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Suppose $V=\{p(x) : p(x) \text{ is a polynomial so that its degree is less or equal to 2}\}$ and $W=\{p(x) : p(x) \text{ is a polynomial so that its degree is less or equal to 1}\}$ are two vector spaces. If $D: V \to W$ is a linear transformation given by $D(p(x))=p'(x)$ (the derivative). Find the matrix D related to the bases as stated in: $$D_{\{1,x,x^2\}}^{\{1,x\}}$$

My proposal for the matrix is \begin{equation*} D = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \end{equation*} for which I define it as a two row, because of the matrix multiplication as the outcome will have two rows too, also that $D(x^k)=kx^{k-1} $. Nevertheless, I am not a hundred percent sure this matrix works or if there is another argument that I still have to show so that this proof is complete, because when proving the general case with the standard basis for polynomials, the same base is used. Link to what I am refering to in the last line: Derivative matrix for n-th degree polynomial

zastenAmir
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1 Answers1

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You're answer is correct: you can easily check that $x^2 = (0,0,1)^t$ is being sent to

\begin{equation} A(0,0,1)^t = (0,2)^t \end{equation}

and $(0,2)^t$ represents the polynomial $2x$. So $x^3$ is being sent to the right polynomial. Similarly you can check that $1$ and $x$ are being sent to the right polynmoial. Since the basis is being sent to the right polynomials and differentiation is a linear transformation, this operator is precisely the derivative operator.

General Grievous
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