- #1
hkus10
- 50
- 0
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
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