0
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ What you quote as "Ans 2" is saying exactly what you want. First, only square matrices have inverses. Then "Ans 2" is showing that for $C=AB$, there is a matrix $D$ such that $CD=DC=I$. That matrix $D$ is $B^{-1}A^{-1}$. The computation is showing that that choice works. It requires using that matrix multiplication is associative. $\endgroup$ – user566930 Jul 7 '18 at 14:18
  • 2
    $\begingroup$ I agree that it is written in the wrong order of the inference. You can order it as $CD=(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I$. And then you can also compute $DC$ if you want and again get $I$. $\endgroup$ – user566930 Jul 7 '18 at 14:22
  • $\begingroup$ @minghan Ohh, okay. Makes a lot more sense after seeing it reordered. Thanks. $\endgroup$ – user2719875 Jul 7 '18 at 14:26
0
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Oh, okay. So isn't "C is invertible iff for all y there is some x such that Cx = y." non-trivial, and thus, have to be proven? It is not in the definition an what an invertible matrix is, right? $\endgroup$ – user2719875 Jul 7 '18 at 14:29
  • $\begingroup$ Sure, it is not exactly obvious/trivial and some argument/s need to be given to shiw it (as in the answer), starting from the usual definition of invertibility $\endgroup$ – AnyAD Jul 7 '18 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.