Let C be an nxn matrix, then the polynomials in C (with appropriate coefficients) form an algebra of commuting matrices. I feel that I should know if the converse is true but I do not. So my first question (adding several conditions) is: Suppose A and B are symmetric integer matrices with AB=BA, must there be a matrix C such that both A and B are polynomials in C? This would imply that the algebra which A and B generate has dimension n or less. In general: Let **A** be an algebra of pairwise commuting nxn matrices over an integral domain D, Under what conditions (if any) on D and **A** can we conclude that there is a matrix C so that **A** is contained in the D-algebra generated by C? I am thinking of conditions such as D is an algebraically closed field, D has characteristic zero, the matrices are symmetric. etc.



When is an algebra of commuting matrices (contained in one) generated by a single matrix?
Thanks for the answers. Just to wrap up a bit, here are a few examples.


1. Sometimes an ACM (algebra of commuting matrices) is sure to be generated by one of its members
2. Other times it has dimension too large to possibly be (embedded in) an ACM with a single generator.
3. An ACM might be generated by $2$ matrices, not generated by any of its members, but embed in a larger ACM which does have a single generator.
4. An ACM might be generated by $2$ matrices, not generated by any of its members, but **not** embed in a larger ACM which does have a single generator (even in the $3\times 3$ case).
5. If the matrices are all normal then they can be simultaneously diagonalized. This reduces the problem to an algebra of diagonal matrices, which is easy to understand. Such an algebra is actually generated by one of its members.
6. The $5$ dimensional algebra ${\cal{A}}\_5$ mentioned by Mariano ($4\times 4$ matrices with $2 \times 2$ blocks $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ has dimension too large to be generated by a single matrix. Furthermore, each member M generates only the $2$ dimensional algebra of matrices $jI+kM$ so no subalgebra of dimension $3$ or $4$ has a single generator.
7. Consider the subalgebra ${\cal{A}}\_3$ of ${\cal{A}}\_5$ generated by $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&B\\0&0\end{smallmatrix}\right)$ with $A$ and $B$ both $2 \times 2$ invertible matrices (neither a scalar multiple of the other). As mentioned, we can't embed ${\cal{A}}\_3$ in a singly generated $4$ dimensional subalgebra of ${\cal{A}}\_5$
However it also embeds in other $4$ dimensional algebras.


For example a $4 \times 4$ matrix $\left(\begin{smallmatrix}BA^{-1}&C\\0&A^{-1} B\end{smallmatrix}\right)$ will generate an ACM which commutes with everything in ${\cal{A}}\_3$. I guess in this case it would automatically contain ${\cal{A}}\_3$. I certainly verified that randomly filling in the C does this in several cases. In many cases I tested one can get away with one or both of $A$ and $B$ having rank $1$... but not always. The two $4 \times 4$ matrices made from matrices with $A=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ give an example of that. One can shrink this to $3 \times 3$ (*in the case that the underlying ring is $\mathbb{Z}\_2$ as noted by Martin and missed by me*), so I will:


4. The two $3 \times 3$ matrices $\left(\begin{smallmatrix}0&1&0\\0&0&0\\0&0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0&1\\0&0&0\\0&0&0\end{smallmatrix}\right)$ generate an algebra **A** with *eight* members $\left(\begin{smallmatrix}a&b&c\\0&a&0\\0&0&a\end{smallmatrix}\right)$.  

**A** is maximal but not generated by any of its members as each member generates a $2$ dimensional (or smaller) subalgebra.


