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Given a matrix $A$ over a field $F$, does there always exist a matrix $B$ such that $AB = BA$? (except the trivial case and the polynomial ring?)

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There's always the identity matrix $I$, and $aI$ for $a \in F - \{0\}$. These commute with everything. Do you want something more sophisticated? –  Dylan Moreland Dec 18 '11 at 9:42
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This is trivial - take $B=I$. Maybe after adding some other conditions for $B$ the question becomes more interesting. –  Martin Sleziak Dec 18 '11 at 9:43
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A itself will do, too :-). –  user20266 Dec 18 '11 at 9:44
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Or $A = B$? This works for every ring, module, group, whatever. –  Noud Dec 18 '11 at 9:45
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Or any sum of the form $$B=\sum_{k=0}^na_kA^k\;,$$ where each $a_k\in F$. –  Brian M. Scott Dec 18 '11 at 9:58
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5 Answers 5

up vote 38 down vote accepted

A square matrix $A$ over a field $F$ commutes with every $F$-linear combination of non-negative powers of $A$. That is, for every $a_0,\dots,a_n\in F$,

$$A\left(\sum_{k=0}^na_kA^k\right)=\sum_{k=0}^na_kA^{k+1}=\left(\sum_{k=0}^na_kA^k\right)A\;.$$

This includes as special cases the identity and zero matrices of the same dimensions as $A$ and of course $A$ itself.

Added: As was noted in the comments, this amounts to saying that $A$ commutes with $p(A)$ for every polynomial over $F$. As was also noted, there are matrices that commute only with these. A simple example is the matrix $$A=\pmatrix{1&1\\0&1}:$$ it’s easily verified that the matrices that commute with $A$ are precisely those of the form $$\pmatrix{a&b\\0&a}=bA+(a-b)I=bA^1+(a-b)A^0\;.$$ At the other extreme, a scalar multiple of an identity matrix commutes with all matrices of the same size.

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Short version: if $p(x)$ is a polynomial, then $p(\mathbf A)$ commutes with $\mathbf A$. Additionally, a matrix commutes with its inverse (if it has one), and with its matrix exponential, and maybe a bunch of other matrix functions... –  J. M. Dec 18 '11 at 10:32
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It might be worth remarking that there are matrices $A$ which commute with nothing other than linear combinations of their powers, –  Geoff Robinson Dec 18 '11 at 10:52
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@J.M., if a matrix $A$ has an inverse then the inverse is a polynomial in $A$ because its minimal polynomial cannot be a multiple of $X$. –  lhf Dec 18 '11 at 13:50
    
@Jack: Oops; that’s what I meant. Thanks. –  Brian M. Scott Dec 19 '11 at 22:41
    
You might add that (at least over infinite fields) most matrices commute only with polynomials in themselves. A sufficient condition for this is that its characteristic polynomial has simple roots; a necessary and sufficient condition is that the degree of the minimal polynomial is$~n$ (so it coincides with the characteristic polynomial). –  Marc van Leeuwen Sep 20 '13 at 9:28
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This doesn't want to be a complete answer, but just a hint to understand the problem a bit better.

Note that the commutativity $AB=BA$ is equivalent (when $B$ is invertible) to $A=BAB^{-1}$.

On the other hand conjugate matrices represent the same endomorphism of the underlying vector space with respect to different basis. Thus $A=BAB^{-1}$ means that the endomorphism corresponding to $A$ has the same matrix representation when the base changed by $B$.

How could that be possible?

If $A$ is diagonalizable, i.e. the vector space admits a basis of eigenvectors, with distinct eigenvalues, the only way that we can modify the basis and leave $A$ as matrix representing the endomorphism is to the effect of replacing the eigenvectors with some multiples. This corresponds to the situation where the only matrices commuting with $A$ are the linear combinations of its powers.

On the other hand, if there is an eigenspace $E_\lambda$ of dimension $\geq2$ we can replace any choice of basis of $E_\lambda$ with any other, and the matrix representing the endomorphism will be left the same. Thus we should expect more matrices commuting with $A$ in this case.

Hope this helps understanding the problem.

