Linear algebra proof with trivial solution

[SpiffyEh]

Homework Statement

Problem:
Prove Ax=b has a solution for each b in R^m if and only if the equation A^T x = 0 has only the trivial solution.

Hint: For the forward direction use theorem 1.4.4 to prove that the dimension of the null space pf A^T is zero

Homework Equations

Theorem 1.4.4: Let A be an mxn matrix. Then the following statements are logically equivalent. That is, for a particular A, either they are all true statements or they are all false.
a. For each b in R^m, the equation Ax=b has a solution
b. Each b in R^m is a linear combination of the columns of A
c. The columns of A span R^m
d. A has a pivot position in every row

The Attempt at a Solution

Can someone please help me through this proof? I'm not even sure how to begin it. Thank you

 

[╔(σ_σ)╝]

Isn't there a simpler way of going about this ?

Correct me if I'm wrong but isn't it true that

B is invertible iff BX = 0 has only the trivial solution ?

Therefore if A^T x = 0 has only the trivial solution. A^T is invertible then so is A.

Which means Ax=b always has a solution .

EDIT

If Ax= 0 had non trivial solutions then A is not invertible, in which case it does not reduce to RREF( reduced row echeol form) and can not have unique solutions.

In which case a row of zero exist somewhere and by putting a 1 on that row in the column matrix of B we can show that there is no solution.

 

[SpiffyEh]

That does make sense, I don't see why it wouldn't work that way. The information above is just what I was given to do this problem. I like your way, its simple and makes sense.

 

[╔(σ_σ)╝]

My only concern is that, you may still need to know perhaps the more difficult way of doing things, for exams or what not.

 

[SpiffyEh]

Yeah, I'm not sure. I have an exam coming up but we usually don't get asked things that are too indepth since the exam only lasts an hour and 20 minutes. Do you understand how to do it the other way? I think it would be good to know just in case.

 

[╔(σ_σ)╝]

Sorry I can't help you there. It's been a while since I took linear algebra. Heck, I can't even remember what it means by A has a pivot position in every row.

 

[SpiffyEh]

lol that's ok. Thanks for your help with the simple proof though