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I'm really struggling with "simple" proofs, and I could only come up with this. I googled for the solution and know how to do it now, but I first want to know if there's anything wrong with my solution? $$ (AB)^{-1}=B^{-1}*A^{-1}. $$ For this equation to hold, $B$ must be invertible and so must $A$. //

Git Gud
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    Stating that equation requires assuming $B^{-1}$ and $A^{-1}$ exist, so that's not a satisfactory proof. – Potato Jul 18 '13 at 23:37

4 Answers4

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First: your solution isn't actually a solution. By writing $A^{-1}$ you are assuming that $A$ is invertible, which is what you want to prove.

One way to do this would be to note that $$ \det(AB) = \det(A)\det(B). $$ If $AB$ is invertible, then $\det(AB)\neq 0$, and so $\det(A)\neq 0$ and $\det(B)\neq 0$.

Another way: If $AB$ is invertible, then there is a $C$ such that $$ (AB)C = I $$ That is $$ A(BC) = I $$ So $BC$ is a right inverse of $A$, and so $A$ is invertible. Likewise $B$ is invertible. (See this question/answers for a bit more: If $AB = I$ then $BA = I$).

Community
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Thomas
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Suppose $AB$ is invertible. Then there is a matrix $C$ such that $C(AB)=I$ and $(AB)C=I$. Rearranging the parenthesis, we have $(CA)B=I$ and $A(BC)=I$, which shows $A$ and $B$ are invertible.

Potato
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  • *which shows that $A4 and $B$ are left and right invertible. The point is not to nitpick (although I do enjoy that), it's just that I don't think the OP will catch the nuance by himself. – Git Gud Jul 18 '13 at 23:42
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    @GitGud It in fact shows they are invertible, since being left or right invertible is equivalent to being invertible. (Posting this for the benefit of the OP.) – Potato Jul 18 '13 at 23:43
  • THanks a lot guys. This forum has helped me so much –  Jul 18 '13 at 23:45
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What makes this proof tick is the following: If $T:V\to W$ is a linear transformation between finite dimensional vector spaces having the same dimension, then $T$ is injective iff $T$ is surjective iff $T$ is bijective. (This result follows immediately from the rank-nullity equation).

Thinking of matrices as linear transformations, the claim is equivalent to: Given finite dimensional vector spaces $U,V,W$, all having the same dimension, and linear transformations $T:U\to V$ and $S:V\to W$, prove that if $S\circ T:U\to W$ is invertible then both $T$ and $S$ are invertible. Here invertible simply means that the function is injective and surjective.

As $S\circ T$ is injective it follows immediately that $T$ is injective. As $S\circ T$ is surjective it follows immediately that $T$ is surjective. The result now follows from the above mentioned result.

Ittay Weiss
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I would like to offer another perspective for how to go about proving this which might provide an intuition that perhaps the other answers might lack for some.

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.

Now that we know that $A$ and $B$ are invertible, suppose that $C = AB$. Then:

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

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

samuel schreiber
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