Surjective polynomial map from $M_n(F)$ to $M_n(F)$

Solution 1:

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 $A$.