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Let $V:=\mathbb R_{\le 3}[x]:=\{p:\mathbb R\to \mathbb R \mid p(x)=a_0 +a_1x +a_2x^2+a_3x^3, a_0,a_1,a_2,a_3 \in \mathbb R\}$ and $W:=\mathbb R_{\le 2}[x]:=\{q:\mathbb R\to \mathbb R \mid p(x)=b_0 +b_1x +b_2x^2, b_0,b_1,b_2 \in \mathbb R\}$ Let $B=(1,x,x^2,x^3)$ a basis of V and $\cal B$$=(1,x,x^2)$ a basis of W. Consider the function of differentation, i.e. $d:V\to W,p\to p'$

$d(1)=0$, $d(x)=1$,$d(x^2)=2x$,$d(x^3)=3x^2$

Then we get the transformation matrix given by

\begin{pmatrix} 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&3 \end{pmatrix}

My question is now, what would be if we considered $\cal V:=\mathbb R_{\le 4}[x]:=\{p:\mathbb R\to \mathbb R \mid p(x)=a_0 +a_1x +a_2x^2+a_3x^3+a_4x^4, a_0,a_1,a_2,a_3,a_4 \in \mathbb R\}$ instead of $V$ with corresponding basis $B=(1,x,x^2,x^3,x^4)$. We would not be able to write the derivation of the monom $x^4$ as a linear combination of elements of $\cal B$. Some enlightenment about this would be appreciated!

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  • $\begingroup$ It is simply not even a function, that's all. $\endgroup$ – TheGeekGreek Mar 21 '18 at 21:47
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Yes we can't, to do that we need a basis $\cal B=(1,x,x^2,x^3)$.

As an alternative you can consider derivation from $\cal V \to \cal V$ using the same basis $\cal B=(1,x,x^2,x^3,x^4)$. Canyou decribe the matrix in this case?

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  • $\begingroup$ I just updated my question. I changed the basis, but this would be still not possible, but why does this violate the definiton. $\endgroup$ – user519338 Mar 21 '18 at 20:28
  • $\begingroup$ @J.DWhy you want change basis? you can work with the same one with a 5-by-5 matrix. $\endgroup$ – gimusi Mar 21 '18 at 20:32
  • $\begingroup$ The transformation matrix is a $5x5$ matrix with every entry zero with diagonal 0,1,2,3,4 $\endgroup$ – user519338 Mar 21 '18 at 20:32
  • $\begingroup$ I changed the basis because otherwise it would be no basis of $R_{\le 4}$ any longer $\endgroup$ – user519338 Mar 21 '18 at 20:34
  • $\begingroup$ @J.D you can use this one with an unique basis $$\begin{pmatrix} 0&1&0&0&0 \\ 0&0&2&0&0 \\ 0&0&0&3&0\\0&0&0&0&4\\0&0&0&0&0 \end{pmatrix}$$ $\endgroup$ – gimusi Mar 21 '18 at 20:36

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