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let V be the vector space of all polynomials in the variable x with real coefficients.in the

space V:

        D(P(X))=P'(x) (the derivative of P(x))

        M(P(x))=xP(x)

prove that both D and M are linear transformations such that DM-MD=I(the identity linear map

on V.

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check that both $M$ and $D$ preserve vector space operations, i.e. sum and scalar multiplication –  Federica Maggioni Apr 22 '13 at 10:38
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for $DM-MD=I$ use the Leibniz rule for the derivative of a product: $D(x\cdot f)=D(x)\cdot f+x\cdot D(f)$ and the fact $D(x)=1$ –  Federica Maggioni Apr 22 '13 at 10:41
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1 Answer 1

up vote 1 down vote accepted

$$(D(aP_1(x)+bP_2(x)))=aD(P_1(x))+bD(P_2(x))$$

and

$$(M(aP_1(x)+bP_2(x)))=x(aP_1(x)+bP_2(x)))=x(aP_1(x))+x(bP_2(x))=$$

$$=aM(P_1(x))+ bM(P_1(x))$$

$$(DM-MD)(P(x))=DM(P(x))-MD(P(x))=D(x(P(x)))-M(P^{'}(x))=$$

$$=P(x)-xP^{'}(x)-x(P^{'}(x))=P(x)+xP^{'}(x)-xP^{'}(x)$$ so $\;DM_MD=I\;$

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