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Problem
Let $V$ be the vector space over $\mathbb{R}$ spanned by Legendre Polynomials:

$P_0(x)=1$,$\space\space$$P_1(x)=x$,$\space\space$$P_2(x)=\frac{1}{2}(3x^2-1)$,$\space\space$$P_3(x)=\frac{1}{2}(5x^3-3x)$

Consider the map $H:V\rightarrow V, f\mapsto f^{\prime\prime}$, where $f^{\prime\prime}$ denotes the second derivative of f. You may assume $H$ is a linear transformation.

$(a)$ Show that $\beta:=\{P_0,P_1,P_2,P_3\}$ is a basis for V.
$(b)$ Determine the matrix representation of $H$ with respect to basis $\beta$.

Attempt
$(a)$ So getting the linear combination, $p(x)=a_0+a_1x+a_2x^2+a_3x^3\in P_3$ and finding co-efficients $c_0$, $c_1$, $c_2$, $c_3$, I ended up with:

$$a_0=2c_0 - c_2$$ $$a_1=2c_1-3c_3$$ $$a_2=3c_2$$ $$a_3=5c_3$$

So when $a_0=a_1=a_2=a_3=0$, $c_0=c_1=c_2=c_3=0$. This shows the set spans $V$ as it is linearly independent and therefore it is a basis.

$(b)$ I'm not sure how to do this part and I can't find any clear instructions for a question like this. Any help would be greatly appreciated.

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  • $\begingroup$ For (b), you should consider writing the matrix representation of the second derivative operator with respect to the monomial basis first. $\endgroup$ – Chester Aug 13 '15 at 0:44
  • $\begingroup$ $P_{3}''=15x=15P_{1}$, $P_{2}''=3=3P_{0}$, $P_{1}''=0$, $P_{0}''=0$. $\endgroup$ – DisintegratingByParts Aug 13 '15 at 3:58
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Starting with $$ \begin{align} P_{0}'' & = 0P_{0}+0P_{1}+0P_{2}+0P_{3} \\ P_{1}'' & = 0P_{0}+0P_{1}+0P_{2}+0P_{3} \\ P_{2}'' = 3 & = 3P_{0}+0P_{1}+0P_{2}+0P_{3} \\ P_{3}'' = 15x & = 0P_{0}+15P_{1}+0P_{2}+0P_{3}, \end{align} $$ which gives the following matrix representation of $\frac{d^{2}}{dx^{2}}$: $$\left[\begin{array}{cccc}0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 15 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right]$$ This assumes that $\alpha_0 P_0 + \alpha_1 P_1 + \alpha_2 P_2 + \alpha_3 P_3$ is written as $$ \left[\begin{array}{c}\alpha_0 \\ \alpha_1 \\ \alpha_2 \\ \alpha_3\end{array}\right] $$

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  • $\begingroup$ I don't think I could have gotten a clearer explanation, thanks a million! $\endgroup$ – teme92 Aug 13 '15 at 8:26
  • $\begingroup$ @teme92 : You're welcome. $\endgroup$ – DisintegratingByParts Aug 13 '15 at 10:28

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