Displayer102 Posted May 5, 2020 Share Posted May 5, 2020 (edited) 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 Link to comment Share on other sites More sharing options...

taeto Posted May 5, 2020 Share Posted May 5, 2020 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 \(W^{\perp}\)? Link to comment Share on other sites More sharing options...

Displayer102 Posted May 5, 2020 Author Share Posted May 5, 2020 (edited) 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 Link to comment Share on other sites More sharing options...

taeto Posted May 5, 2020 Share Posted May 5, 2020 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. Link to comment Share on other sites More sharing options...

Displayer102 Posted May 5, 2020 Author Share Posted May 5, 2020 (edited) 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 Link to comment Share on other sites More sharing options...

taeto Posted May 5, 2020 Share Posted May 5, 2020 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 \(W^\perp\)? Link to comment Share on other sites More sharing options...

Displayer102 Posted May 5, 2020 Author Share Posted May 5, 2020 (edited) 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 Link to comment Share on other sites More sharing options...

taeto Posted May 5, 2020 Share Posted May 5, 2020 (edited) 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 Link to comment Share on other sites More sharing options...

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