Prove that span ({x}) = {ax: a [tex]\in[/tex] F}

[Beaker]

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.

 

[rasmhop]

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)

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 W=\{ax \,:\,a\in F\}. You want to show W=span(x). First prove that W is a subspace of V. Then prove that W contains x (and therefore \text{span}(x)\subseteq W 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 W \subseteq \text{span}(x)).

 

[Beaker]

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 W=\{ax \,:\,a\in F\}. You want to show W=span(x). First prove that W is a subspace of V. Then prove that W contains x (and therefore \text{span}(x)\subseteq W 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 W \subseteq \text{span}(x)).

Is proving that W is a subspace necessary since a span(x) is the intersection of all subspaces containing x? But maybe I could show that span(x) is a subset of W and W is a subset of span(x) this would set them equal. Whoa, thanks you gave me allot to think about...back to the drawing board.

 

[HallsofIvy]

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!

 

[HallsofIvy]

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!

In fact, it is just as easy to prove much more: if v is any vector in the span of x (other than the 0 vector), not just an integer multiple, then span(v)= span(x).