# Thread: Is this proof a correct one?

1. ## Is this proof a correct one?

I've found proofs for this claim but they were too complicated so I came up with my own, though I don't know if it's enogh, would like to get feedback. thank you.
So there is claim saying that
We have a linear system and a matrix A in the order m x n
There is only one solution for Ax=b if and only if rankA=n (n the number of unknowns)

Proof;

(<=) So we know that the dimension space of the solution for Ax=0 is n-rankA (proved it before so I am relying on this fact)

so it is given that rankA=n so we get the dimension of the solution for homogenous system 0, in other words the only solution for Ax=0 is when x=0

And we know that the solution for the non homogenous system Ax=b is in the form x+u when u is the solution for the homogenous system, but we showed that the only solution for homogenous system is 0 so our only solution for Ax=b is x (x+u=x+0=x)

(=>)So if the linear system has only one solution we have to prove that n=rankA
We will assume that rankA<n

So if rankA<n then we get that n-rankA>0 and we know that n-rankA is the dimension of the solution space of the homogenous system and it's bigger than 0 (According to the assumption)

So we know the solution for Ax=b is in the form x+u when u is the solution for homogenous system. And we showed that there are more vectors than zero vector in the homogenous solution space, so lets take u1, u2 that are not equal to each other A(x+u1)=Ax+Au1=b+0=b and A(x+u2)=Ax+Au2=b+0=b
So we got two different solutions and it's a contraddiction to being only solution and therefore RankA=n

So is this correct proof? Thanks

2. ## Re: Is this proof a correct one?

As I see it, your (<=) proof is missing a key element: you have shown that if THERE IS a solution to Ax = b, it is unique. What you still need to do is show that some (particular) solution x0 EXISTS.

And here we have a problem: b may not be in the image space (column space) of A. To see what I mean, let m = 3, n = 2, and consider the matrix:

$\displaystyle A = \begin{bmatrix}1&0\\0&1\\0&0 \end{bmatrix}$

with $\displaystyle b = (0,0,1)^T$.

Then rank(A) = 2, but there is NO SOLUTION to Ax = b, since A(x,y,z)T = (x,y,0)T.

So what you are trying to "prove" IS NOT TRUE.

What IS true, is the following:

A linear system of m equations in n unknowns has AT MOST one solution if and only if rank(A) = n.

3. ## Re: Is this proof a correct one?

Hmmmm, so what you are saying we don't necessarily have solution for every b... How do you prove that something happens "at most one", how do i correct it?

4. ## Re: Is this proof a correct one?

You don't need to change your proof, you need to change what you are "proving".

5. ## Re: Is this proof a correct one?

So I checked my notebook I wrote the claim wrong. The claim is for homogeneous system, Ax=0 not for Ax=b, In this case it's enough to prove that the dimension of the solution space for homogeneous system is n-rank and continue from there I guess..

6. ## Re: Is this proof a correct one?

In the homogeneous case, yes, the statement is an easy consequence of the rank-nullity theorem. Moral of the story: kernels are important.