Given a matrix, is there always another matrix which commutes with it?

Question

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?)

Answer 1

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.

Answer 2

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.

Answer 3

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 $A.$

Answer 4

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 $A$).