Finding the rank of matrix $A^2$

Question

supose $A$ is a $4\times4$ matrix such that $\operatorname{rank}(A)=4$. Find the rank of the matrix $A^2$. if there is a major rule for the power $k$ and not specially the power $2$.

Answer 1

if $rank(A) = 4$ for $4 \times 4$ matrix, then it is invertible. therefore $A^2$ is invertible and has rank $4$ too.

Answer 2

Hint: for a $4\times 4$ matrix with $\operatorname{rank}(A)=4$ you have a one-to-one and onto function $$\widetilde{A}: K^{4\times 1}\rightarrow K^{4\times 1}, x\mapsto A\cdot x$$.

For $A^2$ you simply get $\widetilde{A}\circ\widetilde{A}$ which also is a one-to-one and onto function.

Answer 3

Just notice that $$\det(A^2)=\det(A)^2\neq 0$$

Answer 4

Use Sylvester's formula: rank(A)+rank(B)-n≦rank(AB)≦min[rank(A),rank(B)] where A and B are of orders m×n and n×p respectively. Setting A=B here with n=4, 4≦rank(A^2)≦4.