The discussion revolves around the invertibility of the product of two matrices, specifically AB, where A is invertible and B is not. Participants explore the implications of B's non-invertibility on the product AB and also consider the case of BA.
The discussion is active, with participants examining the reasoning behind the non-invertibility of AB and BA. Some guidance has been offered regarding the kernel of B and its effect on the products, but no consensus has been reached on alternative methods to demonstrate these properties.
Participants are working under the assumption that A is invertible while B is not, and they are exploring the consequences of these conditions on the products of the matrices.
I see. So then AB has a non-trivial kernel, which means that AB is not invertible.fresh_42 said:If B is not invertible, it has a non-trivial kernel. Take a vector from it and apply AB.
The same. Since A is invertible, we can find a y to an element x of B's kernel, such that x=Ay. Now Bx=0=BAy and y is in the kernel of BA.Mr Davis 97 said:I see. So then AB has a non-trivial kernel, which means that AB is not invertible.
What about if we wanted to show that BA is not invertible, given that B is not invertible?
Ah! Makes perfect sense. Thanksfresh_42 said:The same. Since A is invertible, we can find a y to an element x of B's kernel, such that x=Ay. Now Bx=0=BAy and y is in the kernel of BA.