Vector spaces span

Can someone explain what span actually is. Ive looked at definitions but still dint get it.

The span of vectors (1,0,0) and (0,1,0) is the subspace consisting of vectors of the form (a,b,0). More generally, the span of {v1, v2, …, vn} is the space of vectors {a1 v1 + a2 v2 + … + an vn}.

So its just the constants being multiplied to each vector in the set?
Could you help me on this question please. I dont know what an ordered basis is.

The eg question

Posted from TSR Mobile

Your "proof" that the span is R^3 isn't really enough - you'd have to show that for every a,b,c it is possible to find the $\lambda_1, \lambda_2,\lambda_3$. However, because we're in a finite-dimensional vector space, all we need to do is show one of spanning or linearly independent; then because the space is of dimension 3, and we have three spanning/LI vectors, they must necessarily be LI/spanning.

The basis is ordered if it has an ordering. That is, you just impose an order on a basis and it becomes an ordered basis. That's all there is to it.

You may also be interested to know that you draw your lambdas left-to-right reversed from the conventional way.

I'm sorry but im not understanding what ypu are saying. I did show that it is LI and you said 'all we need to do is show one of spanning or linearly independent'.

And yeah its just a habit.

Posted from TSR Mobile

Yes, you showed it was LI - but before that, you "showed" that it was spanning, and that "showing" was very much incomplete. If you'd just left out that "spanning" bit, I wouldn't have complained!

Oh I get you. Say if I were to just show that it spans R3. What would I do? It kinda made sense to me - what I did. Ive never really seen and eg

To show that it spans R^3, you need to pick an arbitrary vector (a,b,c) in R^3, and show that it can be made as a linear combination of the spanning vectors. That is, you need to find the $\lambda_i$ such that $\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 = (a,b,c)$.

Like that?

Yep, exactly. (I'm assuming your arithmetic is correct.) As you can see, linear independence is often easier than spanning to prove.

Thanks. Lastly how do I impose an order on the basis? What does that mean?

Posted from TSR Mobile

You just announce that "the first vector in the basis is <this one>, the second is <this one> and the third is <this one>". That's it.

You just announce that "the first vector in the basis is <this one>, the second is <this one> and the third is <this one>". That's it.

Hi sorry to bother again but what does it mean by extend? The span of those two vectors give already equals R^4 doesnt it.

Posted from TSR Mobile

Certainly not. $\mathbb{R}^4$ is four dimensional, so any set of vectors that spans it must contain at least four vectors.

Totally forgot that. So if I extend it by adding for eg (1,1,1,1) and (2,2,2,2) to the given set, would that work?

Posted from TSR Mobile

No. The resulting set will not be linearly independent, and hence it won't span. (Can you see why?)
You can make your job a lot easier, because you already know of a collection of four basis vectors. It's likely that two of them will be independent of the two already given. (It's not certain, I believe, but it's likely.)

Oh damn. Because 2 (1111) is 2222. I'm guessing (1110) and (0111) would work after considering if they can be turned into the ones given. Pretty sure they cant because of those zeros.