Displayer102

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.

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]

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]

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?