Re: Rank A and Basis of Rn

Hey Kyo.

You can use the fact directly that det(A) <> 0 (and A is square) shows that all vectors are linearly independent and since det(A) = det(A^t) then A^t also has the same property.

I'm not not sure how they expect you to assert det(A) = 0, but once you do that the rest follows.

Re: Rank A and Basis of Rn

Thanks chiro, but are you referring to AB being square? My A matrix is actually an m x n matrix.

Re: Rank A and Basis of Rn

Whatever matrix you are using as your basis (or trying to prove that a basis exists).