An easier proof is to notice that the linear transformation represented by is injective; hence the only solution to is . Since , must take everything to the zero vector.
A is an c x d matrix. B is a d x k matrix.
If rank(A) = d and AB = 0, show that B = 0.
My textbook has a solution but I don't understand it:
The rank of A is d, therefore A is not the zero matrix. (I asked my prof why d can't be equal to zero, he said it just couldn't...?)
If you left multiply A by some elementary matrix to bring it to row echelon form, you get a matrix that looks like:
[ 1 * * * ... *
0 1 * * ... *
0 0 1 * ... *
0 0 0 0 ... 0] (NOTE: * are arbitrary numbers)
And we will write B as a column (1 x k), consisting of [B1, ... , Bd]T
Multiply A and B together, and you get a column that looks like [R1, R2, ... 0, 0, 0]T
For AB = 0, then Ri = 0. Then since A is not zero, B is 0.
This proof seems to make no sense. Why are we writing B as 1 x k, when the question says B is d x k?
And I don't really get how this actually proves anything... =\
Good! Then you are on the right track.
I think that when they write as , the elements are actually the columns of . See if that helps...
What is your textbook? Probably not the best. In my opinion there is no reason to see it in any other way than the one I stated, because it really points out what is actually going on theoretically underneath. The proof you quote is ugly and unilluminating, in my opinion.