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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.

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    $\begingroup$ This comment does not affect the answers given but I feel that it should be mentioned. In questions of when matrices are functions of a given one, the relevant concept is that of the bicommutant. Thus if $A$ and $B$ are two square matrices (of the same order), then $A$ is a function of $B$ if and only if $A$ is in the bicommutant of $B$. This is a standard result and can be proved using the Jordan normal form (compute the bicommutant of a Jordan block). (I am assuming the the underlying field is that of the complex numbers). $\endgroup$
    – jbc
    Nov 3, 2012 at 10:10

4 Answers 4

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Consider the subalgebra of $M_{2n}(k)$ spanned by the matrices of the form $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ (all blocks are $n\times n$) together with the identity, which is commutative. Its dimension is larger than $2n$ when $n$ is sufficiently large, so it is not generated by a single matrix.

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  • $\begingroup$ That is a good point! To slightly recover my dignity I'll ask: what about 3x3 matrices? Symmetric matrices? $\endgroup$ Aug 3, 2010 at 2:22
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    $\begingroup$ Aaron, the point is that if matrices aren't semisimple, bad stuff happens. "Symmetric" doesn't help in general (because it doesn't imply s.s. except over $\mathbb{R}$), but 3x3 matrices are small enough that you can classify commutative subalgebras by hand (work with joint eigenvectors, etc). $\endgroup$ Aug 3, 2010 at 4:32
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    $\begingroup$ It may be useful to look at Jacobson's paper ams.org/mathscinet/pdf/…. Where he proves that maximum dimension of commutative subalgebras in $M_n(\mathbf C)$ is $[n^2/4] +1$. Which essentially matches with Mariano's examples with an additional identity matrix. $\endgroup$ Aug 3, 2010 at 6:28
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For symmetric matrices over the reals, the answer is yes: if $A$ and $B$ commute, then they can be simultaneously diagonalized. That is to say, there exists an orthogonal matrix $U$ ($U^TU = I$) such that $U^TAU$ and $U^TBU$ are both diagonal. It follows that for any diagonal matrix $D$ with distinct diagonal entries, both $A$ and $B$ are polynomials in $C = UDU^T$.

In general it is instructive, and loses no generality, to assume that $A$ is in a normal (Jordan, Smith, etc.) form. Over a field, for example, the dimension of the space of polynomials in $A$ is the same as the degree of the minimal polynomial of $A$, which is the sum, over all eigenvalues $\lambda$, of the size of the largest Jordan block for $\lambda$ in the Jordan normal form of $A$ over the algebraic closure of the field. It's not too hard to verify that every matrix commuting with $A$ is a polynomial in $A$ if and only if the minimal polynomial is the characteristic polynomial, that is if every eigenvalue has a single Jordan block (as happens automatically for example if there are $n$ distinct eigenvalues).

I can't remember the precise statement of something that surprised me, some unsolved problem related to the dimension of a commuting set in terms of Jordan block sizes. Someone please comment if you are familiar with this open problem and can state it precisely.

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  • $\begingroup$ I agree that normal form is the way to go, I'd thought about Jordon form but hadn't considered any other. I wondered what is known. It is not true that the algebra generated by three commuting matrices always embeds in one generated by one matrix ( shown by Mariano's example for 4x4 with a little work). I don't think that an algebra generated by two commuting matrices can have dimension greater than n (at least over an algebraically closed field). That does not mean that it embeds, but at least it blocks a dimension proof. $\endgroup$ Aug 3, 2010 at 5:38
  • $\begingroup$ And see this answer to another question. Notably it is indeed a theorem (Gerstenhaber 1961) that an algebra generated by two commuting matrices cannot have dimension greater than n (but it can be of dimension n while not being single-generated), and it is an open question whether an algebra generated by 3 matrices can. – Marc van Leeuwen 0 secs ago $\endgroup$ Mar 11, 2013 at 5:08
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The open problem that surprised me was on the lower bound of dimensions of maximal commutative subalgebras of $M_n(\mathbf C)$. A maximal commutative sublagebra of $M_n(\mathbf C)$ can have dimension strictly lass than $n$. See for example Courter's article, where he gives a 13 dimensional maximal commutative subalgebra of $M_{14}(\mathbf C)$. I think, In general the problem of finding lower bound for the dimension of maximal commutative subalgebras in $M_n(\mathbf C)$ is still open.

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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:

  1. 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.
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    $\begingroup$ 4) is not quite correct: If $X,Y$ are the two matrices, the subalgebra generated by $P=a+bX+cY$ may have dimension $3$ and therefore equals $\langle X,Y \rangle$ since $P^2=a^2+2abX+2acY$. But over the base field $\mathbb{F}_2$ the dimension is $\leq 2$. $\endgroup$ Mar 10, 2013 at 13:14

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