67th Romanian Mathematical Olympiad

a) Consider the matrix $P\in {\mathcal{M}}_n(\mathbb{C})$, $$P=\left(\begin{array}{ccccc}0& 0& 0& \cdots & 0\\ 0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\end{array}\right).$$ It is not difficult to see that $D=A^{-1}PA$. Since $P^2=P$, a standard inductive argument shows that $P^k=P$, $k\in {\mathbb{N}}^{\ast }$. Atunci $D^k=(A^{-1}PA{)}^k=A^{-1}{P}^{k}A=A^{-1}PA=D$, $k\in {\mathbb{N}}^{\ast }$. Therefore, $\mathrm{rank}(D^k)=\mathrm{rank}(D)=\mathrm{rank}(P)=n-1$, $k=1,\dots ,n$.

b) Let $Q\in {\mathcal{M}}_n(\mathbb{C})$, $$Q=\left(\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \cdots & 1\\ 0& 0& 0& \cdots & 0\end{array}\right).$$ Again, it is easy to check that $E=A^{-1}QA$. The equality $E^k=(A^{-1}QA{)}^k=A^{-1}{Q}^{k}A$, $k\in {\mathbb{N}}^{\ast }$ implies $$\mathrm{rank}(E^k)=\mathrm{rank}(Q^k)=n-k,k=1,\dots ,n,$$ hence the conclusion.