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Given a square $n$ by $n$ matrix $A$ over a field $k,$ it is always true that $A$ commutes with any $p(A),$ where $p(x)$ is a polynomial with coefficients in $k.$ Note that in the polynomial we take the constant $p_0$ to refer to $p_0 I$ here, where $I$ is the identity matrix. Also, by Cayley-Hamilton, any such polynomial may be rewritten as one of degree no larger than $(n-1),$ and this applies also to power series such as $e^A,$ although in this case it is better to find $e^A$ first and then figure out how to write it as a finite polynomial.

THEOREM: The following are equivalent:

(I) $A$ commutes only with matrices $B = p(A)$ for some $p(x) \in k[x]$

(II) The minimal polynomial and characteristic polynomial of $A$ coincide

(III) $A$ is similar to a companion matrix.

(IV) if necessary, taking a field extension so that the characteristic polynomial factors completely, each characteristic value occurs in only one Jordan block.

(V) $A$ has a cyclic vector, that is some $v$ such that $ \{ v,Av,A^2v, \ldots, A^{n-1}v \} $ is a basis for the vector space.

See GAILLARD MINIMAL SIMILAR COMPANION

The equivalence of (II) and (III) is Corollary 9.43 on page 674 of ROTMAN

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Theorem: If $A$ has a cyclic vector, that is some $v$ such that $$ \{ v,Av,A^2v, \ldots, A^{n-1}v \} $$ is a basis for the vector space, then $A$ commutes only with matrices $B = p(A)$ for some $p(x) \in k[x].$

Nice short proof by Gerry at Complex matrix that commutes with another complex matrix.

This is actually if and only if, see Statement and Proof

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Note that, as in the complex numbers, if the field $k$ is algebraically closed we may then ask about the Jordan Normal Form of $A.$ In this case, the condition is that each eigenvalue belong to only a single Jordan block. This includes the easiest case, when all eigenvalues are distinct, as then the Jordan form is just a diagonal matrix with a bunch of different numbers on the diagonal, no repeats.

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For example, let $$ A \; = \; \left( \begin{array}{ccc} \lambda & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{array} \right). $$

Next, with $r \neq s,$ take $$ B \; = \; \left( \begin{array}{ccc} r & 0 & 0 \\ 0 & s & t \\ 0 & 0 & s \end{array} \right). $$

We do get

$$ AB \; = \; BA \; = \; \left( \begin{array}{ccc} \lambda r & 0 & 0 \\ 0 & \lambda s & \lambda t + s \\ 0 & 0 & \lambda s \end{array} \right). $$ However, since $r \neq s,$ we know that $B$ cannot be written as a polynomial in $A.$

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There are some very nice proofs of this fact: if you extend the field of consideration (such as the reals extended to the complexes) the minimal polynomial does not change!! See math.stackexchange.com/questions/136804/… –  Will Jagy Dec 24 '13 at 20:49
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Another example is the adjoint of $A$: $$ A \operatorname{adj}(A)= \operatorname{adj}(A) A = \det(A)I $$ (but for invertible matrices it is equal to the scalar $\det(A)$ multipliying the inverse of $A$, so is trivial that commutes. with $A$).

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Let $A$ be a $n\times n$ matrix over a field $\mathbb{F}$. Let $C_A = \{B\in M_{n\times n}(\mathbb{F}) \mid AB=BA\}$. Then the minimal dimension of $C_A$ over $\mathbb{F}$ is $n$, and this is obtained precisely when the minimal polynomial and characteristic polynomial of $A$ coincide.

The idea of proof is interpreting $C_A$ as an $\mathbb{F}[x]$-endomorphism algebra of the $\mathbb{F}[x]$-module $M^A$ (Here $M^A$ is the $\mathbb{F}[x]$-module structure of $\mathbb{F}^n$. Use the Rational Canonical Form-Primary decomposition. We have the following formula for $\textrm{dim}_{\mathbb{F}} C_A$. $$\textrm{dim}_{\mathbb{F}} C_A=\textrm{dim}_{\mathbb{F}}\textrm{ End}_{\mathbb{F}[x]}M^A=\sum_p(\deg p)\sum_{i,j} \min\{\lambda_{p,i}, \lambda_{p,j}\}, $$ where the first sum is over all irreducible polynomials $p$ that divides the characteristic polynomial of $A$, and the indices $i, j$ of second double sum is from the partition $\lambda_p=\sum_i \lambda_{p,i}$ that indicates the powers of $p$ in $p$-primary part of $M^A$.

Thus, if minimal polynomial does not coincide with characteristic polynomial, then the dimension of $C_A$ is greater than $n$.

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