Question

proof rank(AB)= 13 years ago

Answer 1

Not sure what properties you're allowed to use here, but this comes from the Fundamental Theorem of Linear Algebra.

rank(AB) = dim(R(AB)), where R denotes the range of AB (this is also known as the column space). dim(R(AB)) <= dim(R(A)), because AB is a linear combination of the columns of A. You can say the same thing about the rows of B, which imply dim(R(AB)) <= dim(R(B)). So, clearly, dim(R(AB) <= min(dim(R(A)),dim(R(B))), and the result is proved.

If you'd like to state this without the use of row space, you can also note that rank(AB) = rank((AB)^T) = rank(B^TA^T). From here, you can use the same argument as you did for A, where B^TA^T is a linear combination of the columns of B^T. But, dim(R(B^T)) = dim(R(B)), so rank(AB) = rank(B^TA^T) <= rank(B^T) = rank(B).

Two similar ways to approach the problem. I prefer the second way as you don't have to make use of the row space at all to achieve the result.