Proving span of a Set with Scalar attached to First Element

[shen07]

hi Guys, i Needed your help to prove out the following, thanks in advance;

Let u₁,u₂,...,u_t be vectors in $\Re^n$ and $k\in\Re ,k\neq0.$ Prove that

$Span\{u_1,u_2,...,u_t\}=Span\{ku_1,u_2,...,u_t\}$

 

[shen07]

hi Guys, i Needed your help to prove out the following, thanks in advance;

Let u₁,u₂,...,u_t be vectors in $\Re^n$ and $k\in\Re ,k\neq0.$ Prove that

$Span\{u_1,u_2,...,u_t\}=Span\{ku_1,u_2,...,u_t\}$

i started working like this, hope you can help me further,

let $S=\{u_1,u_2,...,u_t\}\\\\\,S_k=\{ku_1,u_2,...,u_t\}$
then there exist Vector Spaces $V=L(S)\\\\\&\\\\W=L(S_k)$

Now let $v \in V,\alpha\in\Re$
$v=\alpha_1u_1+\alpha_2u_2+...+\alpha_tu_t$let $w \in W,\gamma\in\Re$
$w=\gamma_1(ku_1)+\gamma_2u_2+...+\gamma_tu_t$Now am i right by saying that since $k\in\Re,k\neq0\Rightarrow\\\alpha_1=\gamma_1k \Rightarrow\\v=w\Rightarrow\\\\L(S)=L(S_k)$

 

[I like Serena]

hi Guys, i Needed your help to prove out the following, thanks in advance;

Let u₁,u₂,...,u_t be vectors in $\Re^n$ and $k\in\Re ,k\neq0.$ Prove that

$Span\{u_1,u_2,...,u_t\}=Span\{ku_1,u_2,...,u_t\}$

i started working like this, hope you can help me further,

let $S=\{u_1,u_2,...,u_t\}\\\\\,S_k=\{ku_1,u_2,...,u_t\}$
then there exist Vector Spaces $V=L(S)\\\\\&\\\\W=L(S_k)$

Now let $v \in V,\alpha\in\Re$
$v=\alpha_1u_1+\alpha_2u_2+...+\alpha_tu_t$let $w \in W,\gamma\in\Re$
$w=\gamma_1(ku_1)+\gamma_2u_2+...+\gamma_tu_t$Now am i right by saying that since $k\in\Re,k\neq0\Rightarrow\\\alpha_1=\gamma_1k \Rightarrow\\v=w\Rightarrow\\\\L(S)=L(S_k)$

Hi shen07!

To prove that V and W are the same set, you should prove that any vector in V is also in W, and any vector in W is also in V.

Any vector in V can indeed be written as
$$v=\alpha_1u_1+\alpha_2u_2+...+\alpha_tu_t$$
Now is this vector v also in W?

We can rewrite it as:
$$v=\frac {\alpha_1}{k} (ku_1)+\alpha_2u_2+...+\alpha_tu_t$$
This is a linear combination of the basis of W, so yes, this vector is in W.Note that there is no need to introduce $w$, or $\gamma_i$ at this time.
But if you do, you can't just say that you already know that $\alpha_1=\gamma_1 k$.
What you could do, is say:
Pick $\gamma_1=\frac{\alpha_1}{k}$ and $\gamma_i=\alpha_i$ for $i \in \{2,...,n\}$.
Then $v=w$.Anyway, we're not completely done yet, since we still need to prove that any vector in W is also in V.
Well, any vector in W can be written as:
$$w=\gamma_1 (ku_1)+\gamma_2u_2+...+\gamma_tu_t$$
which is the same as:
$$w=(\gamma_1 k)u_1+\gamma_2u_2+...+\gamma_tu_t$$
Since this vector is a linear combination of the basis of V, it is indeed in V.

This completes the proof.

 

[Deveno]

ILikeSerena post highlights the most important part of this exercise:

$k \neq 0$.

This is why scalars are called "scalars"...all they do is "scale" one or more coordinates (0 is obviously a special case...it annihilates one or more dimensions!).

As you can see, we can "scale" and "un-scale" to our heart's content. Physicists often do this when they change from one scale of measurement to another, like from grams to kilograms. Map-makers also use this property of scaling to fit things on a page by distorting one axis scale slightly. There really isn't anything mysterious about this (linear independence of axes means we can scale the axes independently, and still preserve linear relationships).

Vector spaces make sense, once you can get past the abstraction of the presentation. We are bringing the ability of arithmetic (which is what fields are all about) to bear on geometry (points, lines, planes, and the like).

 

[blue_raver22]

To prove this, we can use the definition of span. The span of a set of vectors is the set of all linear combinations of those vectors. In other words, it is the set of all possible vectors that can be created by multiplying each vector by a scalar and adding them together.

So, for the first set $Span\{u_1,u_2,...,u_t\}$, we can represent any vector in this span as $a_1u_1+a_2u_2+...+a_tu_t$, where $a_1,a_2,...,a_t$ are scalars.

Now, for the second set $Span\{ku_1,u_2,...,u_t\}$, we can represent any vector in this span as $b_1(ku_1)+b_2u_2+...+b_tu_t$, where $b_1,b_2,...,b_t$ are scalars.

But we know that $k$ is a scalar, so we can rewrite the second set as $kb_1u_1+b_2u_2+...+b_tu_t$. This is essentially the same as the first set, just with a different scalar attached to the first vector.

Therefore, we can conclude that $Span\{u_1,u_2,...,u_t\}=Span\{ku_1,u_2,...,u_t\}$, as both sets contain the same set of vectors and can generate the same set of vectors through linear combinations.