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

You are watching: (a-b)(a+b)  \$\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. 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  Thanks for contributing an answer to neurosoup.orgematics stack Exchange!

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