prove or disprove: statement about Span

**Stormey** · December 28th 2012, 04:18 AM

$V$ is a linear space. $A,B\in V$ are sets, and $W$ is subspace:

$Span(A\cup B\cup W)=(Span(A)+Span(B))\cup W$

i just need help to figure out if it's true or false.

thanks in advanced!

**HallsofIvy** · December 28th 2012, 04:31 AM

Do you know what "span" means? It looks to me like a straight forward case of applying the definition.

What have you tried?

**Deveno** · December 28th 2012, 10:25 AM

suppose V is the vector space R³, A = {(0,0,1)} B = {(0,1,0)} and W is the x-axis (vectors of the form (x,0,0)). what is the LHS and RHS then?

**Stormey** · December 29th 2012, 01:17 AM

Hi guys, and thank you.

Denovo, that's the first example that crossed my mind, by this example the LHS and RHS are equal:
$Span\left \{ (0,1,0), (0,0,1), (x,0,0)|x\in \mathbb{R} \right \}=\left \{ (x,y,z)|x,y,z\in \mathbb{R}\right \}=\mathbb{R}^3$
is equal to:
$Span(A)=\left \{ (0,y,0)|y\in \mathbb{R}\right \}+Span(B)=\left \{ (0,0,z)|z\in \mathbb{R}\right \}\cup \left \{ (x,0,0)|x\in \mathbb{R}\right \}=\left \{ (x,y,z)|x,y,z\in \mathbb{R}\right \}=\mathbb{R}^3$

but my intuition misled me so many times, and i wasn't sure that proves it.

**Deveno** · December 29th 2012, 03:04 AM

no, they are not equal. the point (3,4,5) is in Span(AUBUW), since:

(3,4,5) = 4(0,1,0) + 5(0,0,1) + (3,0,0), and (0,1,0) is in A, so in AUBUW, (0,0,1) is in B, so in AUBUW, and (3,0,0) is in W, so is in AUBUW.

is (3,4,5) in Span(AUB)UW? if so, it would have to be in either Span(AUB), or W, or both. is this true?

**Stormey** · December 29th 2012, 08:28 AM

my bad.

$(3,4,5)$ is neither in $(Span(A)+Span(B))$ nor in $W$ , and therefore is not in $(Span(A)+Span(B))\cup W$ .

thanks.
this disproves it.

this is all so confusing...

**HallsofIvy** · December 29th 2012, 08:53 AM

The best way to avoid confusion in mathematics is to learnthe definitions well!

**Deveno** · December 29th 2012, 09:28 AM

and the moral of our story is: span and union don't play well together, and should therefore sleep in separate rooms.

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