$\text{rank}(AB^2)=\text{rank}(AB)$

Question

Let $A$ and $B$ be real $n×n$ matrices such that $$AB=BA,~~\text{rank}(A^2)=\text{rank}(A),~~\text{rank} (B^2)=\text{rank}(B)$$

Show that: $$\text{rank}(AB^2)=\text{rank}(AB)$$

Answer 1

Using Sylvester Theory: $$ Rank(AB^{2}) = Rank(B^{2}) - dim(Im(B^{2}) \cap Ker(A) ) $$ and $$ Im(B^{2}) = Im(B) $$ since $$ Rank(B^{2}) = Rank(B) $$

Hence ,we can obtain \begin{array} RRank(AB^{2}) &=& Rank(B^{2}) - dim(Im(B^{2}) \cap Ker(A) )\\ &=& Rank(B) - dim(Im(B) \cap Ker(A))\\ &=& Rank(AB) \end{array} From the above process,we can see that you have given much more conditions with the redundant terms $$ Rank(A^{2}) = Rank(A) \quad AB = BA $$

Answer 2

Let $V =\mathbb{R}^n$ denote the underlying space, so that $A, B:V \to V$. The spaces $B(V)$ and $B^2(V)\subset B(V)$ have the same dimension and are thus equal. Hence $$AB^2(V) = A(B^2(V)) = A(B(V)) = AB(V).$$ Taking dimensions gives the result.