Linear Algebra: Understanding Spans and Proving Inclusion in Vector Spaces

[RogerDodgr]

Homework Statement

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Homework Equations

I think I get the basic concept of spans (all possible combinations of vectors with all possible scalers).

The Attempt at a Solution

It seems obvious that the span of S would have to be in the span of T, I don't understand what is left to "prove". I have not done a lot of proofs. I don't know where to begin with this problem.

 

[quasar987]

To prove a set A is a subset of some other set B, you must show that every element of A is also an element of B.

In your case, this will follow very easily from the definitions of span(S) and span(T).

 

[RogerDodgr]

http://www.sudokupuzzles.net/IMG_0035.jpg
Thank you quasar987,
I don't know how I could better state what seems obvious.

Maybe it should say 'All possible linear combinations...'

I won't be including the definition (or my misspelling, the definition was copied from my textbook). Sorry if this seems dumb; maybe I'm not clear what proofs are about; the premise of the question almost seems like proof.

 

[quasar987]

I understand how you feel perfectly. Feels like I was there yesterday!

But stick to it... as you read more proofs and attempt to write some yourself, you will eventually see a pattern in the techniques used and you will pick up the proper vocabulary for writing proofs.

In the meantime, I give you this to feed on. Compare my proof to yours.

Proof: Let s=c_1v_1+...+c_kv_k be an element of span(S).

We want to show that s is an element of span(T) also.

Recall that span(T) is the set of all elements of the form t=d_1v_1+...+d_mv_m. In particular, for d_1=c_1,...,d_k=c_k and d_{k+1}=...=d_m=0, we get that c_1v_1+...+c_kv_k+0v_{k+1}+...+0v_m=c_1v_1+...+c_kv_k=s is an element of span(T).

Since the element s is arbitrary, it follows that all elements of span(S) are in span(T); that is to say, span(S) is a subset of span(T).