(1)(1+1)=p(0)+2 p(1)=2 \mathrm{p}(1)
\end{array}
$$
since we are looking for polynomials such that $\mathrm{p}(0)=2 \mathrm{p}(1),$ and with the definition of $\mathrm{P}^1$ all
polynomials $a x^2+b x+c$ such that $c=2(a+b+c),$ so the numbers a,b,c with 2a+2b+c=0. In terms of linear algebra and the null space of $A=[2,2,1]$ which is dimension 2 and generates the vectors
$\begin{bmatrix}1\\0\\-2\end{bmatrix}$ and $\begin{bmatrix}0\\1\\-2\end{bmatrix}$

Which converts back into polynomials to get
$W^{\perp}=\left\{\mathrm{x}^2-2, \mathrm{x}-2\right\}$
Did I solve this question correctly?

Edited May 5, 2020 by Displayer102

[taeto]

You can answer question (a) easily, I assume?

And for any \(q(x) \in P_2\) you know the condition for determining whether \(q(x)\) is in \(W^{\perp}\)?

[Displayer102]

30 minutes ago, taeto said:

You can answer question (a) easily, I assume?

And for any q(x)∈P2 you know the condition for determining whether q(x) is in W⊥ ?

Can you check my edit to see? 

 

Isn't the basis just the Nul(A) which is the same as Row(A)?

So I have a basis of [1,0,-2] and [0,1,-2]

Edited May 5, 2020 by Displayer102

[taeto]

It looks like you are doing more work than you need to. And some mistakes.

Towards the end of your long calculation of \(\langle p, x+1\rangle\) you assume \(p(0)=0?\) But if \(p(x)=ax^2+bx+c,\) then you have \(p(0)=c.\) I guess you just forgot to type it in.

So your conclusion that \(p(x)=ax^2+bx+c \in W^\perp\) if and only if \(2a+2b+c=0\) is correct.

And you correctly found two linearly independent solutions to this equation.

Your solution is correct.

[Displayer102]

1 hour ago, taeto said:

It looks like you are doing more work than you need to. And some mistakes.

Towards the end of your long calculation of 〈p,x+1〉 you assume p(0)=0? But if p(x)=ax2+bx+c, then you have p(0)=c. I guess you just forgot to type it in.

So your conclusion that p(x)=ax2+bx+c∈W⊥ if and only if 2a+2b+c=0 is correct.

And you correctly found two linearly independent solutions to this equation.

Your solution is correct.

So after changing that little type, I have correctly shown a and b?

a)p(x)=ax2+bx+c∈W⊥ if and only if 2a+2b+c=0 

 

b) Basis is [2,2,0]

Edited May 5, 2020 by Displayer102

[taeto]

2 hours ago, Displayer102 said:

a)p(x)=ax2+bx+c∈W⊥ if and only if 2a+2b+c=0 

b) Basis is [2,2,0]

The elements of the basis are polynomials. You correctly found that the polynomials in \(W^\perp\) have the form \(ax^2+bx+c,\) where \(2a+2b+c=0.\)

The basis of \(W^\perp\)  has to be a set of two polynomials which satisfy this. You could choose \(\{x^2-2,x-2\}.\) You said this already. Do you know how to prove that the set of these two polynomials make a basis for \(W^\perp\)?

[Displayer102]

11 minutes ago, taeto said:

The elements of the basis are polynomials. You correctly found that the polynomials in W⊥ have the form ax2+bx+c, where 2a+2b+c=0.

The basis of W⊥   has to be a set of two polynomials which satisfy this. You could choose {x2−2,x−2}. You said this already. Do you know how to prove that the set of these two polynomials make a basis for W⊥ ?

Don't you add them together and set it equal to zero. Then make it a matrix and row reduce them. then after all that its left with have c_1=0, c_2=0 and 0+0

 

But, I actually got those two polynomials from someone else and not 100% sure how to get to those and the person I got it from won't respond to me.

Edited May 5, 2020 by Displayer102

[taeto]

14 minutes ago, Displayer102 said:

Don't you add them together and set it equal to zero. Then make it a matrix and row reduce them. then after all that its left with have c_1=0, c_2=0 and 0+0

You are doing too much work. 

To show that \(x^2-2\)  and \(x-2\) are linearly independent polynomials, you have to show that the equation \(c_1 (x^2-2) + c_2 (x-2) = 0,\) as a polynomial equation, has \(c_1=c_2=0\) as the only solution. This is very easy, because \(c_1 (x^2-2) + c_2 (x-2) = c_1x^2 + c_2 x - 2(c_1 + c_2) = 0x^2+ 0x+0\) immediately tells you that \(c_1=0\) and \(c_2=0,\) just from looking at the coefficients of the powers of \(x\).

All that is left is to argue why \(W^\perp\) does not have additional polynomials in its basis other than these two.

Edited May 5, 2020 by taeto