How can i prove that $AB=BA$?

Question

In given $A,B$ in size of $n\times n$, and, $A=I-AB$.

(I know how to prove that A is invertible)

How can i prove that $AB=BA$?

Answer 1

Multiply by $A^{-1}$ from the left to get $I=A^{-1}-B$ or in other words $B=A^{-1}-I$. Now you can compute $AB$ and $BA$ explicitly and see that they come out equal.

Answer 2

We have $\color{blue}{AB=I-A}$ \begin{eqnarray*} ABA=(\color{blue}{AB})A=\color{blue}{(I-A)}A=A-A^2=A\color{blue}{(I-A)}=A\color{blue}{A B}=A^2B. \end{eqnarray*} Now left multiply by $A^{-1}$.

Answer 3

$A=1-AB,$ therefore $A(1+B)=1.$ If $A$ is invertible then we have $B=A^{-1}-1.$ Thus $$[A,B]=AB-BA=A(A^{-1}-1)-(A^{-1}-1)A=1-A-(1-A)=0.$$

Answer 4

Let's assume we know the following: for matrices $X$ and $Y$ of dimension $n\times n$, if $XY = I$ then $YX=I$ also. In fact we only need $XY=YX$ from that.

Now: $A=I-AB \implies A + AB = I \implies A(I+B) = I \implies A(I+B) =(I+B)A$ (by the result we accepted). So $A + AB = A+BA$, and $AB=BA$.