What's does the span mean?

Sander1337

Hi there!

My name is Sander, and I'm a (dutch) Econometrics student at the Rotterdam Erasmus University. This is my first year and I'm getting an introduction to "Matrix Algebra", which is (I think, shoot me if I'm wrong..) quite the same as Linear Algebra (since the book we use is called Linear Algebra, by David Poole).

The problem I encountered is the word span, I don't quite get what is means, and even more important; what it's results would be if I would "calculate" the span of some vectors/matrices.

The given definition of Span from my book is;

If S={v1,v2,.....,vk} is a set of vectors in R^n, then the set of all linear combinations of v1,v2,......,vk is called the span of v1,v2,.....,vk and is denoted by the span(v1,v2,.......,vk) or span(S). If span(S)=R^n, then S is called a spanning set for R^n.

Now as I said I don't quite get what they are trying to explain, this definition doesn't give me a clear view of what this means, and so I haven't got a clue of what span means and what answer(s) span(S) would get me.

I hope you can help me out, and give me a more easy-to-use explanation of the term "span" and what the span delivers me. I've already searched with google but that didn't bring in any results.

Really hope you can help me out!

Greets,

Sander

PS: Sorry for the bad English, like Linear Algebra, following education in English is new for me as well...

VeeEight

Do you know what a linear combination? If not then you surely need to review that definition.

The space 'spanned' by some vectors is just the space that is 'created' if we take all linear combinations of those vectors. It is easier to write down a space by it's spanning set so that you don't have to write out infinitely many terms each time, which would be impossible!

For example, S= {(1,0), (0,1)} 'creates' R x R (or R^2) since any point in R^2 can be thought of as a linear combination of elements in the set S.

Mark44

To extend what VeeEight said, the set S = {(1,0), (0,1)} spans R2 because any point (vector) (x, y) in R^2 can be written as the sum of scalar multiples of the two vectors in S. "Linear combination of something" means the sum of scalar multiples of of something.
Here is (x, y) written as a linear combination of the vectors in S:
(x, y) = c_1(1,0) + c_2(0,1)

Sander1337

VeeEight and Mark44 many thanks for your response.

I do know what a linear combination is, at least, I think I know. My book has learned me that if vector X and vector Y are linear combinations, aX+bY should give me the 0-vector, or if Z is a linear combination of X and Y then aX + bY should give me Z. (Am I right with that?)

So if I get the picture, the answer to a question like; what's the span of S, where S is a set of vectors v1.....vk, S={v1,v2,.....,vk} and each vector vk is a column-vector built up out of 5 elements each (i.e. {2,0,4,6,7}), then the answer'd be "in what space these linear combinations could ultimately be", so in this case the answer'd be R^5 (since you can ultimately end op with a 5-dimensional matrix?)? Do I get this right or am I interpreting your explanation wrong?

If I am; this is my so-far knowledge of Linear Algebra;
1) Linear combinations
2) DOT products
3) Matrix multiplication (block and/or element-wise multiplications)
4) Special (elementairy) Matrices (like the 0-matrix, or the I-matrix (like I^3=[1 0 0 , 0 1 0 , 0 0 1]) and multiplication with these special matrices (like A*I=A)
5) Row-Echelon forms
6) Gauss-Jordan eliminations
7) Tranlations and "invertions" (don't know what the english word for "inverteren" is, it means ^-1)
8) And ofc. vectors

This is my knowledge so far, hope you can help me out!

Thanks so far!

Greets,

Sander

Mark44

The first part isn't right. It should say: If vectors X and Y are linearly independent, then the equation aX + bY = 0 has only one solution--a = b = 0. If X and Y are linearly dependent, the same equation will still have a = b = 0 as a solution, but there will be other, nonzero values for a and b. This is a very important concept, so it's crucial that you understand it.
If your set S has 5 vectors in it, and each has 5 components, the set might span R^5 or it might not. If these vectors are linearly independent, then they will span R5, which means that any arbitrary vector in R^5 is some linear combination of these vectors. A set of n vectors (each with n components) that is linearly independent spans R^n and is a basis for that vector space. If a different set of n vectors is linearly dependent, this set cannot span R^n, and therefore is not a basis for R^n. (This set might be a basis for some subspace of R^n, however.)
inverteren = inverses

Sander1337

Ah your right, I knew that, my mistake, sorry about that :P ... It's just that the whole Linear Algebra-thing is new to me, so don't bother to tell me when I'm making a little mistake, it's gives me opportunity for me to learn from this again and again (and that's the reason why I'm here!)!

So they will only span R^(number of elements in the biggest vector) if they are linearly dependent. If they are independant you'd have a (as we call it) free variable (a variable with a value assigned by an other variable, i.e. x=3y) which makes them span in R^n. Am I getting that correct or am I doing something wrong?

Thanks anyway for your help so far!

Ah okay, makes sense. You speak Dutch?

Sander1337

Is it by the way alright to say that if vector X and vector Y are linearly dependent, then the 0-vector is a linear combination of vector X and vector Y? Or that if vector X, vector Y and vector Z are linearly dependent, that Z is a linear combination of X and Y, from which automatically follows that the 0-vector is a linear combination of vector X, vector Y and vector Z.

HallsofIvy

The 0-vector is a linear combination of ANY vectors: 0X+ 0Y+ 0Z+ ...= 0 for any list of vectors X, Y, Z, .... If X and Y are linearly independent then that is the ONLY linear combination of X and Y that will give the 0-vector. If they are not, for example if Y= 2X, then (-2)X+ (1)Y= 0 as well as 0X+ 0Y= 0.

Again, there is ALWAYS a linear combination of vectors X, Y, Z equal to the 0-vector: 0X+ 0Y+ 0Z= 0. If they are dependent then there exist some OTHER linear combination that is equal to 0.

Saying that a set, S, of vectors "spans" a space V means that every vector is V can be written as a linear combination of vectors in S. Saying the vectors in S are independent mean any vector that can be written as a linear combination of vectors in S can only be written in one way.

"Span" means the existance of such a linear combination, "independent" means that linear combination, if it exists, is unique. Of course, a set of vectors that both spans the space and is linearly independent is especially nice!

Sander1337

Ah right that makes sense. So would my statement be correct if I'd add that the "linear combination scalars" may not be 0, with that I mean that aX+bY+cZ=0, and not a nor b nor c are 0. Then the vectors would be linearly dependent making them a linear combination with eachother as well... Am I getting that right?

[/quote]
Saying that a set, S, of vectors "spans" a space V means that every vector is V can be written as a linear combination of vectors in S. Saying the vectors in S are independent mean any vector that can be written as a linear combination of vectors in S can only be written in one way.

"Span" means the existance of such a linear combination, "independent" means that linear combination, if it exists, is unique. Of course, a set of vectors that both spans the space and is linearly independent is especially nice![/QUOTE]

So if i'm getting this right; if I can prove that the vectors in S are linearly independent then the answer to the question what Span(S) might get me, would be R^k in which k is an integer larger then 0 and is the value of the numer of elements in my largest vector in S? Say R^5, in which case the answer says that the largest vector in S is 5-dimensional. If S is linearly dependent then the answer to the question what Span(S) can only be be is R^n, since there is an infinite number of linear combinations in S...

If I'm still not getting it right, I'll ask my teacher tomorrow and'll show him this, maybe along with this I can sort it all out, still thanks to your help!

Thanks so far anyway, if anyone is still willing to help me out; gladly!

Greets,

Sander