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For a set $N$ let $id_N:N \rightarrow N$ be the identical transformation. Be $V:=\mathbb{R}[t]_{\le d}$. Determine the matrix representation $A:=M_B^A(id_V)$ of $id_V$ regarding to the basis $A=\{1,t,...,t^d\}$ and $B=\{1,(t-a),...,(t-a)^d\}$.

I know, that i have to write the $t^i$ as a linear combination of $(t-a)^j$. So for

$t^0 = 1*(t-a)^0$

$t^1 = a*(t-a)^0 + 1*(t-a)^1$

$t^2 = (-a^2 + 2a^2) * (t-a)^0 + 2a*(t-a)^1 + 1*(t-a)^2$

$t^3= a^3(t-a)^0 -a^2(t-a)^1+a(t-a)^2+1(t-a)^3$

$...$

How can i figure out a system for the general case?

So the matrix representation is \begin{pmatrix} 1 & 0 & 0 & 0 & 0 &\dots & 0\\ a & 1 & 0 & 0 & 0 &\dots & 0\\ a^2 & \binom{2}{1} a^{2-1} & 1 & 0 & 0 &\dots & 0\\ \vdots & \dots & & \ddots & \vdots\\ a^d & \binom{2}{1} a^{d-1} &\dots &\binom{d}{i} a^{d-i} &\dots & &1 \end{pmatrix} ? Can someone please tell me, if this is correct?

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Do the binomial expansion $$ (z + a)^{k} = a^{k} + \binom{k}{1} a^{k-1} z + \dots + \binom{k}{i} a^{k-i} z^{i} + \dots + z^{k}, $$ and then substitute $t = z + a$.

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How to subitute $a^{k-1}z$? For $a^k$ i've got $(t-z)^k$, and for $a^{k-1}z$ i've got $(t-z)^k*\frac{z}{a}$ – fear.xD May 18 '14 at 14:06
    
No, no, you have to substitute $t = z+a$ in the LHS, and $z = t - a$ in the RHS. – Andreas Caranti May 18 '14 at 14:22
    
So $t^k=a^k+ \binom{k}{1} a^{k-1} (t-a) + \dots + \binom{k}{i} a^{k-i} (t-a)^{i} + \dots + (t-a)^{k}$? – fear.xD May 18 '14 at 14:31
    
Yes, this is it. – Andreas Caranti May 18 '14 at 14:36

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