Putnam and Beyond: Problem 87 (Matrices and algebra)

Question

Below is the problem as stated in Putnam and Beyond:

Let $A$ and $B$ be tw0 $n\times n$ 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!

Answer 1

Your argument is right when $p\ge q$ because $$B^q=0\to B^p=0 \text{ and } (-B)^p=0$$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 $$(A+B)^{-1}=\left((A^{q-1}-\cdots \pm B^{q-1})(A^{q-p})^{-1}\right)^{-1}=A^{q-p}(A^{q-1}-\cdots \pm B^{q-1})^{-1}$$

Answer 2

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 $(A^{-1}B)^k = A^{-k}B^k$, and so, with $C= -A^{-1}B$, 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 $(A+B)^{-1} = \sum_{k=0}^{q-1} (-B)^k A^{p-k}$.