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This is pretty much a check-if-I'm-right sort of question.

Let $φ:K[x]_{\le{n}}\to K[x]_{\le{n-1}}$ with $\varphi$ the linear transformation defind by $\varphi(f)=f'$. Select a base and find the matrix of the linear transformation.

I took the standard basis for grade-$n$ polynomials:

$B =<1,x,x^2,\dots,x^n> ,φ(B)=φ(1,x,x^2,...,x^n) = (0,2x,...,nx^{n-1})$

So, the matrix is $\begin{bmatrix}0\\2x\\\dots\\nx^{n-1} \end{bmatrix}$?

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  • $\begingroup$ The entris of the matrix are supposed to be elements of $K$. $\endgroup$ – José Carlos Santos Oct 1 '17 at 15:08
  • $\begingroup$ Aren't $<1,x^2,...,x^n>$ elements of $K[x]_{\le{n}}$? $\endgroup$ – Alex Matt Oct 1 '17 at 15:30
  • $\begingroup$ Yes. Is that supposed to be relevant to what I wrote? $\endgroup$ – José Carlos Santos Oct 1 '17 at 15:36
  • $\begingroup$ Sorry, I meant $<0,2x,...nx^{n-1}>$. $\endgroup$ – Alex Matt Oct 1 '17 at 15:41
  • $\begingroup$ Again, is that supposed to be relevant to what I wrote? $\endgroup$ – José Carlos Santos Oct 1 '17 at 15:46
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Since $\varphi(1)=0$, $\varphi(x)=1$, $\varphi(x^2)=2x$, and so on, the matrix is$$\begin{bmatrix}0&1&0&0&\ldots&0&0\\0&0&2&0&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&0&n\end{bmatrix}.$$Note that this matrix has $n(=\dim K[x]_{\leq n-1})$ lines and $n+1(=\dim K[x]_{\leq n})$ columns.

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  • $\begingroup$ I don't really understand why it's constructed like this. From the 2nd column forward every pivot of that submatrix is the coefficient of each polynomial bu, I don't get why the first column is like so. $\endgroup$ – Alex Matt Oct 1 '17 at 16:59
  • $\begingroup$ $\varphi (1)=0$ $\endgroup$ – José Carlos Santos Oct 1 '17 at 17:02

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