The original question reads:
Let $H : P_3 → P_1$ be the linear mapping defined by $H(p(x)) = \frac{d^2p}{dx^2}$. Find the matrix of $H$ with respect to the basis {$1, x, x^2, x^3$} of $P_3$ and the basis {$1, x$} of $P_1$.
My answer was as follows:
$H(1) = 0, H(x) = 0, H(x^2) = 2, H(x^3)=6x$.
Hence forming the matrix: $$ \begin{bmatrix} 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
Is this correct?
Secondly, I need help proving that this transformation $H$ is surjective. Any pointers/ help on how to do this? Many thanks.
No, it is not correct. It is a map from a space with dimension $4$ into a space with dimension $2$, and so the answer cannot be a $4\times4$ matrix. The answer is:$$\begin{bmatrix}0&0&2&0\\0&0&0&6\end{bmatrix}.$$And, yes, $H$ is surjective. Since the range of $H$ contains both $2$ and $6x$, it contains all of their linear combinations. And the set of all their linear combinations is $P_1$.