Question: Does there exist $A \in M_{2}(\mathbb{Z})$ such that every element of $M_{2}(\mathbb{Z})$ can be represented as a linear combination of powers of $A$ with integer coefficients? In other words,

$$\exists A \in M_{2}(\mathbb{Z}) \, \,s.t. M_{2}(\mathbb{Z})=\left.\left\{\sum_{i}a_{i}A^{i} \;\right| \; a_{i} \in \mathbb{Z}, a_j=0\text{ for almost all }j\right\}$$

Motivation: I want to construct a surjective ring homomorphism $\varphi: \mathbb{Z}\left[x\right] \rightarrow M_{2}(\mathbb{Z})$ by letting $\varphi: \displaystyle \sum_{i}a_{i}x^{i} \mapsto \sum_{i}a_{i}A^{i}$. The above property is sufficient for surjectivity.

What I've tried: It's sufficient to find four linear combinations that sum to $\left\{ \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] \right\}$. I haven't been able to find a magic $A$ that does this. For example, powers of $\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right]$ generate Fibonacci numbers, except that off diagonal elements are the same, so this will never work.

This is not homework. I have a feeling it may be impossible, but I'm not sure why. If for some reason it is impossible over $\mathbb{Z}$, I would be interested to know if it's possible over $\mathbb{Z}_{p}$ for a prime $p$. Thanks for any help.


You cannot construct a surjective morphism $\mathbb{Z}[x] \to \mathcal{M}_2(\mathbb{Z})$ because the former is commutative and the latter is not. No need to appeal to Cayley-Hamilton. This is true when $\mathbb{Z}$ is replaced by any commutative ring.

  • $\begingroup$ This is also a nice answer, simpler and a little more general. I'd accept both if I could. $\endgroup$ – Jackson Walters Jun 9 '12 at 20:58
  • $\begingroup$ D'oh! Of course. $\endgroup$ – Arturo Magidin Jun 9 '12 at 21:01
  • $\begingroup$ @ArturoMagidin haha yea I facepalmed as soon as I read his answer. $\endgroup$ – Jackson Walters Jun 9 '12 at 21:12
  • $\begingroup$ Arturo's answer does show more, though: it shows that the image of a morphism $F[x] \to \mathcal{M}_n(F)$ has dimension at most $n$ (working over a field for simplicity). It is interesting to ask whether there exists a commutative subalgebra of $\mathcal{M}_n(F)$ of dimension more than $n$, and in fact one can find examples of dimension quadratic in $n$: see mathoverflow.net/questions/29087/commutative-subalgebras-of-m-n . $\endgroup$ – Qiaochu Yuan Jun 9 '12 at 22:10

No, you cannot do it over (either over $\mathbb{Z}$ or over $\mathbb{Z}_p$).

Note that any $n\times n$ matrix satisfies its own characteristic polynomial by the Cayley-Hamilton Theorem. In the case of a $2\times 2$ matrix, the characteristic polynomial is just $x^2 - \mathrm{trace}(A)x + \det(A)$, so if $A$ has integer coefficients, then the polynomial has integer coefficients.

That means that for any $A\in M_2(\mathbb{Z})$, there exist integers $t$ and $d$ such that $$A^2 = tA + dI_2.$$

Thus, any polynomial expression in $A$ can be simplified, by applying the identity above, to an expression of the form $c_0I_2 + c_1A$.

Thus, if $$A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right),$$ then every element of your set is of the form $$\left(\begin{array}{cc} r+sa & sb\\ sc & r+sd \end{array}\right),\qquad r,s\in\mathbb{Z}.$$

Now let $(x,y)$ be any element of $\mathbb{Z}^2$ that is not in the subgroup generated by $(b,c)$. Then the matrix $$\left(\begin{array}{cc} 0 & x\\ y & 0 \end{array}\right)$$ cannot be obtained as a polynomial expression in $A$, because every matrix that is a polynomial expression of $A$ will have that the vector made up of the $(1,2)$ and $(2,1)$ entries of the matrix lies in the subgroup generated by $(b,c)$.

The same argument works over $\mathbb{Z}_p$, except that now you want to take $(x,y)$ a vector not in the linear span of $(b,c)$ in the vector space $\mathbb{Z}_p^2$ over $\mathbb{Z}_p$.

  • $\begingroup$ Very nice answer, thank you! $\endgroup$ – Jackson Walters Jun 9 '12 at 20:39

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