These were some text book questions with no solutions, and I'm still trying to grasp the idea of inner products and orthogonal basis, so if anyone could tell if if I did something wrong, or confirm that I'm on the right track, that would be amazing!
Question: Consider $P_2(\mathbb{R})$ with inner product $\langle p(x),q(x) \rangle = p(−1)q(−1) + p(0)q(0) + p(1)q(1).$
Where $(P_2(\mathbb{R})$ is the set of all polynomials of at most degree two with real coefficients)
a) Find the set of all vectors in $P_2(\mathbb{R})$ which are orthogonal to $x + x^{2}$.
b) Determine the coordinates of $1 + x^{2}$ with respect to the orthogonal basis $B = \{ 1 − x − 2x^{2}, 2 − x^{2}, 2 + 3x − 4x^{2} \}$
a) Let $a+bx+cx^2 \in P_2(\mathbb{R})$. Then we have
$\langle a+bx+cx^{2}, 0 \rangle =0$
$(a-b+c)(0) + (a)(0) + (a+b+c)(2) = 0$
$2a+2b+2c=0$
Thus we need $a+b+c=0$.
Thus, we get
$a = -b -c$, and as a result, the general vector becomes
$-b-c +bx + cx^2$
$= b(-1+x)+c(-1+x^2) $
Here I'm kind of confused how to describe this result as a set?
so all forms of polynomials in $P_2(\mathbb{R})$ that will be orthogonal to $x+x^{2}$ will have the from $b(-1+x)+c(-1+x^2)$
So does the set, { $b(-1+x)+c(-1+x^2) \in P_2(\mathbb{R}) | b,c \in \mathbb{R}$} , make sense?
b) For b), I just used the formula given in class and got the vector:
$\left(\frac{-3}{5}, 1, \frac{-1}{5}\right)$.
Can someone help me come up with the set for part a) and tell if I'm on the right track for parts a) and b)?
Thank you so much!
Regarding a):
We want to find all vectors in $P_2(\mathbb{R})$ which are orthogonal to $x+x^2$. By definition, this means we need to find $a+bx+cx^2$ with $a,b,c \in \mathbb{R}$ such that $$ \langle a+bx+cx^2 , x+x^2 \rangle = 0.$$ (You wrote $\langle a+bx+cx^2 , 0 \rangle = 0$, I assume that it was just a typo.)
Let us denote $a + bx + cx^2$ by $p(x; a,b,c)$ and $x+x^2$ by $q(x)$. Then we have $$p(-1; a,b,c) = a - b + c, \quad p(0; a,b,c) = a, \quad p(1; a,b,c) = a + b + c$$ and $$q(-1) = 0, \quad q(0) = 0, \quad p(1) = 2$$. Therefore, we obtain $$ \langle p(x; a,b,c) , q(x) \rangle = (a-b+c)\cdot 0 + a\cdot 0 + (a + b + c)\cdot 2 = 2a + 2b + 2c,$$ so any polynomial $p(x;a,b,c)$ which is orthogonal to $x+x^2$, i.e. $\langle p(x; a,b,c) , q(x) \rangle = 0$, must satisfy the condition $2a+2b+2c = 0$ which is equivalent to $a+b+c=0$ as you have shown us.
Your computation of the set of solutions $$ \{ b(-1+x) + c(-1+x^2) \in P_2(\mathbb{R}) \, | \, b,c \in \mathbb{R} \}$$ is correct. I would like to add that this is exactly the subset of $P_2(\mathbb{R})$ generated by $-1+x$ and $-1+x^2$, i.e. $\operatorname{span}(-1+x, -1+x^2)$.
Regarding b):
Determining the coordinates of $1+x^2$ with respect to the orthogonal basis $B = \{1-x-2x^2, 2-x^2, 2+3x-4x^2\}$ means that we need to find the unique real coefficients $c_1,c_2,c_3 \in \mathbb{R}$ such that $$x+x^2 = c_1 (1-x-2x^2) + c_2 (2-x^2) + c_3 (2+3x-4x^2). $$
Because $B$ is an orthogonal basis, we will make use of the following
Accoding to the Lemma, it must be $$\begin{align} c_1 &= \frac{\langle x+x^2 , 1-x-2x^2 \rangle}{\langle 1-x-2x^2 , 1-x-2x^2 \rangle} = \frac{0 \cdot 0 + 0 \cdot 1 + 2 \cdot (-2)}{0^2+1^2 + (-2)^2} = -\frac{4}{5},\\ c_2 &= \frac{\langle x+x^2 , 2-x^2 \rangle}{\langle 2-x^2 , 2-x^2 \rangle} = \frac{0 \cdot 1 + 0 \cdot 2 + 2 \cdot 1}{1^2+2^2+1^2} = \frac{2}{6} = \frac{1}{3}, \\ c_3 &= \frac{\langle x+x^2 , 2+3x-4x^2 \rangle}{\langle 2+3x-4x^2 , 2+3x-4x^2 \rangle} = \frac{0 \cdot (-5) + 0 \cdot 2 + 2 \cdot 1}{(-5)^2+2^2+1^2} = \frac{2}{30} = \frac{1}{15}, \end{align}$$ so the coordinate vector would be $(-\frac{4}{5}, \frac{1}{3}, \frac{1}{15})$ which is different from your solution.
Unless I used a wrong vector, my computation should be correct according to WolframAlpha.