# I have a midterm tomorrow, just a conceptual question needs to be answered immediatel

flyingpig

## Homework Statement

I am just wondering if I have Span{v1,v2} where v1 is not a scalar multiple of v2, then it is known that they span a plane, in fact it is R2

So the question is, how? There can be at most one parameter right? One parameter means the solution is a line, we need two to make a plane?

## Answers and Replies

Theorem.

if the span were just a line, then v2 would have to be a scalar multiple of v1, and that the line would basically be equivalent to all scalar multiples of just one of the vectors. i.e. if it this were the case, we would have span{v1}=span{v2}=span{v1, v2}. since v1 is not a scalar multiple of v2 in your case, it is impossible that they lie on the same line, thus they span a plane defined by all combinations cv1+dv2, where c and d are real numbers. Also, you have provided nothing that would suggest the vectors are two dimensional, so I wouldn't say they span R2. They span some plane, yes, but not necessarily R2

flyingpig

I thought Span is the linear combination of all vectors in that set and hence it must be R^2?

Theorem.

I thought Span is the linear combination of all vectors in that set and hence it must be R^2?

your set consists of two vectors, v1 and v2. if v1 and v2 are both elements of R^2, then they will span all of R^2. nowhere have you stated that v1 and v2 are both elements of R^2, they could be vectors in R^3.

Unit

your set consists of two vectors, v1 and v2. if v1 and v2 are both elements of R^2, then they will span all of R^2.
Assuming v1 and v2 are in R^2, if v1 = c v2 for some c in R, then no, span{v1, v2} is not equal to R^2.

Theorem.

Assuming v1 and v2 are in R^2, if v1 = c v2 for some c in R, then no, span{v1, v2} is not equal to R^2.

The OP specificially stated that v1 is not qual to a scalar multiple of v2

flyingpig

your set consists of two vectors, v1 and v2. if v1 and v2 are both elements of R^2, then they will span all of R^2. nowhere have you stated that v1 and v2 are both elements of R^2, they could be vectors in R^3.

Does it matter how many entries I have in my vectors? I can have three entries in my vectors, but with two vectors, they will always span R2.

Is there even any relations to having free variables (parameters) with the concept of Span?

Unit

My apologies, Theorem; I was hasty. You are right.

Theorem.

My apologies, Theorem; I was hasty. You are right.

No worries : ) hopefully the OP understands what i meant

flyingpig

No worries : ) hopefully the OP understands what i meant

I don't unfortunately =(

flyingpig

Please I gotta sleep soon lol

Homework Helper

Does it matter how many entries I have in my vectors? I can have three entries in my vectors, but with two vectors, they will always span R2.

Is there even any relations to having free variables (parameters) with the concept of Span?
Two independent vectors will always span a plane. Whether or not that plane is R2 or not depends upon the vectors and what you mean by R2. The two vectors <1, 0, 1> and <0, 1, 0> will have span a<1, 0, 1>+ b<0, 1, 0>= <a, b, a> which can be expressed as <x, y, z> with z= x, or the plane z= x.
If the two vectors are <1, 0, 0> and <0, 1, 0> then they span <1, 1, 0>, the xy- plane. Do you consider that to be R2? Some people think of R2 as a subset of R3, some do not.

I'm not sure what you mean by "free variables (parameters) with the concept of Span" but in your first post you asked:
So the question is, how? There can be at most one parameter right? One parameter means the solution is a line, we need two to make a plane?
The definition of "span" of a set of vectors is the set of all linear combinations of the vectors. In particular, the span of the two vectors {u, v} is all vectors of the form au+ bv for any scalars a and b. Both a and b are, I think, what you are calling "parameters". Where did you get the idea that there could be "at most one parameter"? The span of a set of n vectors will involve n parameters and, if the vectors are independent, the span will be an n-dimensional subspace.