Vtyjkyuk

Below is the problem as stated in Putnam and Beyond:

Let $A$ and $B$ be tw0 $ntimes n$ matrices that commute and such that for some positive integers $p$ and $q$, $A^p=I_n$ and $B^q=O_n$. Prove that $A+B$ is invertible, and find its inverse.

In the following I will give my "solution" and I would be very grateful if you could either verify that it is correct or else tell me where I've gone wrong. I would also greatly appreciate any comments on my proof writing.

We start of by making the assumption $p>q$. Note that in order for this argument to be correct, $p$ and $q$ may need to be interchanged. It follows that $A^p+B^p=I_n$. The expression on the left can be factored as $(A+B)(A^p-1+BA^p-2+...+B^p-1)= I_n$. From this we see clearly that $A+B$ is invertible, its inverse being $(A^p-1+BA^p-2+...+B^p-1)$. My argument is concluded.

Thank you!

Your argument is right when $pge q$ because $$B^q=0to B^p=0 text and (-B)^p=0$$also we know that $A^p-(-B)^p$ is divisible by $A-(-B)=A+B$ so your conclusion is right in this case. Now let $p<q$. Also since all the eigenvalues of $A^p$ are $1$ so all the eigenvalues of $A$ are roots of $1$ and non-zero and $A$ is invertible which means that$$A^q=A^q-p$$therefore $$A^q-p=A^q-(-B)^q=(A+B)(A^q-1-cdots pm B^q-1)$$therefore $$(A+B)(A^q-1-cdots pm B^q-1)(A^q-p)^-1=I$$or $$(A+B)^-1=left((A^q-1-cdots pm B^q-1)(A^q-p)^-1right)^-1=A^q-p(A^q-1-cdots pm B^q-1)^-1$$

Here is marginally different way, note that $A+B = A(I+ A^-1 B)$. Since $A,B$ commute, so do $A^-1, B$ and so $(A^-1B)^k = A^-kB^k$, and so, with $C= -A^-1B$, we have $C^q = 0$.

Since all of $C$'s eigenvalues are $0$, we see that $I-C$ is invertible and it is easy to check that $(I-C)(I+C+C^2+cdots+C^q-1) = I$.

Hence $(A+B)^-1 = (I-C)^-1 A^-1 = (I+C+C^2+cdots+C^q-1) A^-1$.

Since $A^p = I$, we have $A^-k = A^p-k$ and so
$(A+B)^-1 = sum_k=0^q-1 (-B)^k A^p-k$.