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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