Proof: Let A and B be two n x n matrices such that their product is invertible. Prove that both A and B are invertible...

Question

Let A and B be two n x n matrices such that their product is invertible. (1) Prove that both A and B are invertible,and (2) give their respective inverses.

Please do this without assuming what I am proving and without using determinants. Thanks

Answer 1

Given that $A\in R^{n\times n}$ and $B\in R^{n\times n}$, for $AB$ to be invertible,

$rank(AB) = n$ must be true by definition of an invertible matrix.

We know that this inequality is always true:

$rank(AB) \leq \min (rank(A), rank(B))$

this implies that:

$n \leq \min(rank(A), rank(B))$

Therefore, it is self evident that the rank of an $n\times n$ matrix cannot exceed $n$, so we must have the equality that

$n = min(rank(A), rank(B))$ which is the same as saying

$rank(A) = rank(B) = n$

This implies that the matrices $A$ and $B$ are full rank. All full rank matrices are invertible. Hence, $A$ and $B$ must be invertible.

For your second question, Suppose that $C = AB$. Then:

$ AB = C\\ A = CB^{-1}\\ B^{-1} = C^{-1}A= (AB)^{-1}A\\\\ B = A^{-1}C\\ A^{-1} = BC^{-1}\\ A^{-1} = B(AB)^{-1}\\ $

Note that this means that

$ (AB)^{-1} = B^{-1}A^{-1} $

which is something that is useful and pretty interesting if I do say so myself.

I do hope this was useful. If there are any questions or concerns, please let me know. I will leave it as an excerse for you to prove that $rank(AB) \leq min(rank(A), rank(B))$.