CatalinBordea wrote: Consider a natural number, $ n\ge 2, $ and three $ n\times n $ complex matrices $ A,B,C $ such that $ A $ is invertible, $ B $ is formed by replacing the first line of $ A $ with zeroes, and $ C $ is formed by putting the last $ n-1 $ lines of $ A $ above a line of zeroes. Prove that: a) $ \text{rank} \left( A^{-1} B \right) = \text{rank} \left( \left( A^{-1} B\right)^2 \right) =\cdots =\text{rank} \left( \left( A^{-1} B\right)^n \right) $ b) $ \text{rank} \left( A^{-1} C \right) > \text{rank} \left( \left( A^{-1} C\right)^2 \right) >\cdots >\text{rank} \left( \left( A^{-1} C\right)^n \right) $ Can you clarify how the matrix $C$ is formed?

Acridian9 wrote: CatalinBordea wrote: Consider a natural number, $ n\ge 2, $ and three $ n\times n $ complex matrices $ A,B,C $ such that $ A $ is invertible, $ B $ is formed by replacing the first line of $ A $ with zeroes, and $ C $ is formed by putting the last $ n-1 $ lines of $ A $ above a line of zeroes. Prove that: a) $ \text{rank} \left( A^{-1} B \right) = \text{rank} \left( \left( A^{-1} B\right)^2 \right) =\cdots =\text{rank} \left( \left( A^{-1} B\right)^n \right) $ b) $ \text{rank} \left( A^{-1} C \right) > \text{rank} \left( \left( A^{-1} C\right)^2 \right) >\cdots >\text{rank} \left( \left( A^{-1} C\right)^n \right) $ Can you clarify how the matrix $C$ is formed? $ A $ is formed by $ n $ lines $ L_1,L_2,\ldots ,L_n, $ disposed up to down in this order. $ C $ is formed by the lines $ L_2,L_3,\ldots ,L_n,O, $ disposed up to down in this order, where $ O $ is a line with zeroes.

Ok, thanks.

I'll just give a <hint.>