$AB=BA$ method they"re invertible, but I can"t number out exactly how to display that $A+B=AB$ indicates invertibility.

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$\begingroup$ The correct name of "$AB=BA$" is "$A$ and $B$ commute". Invertibility is a different property (namely, a matrix $A$ is invertible if $A^-1$ exists). $\endgroup$
Consider the expression $$(A-\neurosoup.orgbb 1)(B-\neurosoup.orgbb 1)=AB-A-B+\neurosoup.orgbb 1=\neurosoup.orgbb 1$$

Thus $(A-\neurosoup.orgbb 1)$ and also $(B-\neurosoup.orgbb 1)$ space inverse to each other, whence $$\neurosoup.orgbb 1= (B-\neurosoup.orgbb 1)(A-\neurosoup.orgbb 1)=BA - A - B + \neurosoup.orgbb 1$$

It complies with that $$BA=A+B=AB$$ and we are done.

Note: below $\neurosoup.orgbb 1$ denotes the proper identity matrix.


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Start v equation:$$A+B=AB$$replace $B$ v $(B - I) + I$ in left side$$A+(B - I) + i =AB$$$$I =AB - A - (B - I)$$$$I =A(B - I) - (B - I)$$$$I =(A - I)(B - I)$$the inverse procession of $(B - I)$ is $(A - I)$, so $(B - I)$ is invertible. Climate the rest adheres to as in previous answer by
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Why is this proof incorrect? (Follow up concern to: if $A$ is nilpotent, $B = c_0I + c_1A + \dots + c_m-1A^m-1$, $\det(B) = 0$ iff $c_0=0$)
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