Proving Ax=b is consistent iff b is in span(S)

[hkus10]

Let S = {w1, w2, ..., wn} be a set of n vectors in R^n and let A be nxn matix whoise columns are the elements of S. Prove that for all b belong in R^n, Ax = b is consistent if and only if b belongs span(S).

My approach is:
I use contrapositive method to prove both sides
First, I prove that if Ax = b is consistent, the b belongs span(s).

Assume that Ax = b is inconsistent. Let x be [x1 x2 ... xn] ***This is a vectical vector which means x1, x2, and xn lines up vectically since I cannot express it in this way.
This means the last row of all vectors in S are zeros and the last row of b has nonzero integer. Then, b cannot be written as a linear combination of vectors in S since 0x1 + 0x2 + ... + 0xn = 0.
Therefore, b does not belong span(S).

For the other side:
Assume b does not belong span(S).
Then, b cannot be written as a linear combination of vectors in S.
If the last row of b is an nonzero integer, then the last row of all vectors in S must be zeros so that b cannot be written as a linear combination of vectors in S.
By the Matrix-Vector Product written in terms of columns, [v1 v2 ... vn][x] not equal to .
Thus, Ax = b is inconsistent.

My question is that this proof seems quite reasonably for me. However, am I really proving this question. If not, how to approach it instead?

Thanks

 

[Undoubtedly0]

Greetings! I don't see any errors in your logic, but it does seem that using contradiction adds a layer of complexity. Here's another way to go about it:

Let $$\vec{x} = [ x_1 \cdots x_n ]^T$$ and $$A = [ \vec{w_1} \cdots \vec{w_n} ]$$. If $$A \vec{x} = \vec{b}$$ is consistent, then there exists a $$\vec{b}$$ such that $$A \vec{x} = \vec{b}$$. Since $$A\vec{x} = x_1\vec{w_1} + \cdots + x_n\vec{w_n} = \vec{b}$$, and since $$Span\{\vec{w_1}, \cdots, \vec{w_n}\} = c_1\vec{w_1} + \cdots + c_n\vec{w_n}$$ for some $$c_1, \ldots, c_n\in R^n$$, clearly $$\vec{b}\in Span\{\vec{w_1}, \cdots, \vec{w_n}\}$$.

To prove the other direction, let $$\vec{b}\in Span\{\vec{w_1}, \cdots, \vec{w_n}\}$$. Then $$\vec{b} = c_1\vec{w_1} + \cdots + c_n\vec{w_n} = A \vec{c}$$, where $$\vec{c} = [ c_1 \cdots c_n ]^T$$. By calling $$\vec{x} = \vec{c}$$, then there exists an $$\vec{x}$$ such that $$A \vec{x} = \vec{b}$$ is consistent, as desired.