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Find matrix of linear operator $\mathcal{A} : \mathcal{P_5}\rightarrow \mathcal{P_5},\mathcal{A}(p)=-2p+(3x-1)p^{'}$

$\mathcal{P_5}$ is the space of polynomials with degree not greater than $5$.

$\mathcal{A}(p)=(-2p-p^{'})+3p^{'}x+0x^2+0x^3+0x^4+0x^5$

$$\mathcal{A} \begin{bmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \\ \end{bmatrix}= \begin{bmatrix} -2p_1-p_1^{'} \\ 3p_2^{'} \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} $$

How to get matrix of $\mathcal{A}$?

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  • $\begingroup$ You’re not going to get very far with only five components in the vector on the lhs. $\endgroup$ – amd Dec 7 '15 at 23:33
  • $\begingroup$ $\mathcal{P}_5$ is has dimension 6 and i guess $p'$ denotes the derivative of $p$. $\endgroup$ – testman Dec 7 '15 at 23:35
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Let $p(x) = ax^5+bx^4+cx^3+dx^2+ex+f$ be an arbitrary polynomial in $\mathcal{P}_5$. Then $$\begin{align} (\mathcal{A}p)(x)&= -2p(x)+(3x-1)p'(x) \\ &= -2p(x)+(3x-1)(5ax^4+4bx^3+3cx^2+2dx+e) \\ &=13ax^5+(10b-5a)x^4+(7c-4b)x^3\\ &\quad +(4d-3c)x^2+(e-2d)x-2f-e. \end{align}$$

Using the canonical transformation $T:\mathbb{R}^6 \to \mathcal{P}_5, (a,b,c,d,e,f)^T\mapsto p$ you can write $\mathcal{A}p = TBT^{-1}p$ where $$B = \begin{bmatrix} 13 & 0 & 0 & 0 & 0 & 0 \\ -5 & 10 & 0 & 0 & 0 & 0 \\ 0 & -4 & 7 & 0 & 0 & 0 \\ 0 & 0 & -3 & 4 & 0 & 0 \\ 0 & 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & -2 \\ \end{bmatrix}.$$

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