# Thread: AB=A+B;Prove A and B commute

1. ## AB=A+B;Prove A and B commute

If $\displaystyle A$ and $\displaystyle B$ are two square matrices of the same order such that $\displaystyle AB=A+B$ then prove that A and B commute i.e. AB=BA

2. Umm working on it.

$\displaystyle AB = A+B$

$\displaystyle A(BB^{-1}) = (A + B)B^{-1}$

$\displaystyle A = (AB^{-1} + I)$

$\displaystyle BA = BAB^{-1} + B$

...

3. Originally Posted by pankaj
If $\displaystyle A$ and $\displaystyle B$ are two square matrices of the same order such that $\displaystyle AB=A+B$ then prove that A and B commute i.e. AB=BA
If $\displaystyle AB=A+B$ then $\displaystyle (A-I)(B-I)=I$ (I is the identity matrix). So $\displaystyle B-I$ is the inverse of $\displaystyle A-I$ and therefore commutes with it. Thus $\displaystyle (B-I)(A-I) = I$, from which $\displaystyle BA = A+B = AB$.

4. So if $\displaystyle CD=I$ does it always mean that $\displaystyle C$ and $\displaystyle D$ are inverse of each other.What if any one of $\displaystyle C$ or $\displaystyle D$ is singular matrix

5. Originally Posted by pankaj
So if $\displaystyle CD=I$ does it always mean that $\displaystyle C$ and $\displaystyle D$ are inverse of each other.What if any one of $\displaystyle C$ or $\displaystyle D$ is singular matrix
If $\displaystyle CD = I$ then C and D cannot be singular. One way to see this is to notice that the determinant is multiplicative: $\displaystyle \det(C)\det(D) = \det(I) = 1$, so that $\displaystyle \det(C)$ and $\displaystyle \det(D)$ cannot be 0.