Linear Algebra Proof: Rank and Zero Matrix

[jumbogala]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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? It says in the question B is d x k! Also if A is not zero then why can't you say right off the bat that AB = 0 implies B =0?

 

[boboYO]

Also if A is not zero then why can't you say right off the bat that AB = 0 implies B =0?

because these are matrices not numbers. for example

A= [0 1
   0 0]B=[1 0 
   0 0]

AB=0 yet neither A or B are 0.as to why they say 'write B as 1xk', maybe they mean write Bv (i.e. B times an arbitrary vector) as a 1xk?

 

[jumbogala]

But when A is in row echelon form and you multiply it by some B, the because the solutions are zero the entries of B must be zero??