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I have to determine a polynomium $q \in P_3(\mathbb{R})$ where the coordinat vector $[q]_v$ is equal to v where v is a vector given by $$ v = \begin{pmatrix} \alpha \beta \\ -\alpha -\beta \\ 1 \end{pmatrix} $$ My books has a note saying that the coordinat vector $[q]_v$ is given by the vector $$ \begin{pmatrix} a_1 \\ a_2 \\ ... \\ a_n \end{pmatrix} \in P_3(\mathbb{R}) $$ where $$ v = a_1 \cdot q_1 + a_2 \cdot q_2 + a_3 \cdot q_3 $$ where $Q = (q_1,q_2,q_3)$ is a basis given by $Q = (1,X,X^2)$. I am not really sure that I am explaining this correct so I have this is correct so far.

In regards I have then said that the polynomium q will be $$ q = \frac{1}{X^2} + \frac{-\alpha -\beta}{X} + \alpha \beta $$ Is this correct? Or am I doing it completely wrong?

Thanks in advance.

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It is not correct. We have

$$q=\alpha \beta q_1-(\alpha + \beta)q_2+q_3,$$

hence

$$q(X)=\alpha \beta -(\alpha + \beta) X+X^2.$$

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  • $\begingroup$ Oh ye, sure. Makes sense. Thank you! $\endgroup$ – Mathias Mar 12 at 13:28

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