Surjective polynomial map from $M_n(F)$ to $M_n(F)$ Assume that $F$ is a field, and that $f \in F[x]$ is a polynomial. If $f$ is surjective for every sufficient large $n$ when we regard $f$ as a map from $M_n(F)$ to $M_n(F)$, must $f$ be linear? 
 A: A partial answer (see also the more substantial answer below): the following is not true

If $f$ is surjective for a fixed value $n > 1$ when we regard $f$ as a map from $M_n(F)$ to $M_n(F)$, then $f$ must be linear

For instance, with $n = 2$ and $F = \Bbb C$, we find that the polynomial $f(x) = x^3 - x = x(x - 1)(x + 1)$ is surjective as a function on $M_n$.  To show that this is the case: first, we note that $f$, taken as a function over $\Bbb C$, is surjective.  From there, it's clear that the image of $f$ contains all diagonalizable matrices.  It therefore suffices to show that $f$ also contains all matrices similar to the Jordan block
$$
J(\lambda) = \pmatrix{\lambda & 1\\0 & \lambda}
$$
In fact, since the image of $f$ is invariant under conjugation, it suffices to note that the image of $f$ contains a matrix similar to the above, for arbitrary $\lambda \in \Bbb C$.
We note that for any $\mu \neq \pm 1/\sqrt{3}$, if we take
$$
A = \pmatrix{\mu & 1\\0 & \mu}
$$
we find that
$$
f(A) = \pmatrix{f(\mu) & 3 \mu^2 - 1 \\0 & f(\mu)}
$$
is similar to $J(f(\mu))$.  This means that the only problematic eigenvalues are $\lambda = f(\pm \sqrt{3}) = \pm \frac{2}{3\sqrt{3}}$.  We see that 
$$
f(\mu) = \frac{2}{3\sqrt{3}} \implies \mu = -1/\sqrt{3},2/\sqrt{3}
$$
So the matrix $f(J(\frac{2}{\sqrt{3}}))$ is similar to $J(\frac{2}{3\sqrt{3}})$.  Similarly, 
$$
f(\mu) = -\frac{2}{3\sqrt{3}} \implies \mu = 1/\sqrt{3},-2/\sqrt{3}
$$
So the matrix $f(J(-\frac{2}{\sqrt{3}}))$ is similar to $J(-\frac{2}{3\sqrt{3}})$
So, in this case we find that although $f$ is non-linear, $f:M_2(\Bbb C) \to M_2(\Bbb C)$ is surjective.

A more substantial answer:
In fact, we can see that for any $n$, the map $f:M_n \to M_n$ will be surjective. For any $k$ from $1$ to $n$, define $J_k(\lambda)$ to be the $k \times k$ Jordan block
$$
J_k(\lambda) = \pmatrix{\lambda & 1\\&\lambda&1\\&&\ddots&\ddots\\&&&&1\\&&&&\lambda}
$$
We see that for any $\mu \in \Bbb C \setminus \{\pm 1/\sqrt{3}\}$, $f(J_k(\mu))$ is similar to $J_k(f(\mu))$.  To see this: note that $f(J(\mu)) - f(\mu)I$ has rank $k-1$, so long as $f'(\mu) \neq 0$.  Moreover, $f(J(\mu))$ has $f(\mu)$ as its only eigenvalue. Seeing that $[f(J(\mu)) - f(\mu)I]^{n-1} \neq 0$ is enough to reach the desired conclusion.
Since $f:\Bbb C \setminus \{\pm 1/\sqrt{3}\} \to \Bbb C$ is surjective, we conclude that $f$ is surjective.  In particular: if 
$$
A = J_{k_1}(\lambda_1) \oplus \cdots \oplus J_{k_m}(\lambda_m)
$$
(where $\oplus$ is a diagonal direct sum) then we can select $\mu_j \in \Bbb C \setminus \{\pm 1/\sqrt{3}\}$ such that $f(\mu_j) = \lambda_j$, and we see that
$$
f[J_{k_1}(\mu_1) \oplus \cdots \oplus J_{k_m}(\mu_m)] = \\
f(J_{k_1}(\mu_1)) \oplus \cdots \oplus f(J_{k_m}(\mu_m))
$$
is similar to $A$.
