What does $[A,B]=0$ mean in matrix theory?

Question

This is probably very simple but I can't seem to find an answer anywhere.

I'm being asked to prove $e^{tA}e^{tB} = e^{t(A+B)}$ for all $t \in \mathbb{R}$ iff $[A,B]=0$

However, I am unsure what $[A,B]=0$ even means! Any help?

Answer 1

In the arithmetic of the real numbers and many other number systems, multiplication satisfies the commutative property, $ab=ba$ for all elements $a$ and $b$.in these number systems you can always exchange the order of the factors in any multiplication.

In ring theory, linear algebra, operator theory, quantum mechanics, and related fields, we generally still have systems with a notion of multiplication, but in general they need not satisfy the commutative property. We say that two elements $A$ and $B$ commute if $AB=BA$. We can think of "moving $B$ past $A$" if we ever see it multiplying on one side. In these number systems you may only change the order of factors in a multiplication if you know explicitly that the factors in question commute, otherwise you must always be careful to preserve the order of the factors in every multiplication.

Then $[A,B]$ is shorthand notation for $AB-BA$, called the commutator. So two elements commute if and only if $[A,B]=0$, and the commutator can be thought of as a measure of the failure of the elements to commute. A theorem states that commuting normal matrices or operators may be simultaneously diagonalized. Diagonal matrices all commute with each other, and scalar matrices (which have the same element in every diagonal entry, zero elsewhere) commute with every matrix.

In a noncommutative ring or algebra you no longer have the identity $(AB)^2 = A^2B^2$ or $(A+B)^2 = A^2 + 2AB + B^2$ (the latter is instead $A^2 + AB + BA + B^2$). This is why the identity $e^{tA}e^{tB}=e^{t(A+B)}$ fails in general for non-commuting $A,B$. In the expression $e^{tA}e^{tB}$ all the $A$ operators appear to the left of all the $B$ operators, whereas when you look at the Taylor expansion for $e^{t(A+B)}$ you have terms with $A$ to the right of $B$.

But if $A$ and $B$ commute, then the proof of this identity found in real analysis by manipulating Taylor series works unchanged. Just expand the product of the two Taylor series, term-by-term, and commute all the $A$ factors to the left of all the $B$ factors in every term. Then use the binomial theorem. And to prove the converse statement ($e^{tA}e^{tB}=e^{t(A+B)}$ implies $A$ and $B$ commute) you may find it helpful to compare powers of $t$ termwise.

Note that in group theory where subtraction is not available, there is another commutator $[a,b]=aba^{-1}b^{-1}$. They are both measures of failure to commute, but they are not the same, so be aware of the context.

Answer 2

In Lie theory, the brackets $[,]$ are used to define an operation called a Lie bracket, which is a non-associative multiplication. A vector space with a Lie bracket is a Lie algebra, and when there is already a product defined on the space, the natural way to define the Lie bracket is the commutator: $[x,y]=xy-yx$. This is the case with the space of square matrices (of any order).

So, $[A,B]$ is the commutator $AB-BA$ of $A$ and $B$, and $[A,B]=0$ is short for '$A$ and $B$ commute': $AB=BA $

The result then follows easily from the series representation of the matrix exponential.