# Is there a "weak" fundamental theorem of algebra for matrices?

Let $$R$$ be the ring of complex $$n\times n$$ matrices, where $$n>1$$.
Does every nonconstant polynomial in $$R[X]$$ have a root in $$R$$?

Note: The "strong" fundamental theorem of algebra for matrices fails, see here. As far as I can see the "weak" fundamental theorem of algebra would not imply the "strong" fundamental theorem of algebra, due to noncommutativity.

• In your linked question, a commenter points out the existence of a $2\times 2$ matrix $N$ which is not a square. The example given is $N = \bigl(\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr)$: if $N$ had a square root $\sqrt{N}$, then $N$ and $\sqrt{N}$ would commute; but the commutator of $N$ are the matrices of the form $a + Nb$ for $a,b \in \mathbb{C}$, and $(a + Nb)^2 = a^2 + 2ab N \neq 0 + 1N$. For this matrix $N$, $x^2 - N$ does not have a root. Commented yesterday
• @TheoJohnson-Freyd: it's even simpler than that :-), see my answer below.
– M.G.
Commented yesterday
• Perhaps @TheoJohnson-Freyd's example is a little more satisfying because the polynomial is monic. (Although I would say centraliser instead of commutator, which to me has a different meaning.) Commented yesterday
• @LSpice: I agree! His example does not rely on the cheap commutative algebra characterization of invertible polynomials I used :-)
– M.G.
Commented yesterday
• @M.G. I was also trying to point out that the answer was already available in the OP's source material... Commented yesterday

No, for rather trivial reasons. Consider the polynomial $$f(X) = \varepsilon X - 1$$ with $$\varepsilon^2 = 0$$, $$\varepsilon \neq 0$$, in $$R = M_2(\mathbb{C})$$. Then a root of $$f(X)$$ would mean that $$\varepsilon$$ is invertible in $$R$$, which is of course impossible. Alternatively, just take $$\det$$ on both sides.
Although what you are asking for is not true, there is a very interesting related fact that is true. Let $$k$$ be an algebraically closed field of any characteristic. The theory of matrix factorisations shows that every homogeneous polynomial in $$R=k[X_1,\dots,X_n]$$ factorises in a suitable matrix ring $$\operatorname{Mat}_s(R)$$ as a product of linear factors. This follows from Theorem 1.2 of Herzog, Ulrich and Backelin, "Linear maximal Cohen-Macaulay modules over strict complete intersections" (see also Backelin, Herzog and Sanders, "Matrix factorizations of homogeneous polynomials"). To illustrate this theorem, consider the polynomial $$f=X_1^2+X_2^2+X_3^2+X_4^2$$. Clearly, this does not factor into linear factors in $$R$$. But in $$\operatorname{Mat}_4(R)$$ we have $$f.I_4=\begin{pmatrix} X_1&-X_2&X_3&X_4\\X_2&X_1&-X_4&X_3\\-X_3&X_4&X_1&X_2\\-X_4&-X_3&-X_2&X_1\end{pmatrix}\begin{pmatrix} X_1&X_2&-X_3&-X_4\\-X_2&X_1&X_4&-X_3\\X_3&-X_4&X_1&-X_2\\X_4&X_3&X_2&X_1\end{pmatrix}.$$ where $$I_4$$ is a $$4\times 4$$ identity matrix.
The theorem is that if $$f$$ has degree $$d$$ then it factorises as a product of $$d$$ linear factors in a suitable size matrix ring over $$R$$. One consequence of this is that some power of $$f$$ is the determinant of a matrix whose entries are homogeneous linear polynomials (or zero).