0

The approach I am thinking of is examining $AX=I$ and then solving for $X$, but that seems a bit tedious, if I am not missing something. Any help or hints are appreciated!

matte_studenten
  • 305
  • 2
    Does this answer your question? Inverse of an upper triangular matrix with all entries 1 – Anne Bauval Sep 09 '22 at 14:55
  • @AnneBauval Probably. If $A\in T_n$ then we can say $A= I-B$, where B is nilpotent. According to a theorem in my coursebook, for nilpotent matrices B we have $(I-B)^{-1}=I+C$ where C is also nilpotent. $(I+C)\in T_n$ and we are done. Is this approach correct? – matte_studenten Sep 09 '22 at 15:14
  • Not quite ($C$ nilpotent does not imply $I+C\in T_n$), but a correct and shorter proof is in the link above. – Anne Bauval Sep 09 '22 at 15:27
  • @AnneBauval I guess the top answer but I am not sure where the sum $I-N+N^2-N^3+\cdots$ comes from. – matte_studenten Sep 09 '22 at 15:51
  • The problem here seems to be setup in the title, with the entries of upper triangular $n\times n$ matrix $T$ unspecified above the diagonal (unlike the proposed duplicate, @AnneBauval). The Question in its current form needs editing, so that a complete problem formulation appears in the body of the Question (not in the title alone). – hardmath Sep 16 '22 at 22:06

1 Answers1

1

Yes, the top answer. And if $A=I+N$ with $N^k=0$, then $C:=-N+N^2-N^3+\dots+(-N)^{k-1}$ (which is not only nilpotent but, more precisely, uppertriangular with $0$'s on the diagonal, since $N$ is) satisfies: $$A(I+C)=AI-AN+AN^2+\dots=I+N-N-N^2+N^2+\dots=I$$ (by telescoping) and similarly, $(I+C)A=I.$

Anne Bauval
  • 49,005