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Find the matrix of a linear transformation $f:P_2(x,\mathbb R) \to P_2(x,\mathbb R)$ satisfying $p(x) \to p'(x) - 2p(x)$, where $p'(x)$ is the derivative of the polynomial $p(x)$ in the respect of those bases: $ M = \{x^2 + 2x +1, 2x^2 + 1, x^2 -x\}$ and $N= \{x^2 +2, x^2 -3x +1, x^2 + x +3\}$.

My attempt:

$p(x) = (ax^2 + bx + c) \to (2ax + b - 2ax^2 - 2bx - 2c) = p'(x) - 2p(x)$

We need to use the canonical basis: $\{ 1, x, x^2\}$ and find the coordinates of that mapped polynomial.

$p(1) = -2$

$p(x) = 1 -2x$

$p(x^2) = -2x^2 + 2x$

and I put those results into the matrix: $$ f_{KK} = \begin{pmatrix} -2 & 1 & 0 \\ 0 & -2 & 2 \\ 0 & 0 & -2 \\ \end{pmatrix} $$

The wanted matrix of that linear transformation would be $f_{NM}$ and we could get it from this equation:

$ f_{NM} = (1_P)_{NK} * f_{KK} * (1_P)_{KM}$

We can rewrite our bases M and N as:

$$ M = \begin{pmatrix} 1 & 1 & 0 \\ 2 & 0 & -1 \\ 1 & 2 & 1 \\ \end{pmatrix} $$

$$ N = \begin{pmatrix} 2 & 1 & 3 \\ 0 & -3 & 1 \\ 1 & 1 & 1 \\ \end{pmatrix} $$

$$ \begin{pmatrix} 2 & 1 & 3 \\ 0 & -3 & 1 \\ 1 & 1 & 1 \\ \end{pmatrix} * X = E $$

where E is an identity matrix and $(1_P)_{NK} = X$,... $(1_P)_{KM} = M$

$$(1_P)_{NK}= \begin{pmatrix} -2 & 1 & 5 \\ 1/2& -1/2 & 1 \\ 3/2 & -1/2 & -3 \\ \end{pmatrix} $$

after multiplication I got: $$(f)_{NM}= \begin{pmatrix} -12 & -12 & -4 \\ -1& -7 & -9/2 \\ 7 & 7 & 5/2 \\ \end{pmatrix} $$

Am I wrong?

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    Are (2,3) and (3,3) entries of $M$ having wrong sign? – 2017-01-31
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    And why do you suppose you are wrong in the end? Do you have a solution? – 2017-01-31
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    No unfortunately, I am just very insecure in this subject and the matrix has weird numbers. I will edit that last sentence I have written. – 2017-01-31
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    Is it an endomorphism? – 2017-01-31
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    Every linear transformation written in the same basis for input and output, is an endomorphism. Answer me, are (2,3) and (3,3) entries of $M$ correct? Don't you suppose $\{1,x,x^2\}$ basis? – 2017-01-31
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    Yeah, I suppose that canonical basis but I don't see any error. – 2017-01-31
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    $2x^2+1$ is $\{1,0,2\}$, but $x^2-x$ is $\{0,1,-1\}$, why – 2017-01-31
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    I see, I did a mistake. But just here, not in the paper where I was solving this problem. – 2017-01-31
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    Why don’t you check your work by trying a couple of examples? – 2017-01-31

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