Finding a generalized inverse

Question

I am trying to find a generalized inverse for all real $n \times n$ matrices such that it is 1 on the diagonal as well as 1 on the two entries directly above the diagonal, but 0 otherwise. Is there a "slick" way of doing it or does it amount to calculating until a pattern if found?

Answer 1

Yes, denote $J$ the matrix with 1 above the diagonal and $0$ otherwise. $$X=I+J+J^2$$ $$X^{-1}=(I+J+J^2)^{-1}=(I-J)(I-J^3)^{-1}=(I-J)(I+J^3+J^6+\cdots)=I-J+J^3-J^4+J^6-J^7+\cdots$$ Note that $J^n=0$. The sum above has only finite terms which is not $0$. Wish it helps.

Answer 2

You can use the classical method consisting in writing $A=I+N$ where $I$ stands for the identity and $N$ is nilpotent or order $n$ (check it, $N^{n}=0$). Then, Use Newton's binomial expansion on $A^{n}$ to obtain a formula for the inverse by gathering all powers of $A$ on one side and identity on the other side.