# Thread: Matrix Rank Proof - I don't understand the textbook solution

1. ## Matrix Rank Proof - I don't understand the textbook solution

The question:
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.

Solution given:
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... =\

2. An easier proof is to notice that the linear transformation represented by $\displaystyle A$ is injective; hence the only solution to $\displaystyle Ax=0$ is $\displaystyle x=0$. Since $\displaystyle AB=0$, $\displaystyle B$ must take everything to the zero vector.

3. Oh yes, I did notice that. I figured understanding the solution given might come in helpful though! I'd still like to try and understand it

4. Good! Then you are on the right track.

I think that when they write $\displaystyle B$ as $\displaystyle 1 \times k$, the elements $\displaystyle B_1, \dots, B_d$ are actually the columns of $\displaystyle B$. 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.