Is $A$ a surjective map? Let $P_3(\mathbb{R})$ denote the vector space of polynomials of degree at most three. Let $A:P_3(\mathbb{R})\to M_{2}(\mathbb{R})$ be defined as $A(p)= \begin{pmatrix} p(1) & p(2)\\
p(3) & p(4) \end{pmatrix}$. Is $A$ a surjective map?
I don't think that $A$ will be a surjective map. If we consider the zero matrix, then we will get four roots of the polynomial but any polynomial in $P_{3}(\mathbb{R})$ can have at most $3$ real roots.
This question appeared in a test I took yesterday and the answer key is showing that $A$ is a surjective map. Can someone please tell me if my reasoning is correct or am I missing something?
 A: The map $A$ is indeed onto.
The zero matrix is the image of the zero polynomial.
To prove surjectivity, you can use Lagrange polynomials to make a constructive proof.
Another proof, can be based on noticing that $A$ is a linear map between linear spaces of the same dimension. So proving surjectivity is equivalent to prove injectivity or to prove that the kernel is reduced to the zero polynomial. Which is quite immediate: the zero polynomial is the only polynomial of degree less or equal to $3$ vanishing at $1,2,3,4$.
A: You are missing something : the constant polynomial zero.
In fact, you actually have a question relating to vector spaces : note that $A$ is a linear transformation, therefore it is sufficient to take a basis of $M_2(\mathbb R)$ and show that each such basis element is in the range of $A$.
However, let $P(x) = (x-1)(x-2)(x-3)(x-4)$, and consider $p_i(x) = \frac{P(x)}{(x-i)}$ for $i=1,2,3,4$. The $p_i$ belong in $P_3(\mathbb R)$. What are $Ap_i$? Are they forming a basis of $M_2(\mathbb R)$?
Note : one can also prove injectivity of $A$, which is equivalent to surjectivity because the dimensions of the two vector spaces are the same. This follows from the fact that any polynomial of degree three having four zeros is identically zero, as you point out.
A: Let $A = \left[\begin{matrix} x & y \\
z & w \end{matrix}\right]$. Then, we need  $p(x) = a_3x^3+a_2x^2+a_1x+a_0$ such that $$a_3+a_2+a_1+a_0 = x \\
8a_3+4a_2+2a_1+a_0 = y \\
27a_3+9a_2+3a_1+a_0 = z \\
64a_3+16a_2+4a_1+a_0 = w$$
The existence of such a polynomial is subject to a solution for this system of equations, which exists always here (why?).
