# Let $A,B$ be two $n\times n$ matrices each having $\text{rank}=n$. Then $\text{rank}A^3B^2A=n$.

Is this statement true/false?If true prove it if false then give a counter example.

Let $A,B$ be two $n\times n$ matrices each having $\text{rank}=n$. Then $\text{rank}A^3B^2A=n$.

I tried various examples where I am getting the result to be true. Also I am unable to prove the result.What should I do?

There are several essentially equivalent ways to solve this problem. Here are two possible approaches:

1. An $n\times n$ matrix has rank $n$ if and only if its kernel is zero. So it's enough to show that the kernel of $A^3B^2A$ is zero.

2. An $n\times n$ matrix has rank $n$ if and only if its determinant is non-zero. And what is the determinant of $A^3B^2A$ in terms of the determinants of $A$ and $B$?

• $\det A^3B^2A=(\det A)^3(\det B)^2\det A\neq 0$ – Learnmore Aug 16 '16 at 6:26
• Yes, that's correct. – carmichael561 Aug 16 '16 at 15:23

Major hint:

If $A$ is an $n\times n$ matrix of rank $n$ then it is both injective and surjective: see a proof here.

In otherwords, it is bijective. That is to say in more common words, it is an invertible matrix.

Similarly $B$, also being an $n\times n$ matrix of rank $n$, is also an invertible matrix.

In fact, we know that any invertible matrix of size $n\times n$ will also be of rank $n$.

Will the matrix in question, $A^3B^2A$ be invertible? Not invertible? Do we not know?

Is the product of two invertible matrices invertible? If $E$ and $F$ are invertible, what is $(EF)(F^{-1}E^{-1})=$?