CSIR JUNE 2011 PART B QUESTION 35 SOLUTION

Let $W$ be the vector space of all real polynomials of degree at most $3$. Define $T:W \to W$ by $(Tp)(x) = p^{'}(x)$ where $p^{'}$ is the derivative of $p$. The matrix of $T$ with respect to the basis $\{1,x,x^2,x^3\}$, considered as a column vector is given by
1)$\begin{bmatrix}0&0&0&0\\ 0&1&0&0\\ 0&0&2&0\\ 0&0&0&3\end{bmatrix}$,
2) $\begin{bmatrix}0&0&0&0\\ 1&0&0&0\\ 0&2&0&0\\ 0&0&3&0\end{bmatrix}$
3)$\begin{bmatrix}0&1&0&0\\ 0&0&2&0\\ 0&0&0&3\\ 0&0&0&0\end{bmatrix}$
4)$\begin{bmatrix}0&1&2&3\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\end{bmatrix}$
Solution:
We will calculate the image of the basis elements. First column : $$(Tp)(1) = 0 \cdot 1+ 0 \cdot x + 0 \cdot x^2 + 0 \cdot x^3.$$ Second column : $$(Tp)(x) = 1 \cdot 1+ 0 \cdot x + 0 \cdot x^2 + 0 \cdot x^3.$$Third column : $$(Tp)(x^2) = 0 \cdot 1+ 2 \cdot x + 0 \cdot x^2 + 0 \cdot x^3.$$ Fourth column : $$(Tp)(x^3) = 0 \cdot 1+ 0 \cdot x + 3 \cdot x^2 + 0 \cdot x^3.$$
But we have to encode this data as column matrix. Hence the required matrix is (3) $\begin{bmatrix}0&1&0&0\\ 0&0&2&0\\ 0&0&0&3\\ 0&0&0&0\end{bmatrix}$.
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