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A linear transformation between polynomial spaces $f: P_2(\mathbb{R}) \to P_2(\mathbb{R})$ is given by $$f(p(x))=3 \cdot p(1)-x^2 \cdot p(0)+(x-1) \cdot p'(1)$$ Determine the transformation matrix with respect to the monomial basis $(1,x,x^2)$

I tried to approach it this way:

The general form of these polynomials is $ax^2+bx+c$. We have $$P(1)=a+b+c \text{ and } P(0)=c \text{ and } P'(1)=2a+b$$

Therefore $f(p(x))=3(a+b+c)-cx^2+2ax+bx-2a-b$$

I therefore thought that the transformation matrix would be $$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$

What am I doing wrong?

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  • $\begingroup$ When you work out $f(p(x))$, $p'(1)=2a+b$, not $2ax+b$ $\endgroup$ – A. Goodier Dec 9 '17 at 12:24
  • $\begingroup$ Ah I see - I'll edit $\endgroup$ – Alex5207 Dec 9 '17 at 13:07
  • $\begingroup$ @Alex5207 Don't forget, if you feel my answer is satisfactory you can mark the answer as correct. $\endgroup$ – user3002473 Dec 9 '17 at 17:13
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Well, if you have a basis $\beta =\{1, x, x^2\}$, and we need to determine the matrix of $f$ wrt $\beta$, then let's first write out $f$ in the way that makes it the most obvious how it's transforming it's coordinates:

$$ \begin{aligned} f(ax^2+bx+c) = f((a, b, c)) & = 3(a + b + c) - cx^2 + (2a + b)x - (2a + b) \\ & = (-c, 2a+b, 3(a+b+c)-(2a+b)) \\ & = (-c, 2a+b, a+2b+3c) \end{aligned} $$ I'm using $(a,b,c)$ to denote $ax^2+bx+c$ to emphasize the fact that these are coordinates in $P_2(\mathbb R)$. So, if our matrix for $f$ is $$ [f]_\beta = \begin{pmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{pmatrix} $$ then notice that the first coordinate in $[f]_\beta(a,b,c)^\text{T}$ is given by the dot product of the first row with $(a, b, c)$, so $a_{11}a + a_{12}b + a_{13}c = -c$, implying $a_{13}=-1$, and $a_{11}=a_{12}=0$. We can continue this process for each row, since the $i$th coordinate of $[f]_\beta(a,b,c)^\text{T}$ is given by the dot product of the $i$th row with $(a, b, c)$. Hence $$ a_{21}a + a_{22}b + a_{23}c = 2a + b \\ a_{31}a + a_{32} b + a_{33}c = a + 2b + 3c $$ from which we can see that $(a_{21}, a_{22}, a_{23}) = (2, 1, 0)$, and $(a_{31}, a_{32}, a_{33}) = (1, 2, 3)$. Hence $$ [f]_\beta = \begin{pmatrix}0 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & 2 & 3\end{pmatrix} $$ So you were on the right track, you just had your rows in the reverse order.

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