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$P(2)$ is the space of polynomials of degree $\le2$ with the inner product defined by

$$(a_0 + a_1x + a_2x^2 , b_0 + b_1x + b_2x^2) = a_0b_0 + a_1b_1+ a_2b_2$$

and $P(x) = x^2 + x + 1$ and $W = \{ q(x) \text{ is an element of } P(2)\mid q(x) \text{ is perpendicular to } P(x)\}$

How do I prove that $ W$ is a subspace of $P(2)$ with dimension $2$ and what would be the orthogonal basis of $W$?

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Identify $ax^2 + bx + c = (c,b,a)$ in this specific order to make thing easier to work with, then: $P = (1,1,1)$, and $W = \{ (m,n,p) \in \mathbb{R}^3: m+n+p = 0\}$, and this identification means that $P_2 = \mathbb{R}^3$. We now check $W$ a subspace of $\mathbb{R}^3$.

$(m,n,p) + (h,k,l) = (m+h,n+k,p+l) \in W$ since $ m+h+n+k+p+l = (m+n+p)+(h+k+l) = 0 + 0 = 0$,

also $r(m,n,p) = (rm,rn,rp) \in W$ since $rm + rn + rp = r(m+n+p) = r\cdot 0 = 0$.

Thus: $W$ is a subspace of $P_2$.

To extract a basis $\mathcal{B}$ for $W$, we have: $p = -m-n$. Thus:

$(m,n,p) = (m,n,-m-n) = m(1,0,-1) + n(0,1,-1) \Rightarrow \mathcal{B} = \{(1,0,-1),(0,1,-1)\}$ as the basis, and clearly it has $2$ elements, so $\text{dim}(W )= 2$ as claimed.

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  • $\begingroup$ This is the orthogonal basis,right? $\endgroup$ Dec 9, 2014 at 7:24

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