Let $B_{n\times n}$ be a nilpotent matrix. How should I go around proving that $B$ is not invertible?
I was thinking this: if $B*B = 0$ , then if I rank $B$ to echelon form, I will always have two or more rows that are the same, and then because $B$ is square, it is not possible to find another matrix $(B^{-1})$ that will transform $B$ to $I_{n}$…
But I don't know how to write it formally.
Best Answer
Usually, nilpotent means that $B^m=0$ for some $m>1 $, not necessarily $2$.
A direct way to see that $B $ is singular is $$ 0=\det (B^m)=(\det (B))^m, $$ so $\det (B)=0$.
Another way, without using determinants: if $B $ were invertible, then $$B=(B^{-1})^{m-1}\,B^m=0, $$ a contradiction.