- #1
Bipolarity
- 776
- 2
If AB is invertible, then A and B are invertible for square matrices A and B.
I am curious about the proof of the above. I know there is a very straightforward proof that involves determinants, but I am interested in seeing if there is a proof that doesn't use determinants.
In an attempt to proof this, I considered the contrapositive:
If at least one of {A,B} is singular, then AB is singular.
I successfully proved that if B is singular (or if both A and B are singular), then AB is necessarily singular. To do this, I showed that Bx = 0 having nontrivial solutions implies that ABx= 0 has nontrivial solutions.
Unfortunately, I was not able to apply the above step to the case where only A is singular. If A is singular, Ax= 0 has nontrivial solutions. But how can I show that ABx = 0 has nontrivial solutions?
BiP
I am curious about the proof of the above. I know there is a very straightforward proof that involves determinants, but I am interested in seeing if there is a proof that doesn't use determinants.
In an attempt to proof this, I considered the contrapositive:
If at least one of {A,B} is singular, then AB is singular.
I successfully proved that if B is singular (or if both A and B are singular), then AB is necessarily singular. To do this, I showed that Bx = 0 having nontrivial solutions implies that ABx= 0 has nontrivial solutions.
Unfortunately, I was not able to apply the above step to the case where only A is singular. If A is singular, Ax= 0 has nontrivial solutions. But how can I show that ABx = 0 has nontrivial solutions?
BiP