### Span

##### Properties of span

Theorem 1 span{v1, v2, ..., vk} ⊂ span{v1, v2, ..., vk, vk+1}. Moreover, equality holds ⇔ vk+1span{v1, v2, ..., vk}.

Proof In case k = 2, for uspan{v1, v2} we have

u = c1v1 + c2v2 = c1v1 + c2v2 + 0v3span{v1, v2, v3}.

Thus we conclude span{v1, v2} ⊂ span{v1, v2, v3}.

If span{v1, v2} = span{v1, v2, v3}, then

v3 = 0v1 + 0v2 + 1v3span{v1, v2, v3} = span{v1, v2}.

Conversely, if v3span{v1, v2}, then v3 = av1 + bv2, and for uspan{v1, v2, v3} we have

u = c1v1 + c2v2 + c3v3 = c1v1 + c2v2 + c3(av1 + bv2) = (c1 + ac3)v1 + (c2 + bc3)v2span{v1, v2}.

Thus we conclude span{v1, v2, v3} ⊂ span{v1, v2}. Combined with the first part, we have the equality.

This completes the proof for k = 2. The general case is similar.

Theorem 2 span{v1, v2, ..., vm, w1, w2, ..., wn} = span{v1, v2, ..., vm} + span{w1, w2, ..., wn}.

Proof Any vector in span{v1, v2, ..., vm} + span{w1, w2, ..., wn} is of the form v + w, where

v = c1v1 + c2v2 + ... + cmvmH, w = c'1w1 + c'2w2 + ... + c'nwnK.

Thus the vectors in the sum are

v + w = c1v1 + c2v2 + ... + cmvm + c'1w1 + c'2w2 + ... + c'nwn,

which are exactly vectors in span{v1, v2, ..., vm, w1, w2, ..., wn}.

Theorem 3 The three operations do not change the span.

Proof We only consider special cases of the operations on three vectors. The general case is similar.

For the first operation, we will prove span{v1, v2, v3} = span{v1, v2 + rv1, v3}. For uspan{v1, v2, v3}, we have

u = c1v1 + c2v2 + c3v3 = (c1 - rc2)v1 + c2(v2 + rv1) + c3v3span{v1, v2 + rv1, v3}.

This proves span{v1, v2, v3} ⊂ span{v1, v2 + rv1, v3}. Conversely, for uspan{v1, v2 + rv1, v3}, we have

u = c1v1 + c2(v2 + rv1) + c3v3 = (c1 + rc2)v1 + c2v2 + c3v3span{v1, v2, v3}.

This proves span{v1, v2, v3} ⊃ span{v1, v2 + rv1, v3}.

For the second operation, we will prove span{v1, v2, v3} = span{v2, v1, v3}. For uspan{v1, v2, v3} we have

u = c1v1 + c2v2 + c3v3 = c2v2 + c1v1 + c3v3span{v2, v1, v3}.

This proves span{v1, v2, v3} ⊂ span{v2, v1, v3}. The inclusion in the other direction is similar.

For the third operation, we will prove span{v1, v2, v3} = span{dv1, v2, v3} for d ≠ 0. For uspan{v1, v2, v3} we have

u = c1v1 + c2v2 + c3v3 = (c1/d)dv1 + c2v2 + c3v3span{dv1, v2, v3}.

This completes the proof that span{v1, v2, v3} ⊂ span{dv1, v2, v3}. The inclusion in the other direction can be proved similarly.