This discussion revolves around proving that the span of a single vector \( x \) in a vector space is equivalent to the set of all scalar multiples of \( x \), denoted as \( \{ax : a \in F\} \). Participants are exploring the definitions and properties of spans in the context of vector spaces.
The discussion is ongoing, with participants raising questions about the clarity of the original problem statement and the assumptions made in the attempts. There is a recognition of the need for further exploration of the definitions and properties involved, but no consensus has been reached yet.
Participants note potential gaps in the original poster's reasoning and express uncertainty about the implications of their assumptions. There is also mention of the richness of fields beyond integers, indicating a broader context for the discussion.
I'm assuming you let x1=x2=...=xn=x, but this only works when x is of the form a*x where a is an integer, but most fields are much more rich. Also it would only prove that ax is in span(x).The Attempt at a Solution
span(x)={x1+x2+...+xn}
x1=x2=...=xn
x1+x2+...+xn=n*x=a*x a=n
therefore ax is equal to span(x)
rasmhop said:I'm assuming you let x1=x2=...=xn=x, but this only works when x is of the form a*x where a is an integer, but most fields are much more rich. Also it would only prove that ax is in span(x).
Let V be the vector space we work in.
Let [itex]W=\{ax \,:\,a\in F\}[/itex]. You want to show W=span(x). First prove that W is a subspace of V. Then prove that W contains x (and therefore [itex]\text{span}(x)\subseteq W[/itex] since span(x) is the minimal space containing x), and that every element of W is in span(x) since span(x) is a vector space (this proves [itex]W \subseteq \text{span}(x)[/itex]).
It is not at all clear what you are trying to prove! x is some vector in a vector space, yes, but what are x1, x2, x3,...? you say they are equal but if so why call them "x1, x2, ..."? In fact, as rasmhop says, you seem to be assuming they are all equal to x, though you never say so. If that is true, then x1+ x2+ ...+ xn is just "nx". Any vector in the span of x is of the form "ax" for some number a. Any vector in the span of nx is of the form b(nx) for some number b. But b(nx)= (bn)x and bn is a real number- thus b(nx) is in the span of x. Conversely, any ax= (a/n)(nx) and a/n is still a real number. That is, ax is in the span of nx. It's a pretty trivial statement!Beaker said:Homework Statement
This is for any vector x in a vector space.
The Attempt at a Solution
span(x)={x1+x2+...+xn}
x1=x2=...=xn
x1+x2+...+xn=n*x=a*x a=n
therefore ax is equal to span(x)
There may be a few holes in my attempt.
Sorry I posted this thread in the wrong forum twice.
It is not at all clear what you are trying to prove! x is some vector in a vector space, yes, but what are x1, x2, x3,...? you say they are equal but if so why call them "x1, x2, ..."? In fact, as rasmhop says, you seem to be assuming they are all equal to x, though you never say so. If that is true, then x1+ x2+ ...+ xn is just "nx". Any vector in the span of x is of the form "ax" for some number a. Any vector in the span of nx is of the form b(nx) for some number b. But b(nx)= (bn)x and bn is a real number- thus b(nx) is in the span of x. Conversely, any ax= (a/n)(nx) and a/n is still a real number. That is, ax is in the span of nx. It's a pretty trivial statement!Beaker said:Homework Statement
This is for any vector x in a vector space.
The Attempt at a Solution
span(x)={x1+x2+...+xn}
x1=x2=...=xn
x1+x2+...+xn=n*x=a*x a=n
therefore ax is equal to span(x)
There may be a few holes in my attempt.
Sorry I posted this thread in the wrong forum twice.