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Let V be the vector space of all polynomials of degree less than 4 and W those of degree less than 3. Define linear transformation D from V to W by $Dp=p'$. Define linear transformation I from W to v by $I(q)=\int_0^xq(t)dt$. Find the matrices of $D$ and $D\circ I$.

First of all, what is a matrix of a linear map? Is it like $x \to Ax$?

Is the basis of $V=\{1,x,x^2,x^3\}$?

Unfortunately, Thats all I know. If this post is against the guidelines, can you post some resources which will help me solve the question?

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    For your first question: https://en.wikipedia.org/wiki/Transformation_matrix – 2017-02-18
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    @Ant so if use the basis of V (the one I mentioned in the post), will I get the matrix of linear transform? – 2017-02-18
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    @AdityaDev - That is *a* basis for $V$. There are of course many others. But I suspect that is the one you are expected to use. Probably this is specified somewhere in the context of this question (I'd look carefully, to make sure the context doesn't specify a different basis instead). You know how differentiation and integration of polynomials work, so this shouldn't be too hard to figure out. – 2017-02-18

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To simplify the answer, assume that $\{1, x, x^2, x^3\}$ is the ordered basis for $V,$ and $\{1, x, x^2\}$ for $W.$

The matrix of $D$ is a $3 \times 4$ matrix $A$ such that $[Dp] = A[p]$ where $[Dp]$ and $[p]$ are coordinate matrices, which, in this case, are $3 \times 1$ column matrices whose entries are the coefficients of the ordered basis elements of $W.$

We find $A$ as follows: The $i$th column of $A$ is $[Dp_i]$ where $p_i$ is the $i$th element in the ordered basis of $V.$ Hence, $$\begin{align} [Dp_1] & = [D(1)] = [0] = [0 \cdot 1 + 0x + 0x^2] = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix},\\\\ [Dp_2] & = [D(x)] = [1] = [1 \cdot 1 + 0x + 0x^2] = \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix},\\\\ [Dp_3] & = [D(x^2)] = [2x] = [0 \cdot 1 + 2x + 0x^2] = \begin{bmatrix} 0\\ 2\\ 0\end{bmatrix}, \mbox{ and}\\\\ [Dp_4] & = [D(x^3)] = [3x^2] = [0 \cdot 1 + 0x + 3x^2] = \begin{bmatrix} 0\\ 0\\ 3\end{bmatrix}. \end{align}$$ Thus, the matrix of $D$ is $$\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 3\end{bmatrix}.$$

If you go through the same procedure for $D \circ I,$ you find that the matrix of $D \circ I$ is $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix},$$ although if you see that $D \circ I$ is the identity operator on $W,$ you can skip straight to that answer.