# More Linear Algebra Help Needed.

How would this be proven?

Remember that a set of vectors is in its own span, and that the span of a set of vectors is the set of all linear combinations of vectors in that set. That should help you.

Suppose that span {v1,v2,?..vr} = span {w1,w2?.wk}. Since each vector vi in S belongs to span {v1,v2,?..vr}, it must by the definition of span{w1,w2,?..wk} be a linear combination of the vectors in S?. The converse must also hold.

Suppose that each vector in S is a linear combination of those in S? and conversely. Then we can express each vector v as a linear combination of the vectors w1, w2,?wk, so span{v1,v2,?vr} C_ span{w1, w2,?wk}. But conversely we have
span{w1,w2,?wk} C_ span{v1,v2,?vk}, so the two sets are equal.

Jeez Us where did all those question marks come from?

Suppose that span {v1,v2,…,vr} = span {w1,w2…,wk}. Since each vector vi in S belongs to span {v1,v2,…,vr}, it must by the definition of span{w1,w2,…,wk} be a linear combination of the vectors in S’. The converse must also hold.

Suppose that each vector in S is a linear combination of those in S’ and conversely. Then we can express each vector v as a linear combination of the vectors w1, w2,…,wk, so span{v1,v2,…,vr} C span{w1, w2,…,wk}. But conversely we have span{w1,w2,…,wk} C span{v1,v2,…,vk}, so the two sets are equal.

Oh God help me, I meant to hit preview not submit. I wanted to add that the above proof is right from a textbook.