Jump to content

Span

Featured Replies

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

Archived

This topic is now archived and is closed to further replies.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.