$A+B=AB$ does it follows that $AB=BA$?

Question

If $A$, $B$ are two normal operators such that: $A+B=AB$ does it follow that $AB=BA$?

Answer 1

Rearrange the equation to \begin{eqnarray*} (A-1)B=A \end{eqnarray*} Now assume (A-1) is non singular \begin{eqnarray*} B=(A-1)^{-1}A \end{eqnarray*} Geometrically expand $(A-1)^{-1}$ ... so B can be expressed as a sum of powers of A , A will commute with each of these terms, so A commutes with B.

Answer 2

In the finite dimensional case:

We have $(A-I)$ invertible because $$(A-I)(B-I)=AB-B-A+I=AB-(A+B)+I=I.$$ Note that this doesn't show invertibility if our vector space is infinite dimensional. To convince yourself with matrices, take the determinant of the above expression.

Now since $(A-I)B=A$, $$A(A-I)B=A^2=(A-I)BA.$$ Because $A(A-I)=(A-I)A$, this means $$(A-I)AB=(A-I)BA.$$ Invertibility was shown so we cancel to obtain the result.

Answer 3

First, note that if $C$ is a normal operator which has a left inverse $D$ (so $DC=I$), then $C$ is invertible (and thus $D$ is its inverse and $CD=I$ as well). This follows from the spectral theorem: you can identify your Hilbert space with $L^2(X)$ for some semifinite measure space $X$ and $C$ with multiplication by $f$ for some $f\in L^\infty(X)$. If $0$ is in the essential range of $f$, then this means that for any $\epsilon>0$ there exists $g\in L^2(X)$ such that $\|fg\|<\epsilon\|g\|$ (choose $g$ to be supported on the set where $|f|<\epsilon$). Since $DC=I$, we must have $D(fg)=g$ so $\|D\|\geq\|g\|/\|fg\|>1/\epsilon$. Since $\epsilon$ is arbitrary, this is a contradiction. Thus $0$ is not in the essential range of $f$, so $1/f\in L^\infty(X)$, and multiplication by $1/f$ is an inverse for $C$.

(Probably there is a more elementary argument that avoids spectral theory, but this is all I can come up with at the moment.)

Now note that $A+B=AB$ implies $(I-A)(I-B)=I-A-B+AB=I$. Since $I-B$ is normal, by the result above this implies $(I-B)(I-A)=I$ as well. That is, $I-A-B+BA=I$, or $BA=A+B=AB$.

Answer 4

There is a flaw in my first line equation.

First we will solve this problem: Let $X$ and $Y$ be normal and let $XY=I$. We will show $XY=YX$.

We have by normality $$||X^*Yz||=||XYz||= ||Iy|| = ||I^*y||=||(XY)^*y||= ||Y^*X^*z||=||YX^*z||.$$ It follows that $X^*Y$ is normal since normal operators are precisely those that have $||Tx||=||T^*x||$ for all $x$. This means $$(X^*Y)Y^*X=Y^*X(X^*Y),$$ and by swapping the middle terms on each side $$X^*Y^*YX=Y^*X^*XY.$$ Thus, $||(YX-XY)z||=0$ for all $z$ so $XY=YX$.

Next we realize that $(A-I)(B-I)=(B-I)(A-I)$ implies that $AB=BA$ by expanding and canceling and that $X=(A-I)$ and $Y=(B-I)$ are normal operators with $XY=I$. (Expand and cancel to verify this.)