Is A+B invertible?

Clearly $A$ is invertible (indeed, $\det(A)^p = \det(A^p) = \det(I) = 1,$ so $\det(A) \ne 0$ ). Find $k$ such that $2^k \ge q.$ Then $(A+B)(A-B)(A^2+B^2)(A^4+B^4)(\cdots)(A^{2^{k-1}} + B^{2^{k-1}})(A^{-1})^{2^k} = (A^{2^k} - B^{2^k})(A^{-1})^{2^k} = A^{2^k} (A^{-1})^{2^k} = I.$

Let's note that ${A}^{p-1}$ is the inverse of $A$ .

Assume there is an inverse, then $M = {(A+B)}^{-1}$ :

• $M(A+B) = {I}_{n}$
• $MA + MB = {I}_{n}$
• $MA{B}^{q-1} + M{B}^{q} = {B}^{q-1}$
• $MA{B}^{q-1} + {O}_{n} = {B}^{q-1}$
• $MA{B}^{q-1} = {B}^{q-1}$

This holds true if $MA = {I}_{n}$ . Then $M = {A}^{p-1}$ :

• ${A}^{p-1}(A+B) = {I}_{n}$
• ${A}^{p} + {A}^{p-1}B = {A}^{p}$
• ${A}^{p-1}B = {O}_{n}$
• ${A}^{p}B = {O}_{n}$
• $B = {O}_{n}$

That's all i got, obviously not enough so feel free to add on missing parts)

Because $A$ and $B$ commute we heve the usual identity : $a^n-b^n=(a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$

Apply it to $a=A$ and $b=-B$ , note that $(-B)^p=(-1)^p B^p = 0$

$I_n = A^p - (-B)^p = (A+B)(A^{p-1} + A^{p-2}(-B)+ \cdots + A(-B)^{p-2} + (-B)^{p-1})$

we have found the inverse of $A+B$ .