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$V$ is an inner product space and $g: V \rightarrow F$ is a linear transformation.
Find a vector $y$ such that $g(x)=\langle x,y \rangle$ for all $x \in V$
$V=P_2(R)$ with $\langle f,h \rangle=\int_{0}^{1}f(t)h(t)dt$, $g(f)=f(0)+f'(1)$.

I know, by formula, $y=\sum_{i=1}^{n} \bar{g(v_i)}v_i$ where {$v_1,\dots,v_n$}=$\beta$ is an orthonomal basis.
In that problem, the basis $\beta=(1,x,x^2)$ is not orthonormal so I have to use gram-schmidt and normalize it. I tried to calculate them but the answer $y=210x^2-204x+33$ is not achievable.
Of course there can be other ways to solve this, but I want to solve it by standard way (using gram-schmidt). I need your help. How to get the answer?

noname
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1 Answers1

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You can solve the problem directly, albeit you need to invert a $3 \times 3$ matrix:

Let $e_k(x) = x^k$. If $f=\sum_{i=0}^2 f_i e_i$, then $g(f) = f_0+f_1+2 f_2$.

If $y = \sum_{i=0}^2 y_i e_i$, then $\langle f,y \rangle = \sum_{i=0}^2 \sum_{j=0}^2 f_i y_j \int_0^1 x^{i+j} dx = \sum_{i=0}^2 \sum_{j=0}^2 f_i y_j \frac{1}{i+j+1}$.

We want to find $y$ such that $g(e_k) = \sum_{j=0}^2 y_j \frac{1}{k+j+1}$. Note that $g(e_0) = 1$, $g(e_1) = 1$ and $g(e_2) = 2$. If we let $A = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{bmatrix}$, and $b = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$, then the problem can be written as $A \begin{bmatrix} y_0 \\ y_1 \\ y_2 \end{bmatrix} = b$, which has the solution $\begin{bmatrix} y_0 \\ y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} 33 \\ -204 \\210 \end{bmatrix} $.

Hence $y = 33 e_0 -204 e_1 + 210 e_2$.

copper.hat
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