Proving that the product of two invertible matrices is also invertible, without determinants

Question

I want to prove that if there are two matices A and B and A and B are both invertible, then the product A * B is also invertible.

This question is similar: Prove that the product of two invertible matrices also invertible but the answer uses determinants, which I have not learned about yet.

There is an answer provided in the thread above which does not use determinants.

Ans 2 is:

It is that $(AB)^{-1}=B^{-1}A^{-1}$, because $AB(AB)^{-1}=ABB^{-1}A^{-1}=1\!\!1$, but only for $n\times n$ matrices.

But I can't see how this shows that AB = matrix C such that C is invertible (such that there exists a matrix D where CD = DC = I, which is the definition of invertible given to me).

The other answer starts off by saying:

C is invertible iff for all y there is some x such that Cx = y.

I thought C is invertible iff there exists a matrix D such that CD = DC = I. How is it the case that for every y, there is some x such that Cx = y? Is y a matrix in this case? and is x a scalar?

Answer 1

I don’t know if I understood your doubt. However, there is a very simple answer to the first question of your post. Suppose $A$ and $B$ are $n\times n $ matrices, both invertible. Then there are matrices $A^{-1}$ and $B^{-1}$ such that $A^{-1}A=AA^{-1}=I$ and $B^{-1}B=BB^{-1}=I$. We want to find a matrix that satisfies the same kind of relations for $AB$. Well, such matrix is $B^{-1}A^{-1}$. We prove this by direct verification; $$B^{-1}A^{-1} AB=B^{-1}IB=B^{-1}B=I, $$ and, analogously, $$ ABB^{-1}A^{-1}=I.$$

This concludes the proof.

Answer 2

Let $C $ be $n\times n $ and $y=n\times 1$. Then if for every $y $ there is $x $ such that $Cx=y $ we are saying that the columns of $C $ span $\mathbb{R}^n $.

In particular, for each elementary vector $e_i $ ($1$ in ith place, zero elsewhere) there is $x_i $ such that $Cx_i=e_i $. The inverse of $C $ then contains these $x_i $s as its columns.