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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?

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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 $B$?

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  • $\begingroup$ $\det A^3B^2A=(\det A)^3(\det B)^2\det A\neq 0$ $\endgroup$ – Learnmore Aug 16 '16 at 6:26
  • $\begingroup$ Yes, that's correct. $\endgroup$ – carmichael561 Aug 16 '16 at 15:23
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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 $(EF)(F^{-1}E^{-1})=$?

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