 Solution

Expand $(I+B{)}^2$ in components:If $B^2=0$, all powers $B^k$ for $k\ge 2$ vanish. The matrix binomial theorem, valid here because $I$ commutes with $B$, therefore yieldsThese powers of the identity plus a square-zero matrix can also be proved by induction: $(I+nB)(I+B)=I+(n+1)B$. For the displayed example, $B^2=0$ directly, andThis is the repeated function composition of a shear mapping. The argument gives every nonnegative integer $n$; since $(I+B{)}^{-1}=I-B$, the same formula also holds for negative integers.