Understanding Matrix Solutions: AX=B vs. AX=0

[aaaa202]

Let A denote an mxn matrix and let A' denote the row echelon form of it, which has d steps. We then have according to my textbook:
1) If m>d there exists a column such that the set of equation has no solutions.
2) If n>d the matrixequation AX=0 has a set of solution expressed parametrically by (n-d) parameters.
3) If m=n=d there exists a unique solution for every B in the equation AX = B.

Now 1) and 3) I understand. What troubles me is 2). Why to they switch the equation to AX=0 rather than AX=B? Wouldn't that last equation also be dependent on n-d parameters. I'm pretty sure that this has something to do with the fact that the nullspace is quite a unique thing since it forms a linear subspace. Thus we can define its dimension and later use all this to prove the rank nullity theorem. So can someone explain what's going on on a deeper level?

 

[Xiwi]

I assume by steps you mean pivots. In this case I prefer the term "rank" i.e. rank = d.Solution for problem AX = b is X = X_p + X_n.
Where,

AX = A(X_p) + A(X_n) = b + 0 = b.

X_p = particular solution
X_n = solution of Ax = 0 ... (n is not a number here, it is just a symbolic name)(n-d) parameters are the free variables where (n-d) is the dimension of the null space.

What this means is that you have a whole space of dimension (n-d) and each vector in that space is a solution of Ax=0. Hence the name nullspace.

 

[Xiwi]

You use the word "column" in your first claim. By column do you mean the column vector b of Ax = b or a column of matrix A?

You ought to use standard terms here. If the book you're reading is using "column" for "column vector" and "steps" for "pivots", I suggest that you throw it away.

 

[mathwonk]

in part 2, you could use AX=B, but the statement would be more complicated. I.e. you could say, for every B, AX=B either has no solutions or has solutions depending on n-d parameters. but in fact this is implied by the special case AX=B, since if C is a solution of AC=0, and if D is a solution of AD=B, then also A(C+D) = B. So as long as AX=B has at least one solution, then its set of solutions is in one one correspondence with the solutions of AX=0. (I.e. notice that A0=0 is true, so when B=0, the equation Ax=B does always have at least one solution.)