# dimention of group question

• Jun 22nd 2011, 04:20 AM
transgalactic
dimention of group question
there is
v1,..,vk,u,w vectors on space V
the group {v1,..,vk,u} is independant
and
$Sp\{v1,..,vk\}\cap Sp\{u,w\}\neq\{0\}$
find the dimention of sp{v1..vk,u,w}
?
if {v1,..,vk,u} is independant then its span dimention is k+1
and span so {v1,..,vk} independat to and dim sp{v1,..,vk}=k
the dimention of sp{v1..vk,u,w} is
dim(sp{v1..vk,u,w})=dim(sp{v1..vk}and sp{u,w}})=dim(sp{v1..vk}+sp{u,w})=dim(sp{v1..vk})+ dim(sp{u,w})-dim(sp{v1..vk} intersect sp{u,w}})
if v1..vk,u is independant then v1,..,vk is independant too so dim (sp{v1..vk})=k
now regardingthe other two members i dont know
• Jun 22nd 2011, 05:06 AM
hatsoff
Re: dimention of group question
Let $x$ be a nonzero vector in

$\text{span}\{v_1,\cdots,v_k\}\cap\text{span}\{u,w\ }$.

Then there are constants $a_1,\cdots,a_k$ and $b,c$ such that $a_1v_1+\cdots+a_kv_k=x=bu+cw$ and hence

$w=c^{-1}a_1v_1+\cdots+c^{-1}a_kv_k-c^{-1}bu$.

So $w\in\text{span}\{v_1,\cdots,v_k,u\}$ which means $\dim\text{span}\{v_1,\cdots,v_k,u,w\}=\dim\text{sp an}\{v_1,\cdots,v_k,u\}=k+1$.
• Jun 22nd 2011, 05:13 AM
transgalactic
Re: dimention of group question
ok so we need to find
dim(sp{v1,..vk,u,w})=dim(sp{v1,..,vk,u}+sp{w})
dim(sp{v1,..,vk,u}+sp{w})=dim(sp{v1,..,vk,u})+dim( sp{w})- $dim(sp{v1,..,vk,u}\cap sp{w})$=k+1
beacuse $dim(sp({v1,..,vk,u}}) \cap sp{w})=0$
so
dim(sp{v1,..vk,u,w})=k+1
correct?
• Jun 22nd 2011, 12:31 PM
topspin1617
Re: dimention of group question
Quote:

Originally Posted by transgalactic
ok so we need to find
dim(sp{v1,..vk,u,w})=dim(sp{v1,..,vk,u}+sp{w})
dim(sp{v1,..,vk,u}+sp{w})=dim(sp{v1,..,vk,u})+dim( sp{w})- $dim(sp{v1,..,vk,u}\cap sp{w})$=k+1
beacuse dim(sp{v1,..,vk,u}\cap sp{w})=0
so
dim(sp{v1,..vk,u,w})=k+1
correct?

This is extremely unreadable. When you write out proofs in mathematics, you need to make sure that everyone can understand what you are trying to say. We aren't in your head; all we see is what you have written down, so if we can't interpret that, then we are out of luck.

Anyway, hatsoff has given the complete proof, with no steps missing.
• Jun 22nd 2011, 12:50 PM
transgalactic
Re: dimention of group question
hatsoff proved a very important step
it showed that i divided the group in the wring way

i made some mistakes in latex
but its only about one formula for the dimention of the sum of spans

here is the edited:
ok so we need to find

dim(sp{v1,..vk,u,w})=dim(sp{v1,..,vk,u}+sp{w})

dim(sp{v1,..,vk,u}+sp{w})=dim(sp{v1,..,vk,u})+dim( sp{w})- $dim(sp{v1,..,vk,u}\cap sp{w})$=k+1

beacuse $dim(sp({v1,..,vk,u}}) \cap sp{w})=0$

so

dim(sp{v1,..vk,u,w})=k+1

correct?
• Jun 22nd 2011, 01:03 PM
topspin1617
Re: dimention of group question
Quote:

Originally Posted by transgalactic
hatsoff proved a very important step
it showed that i divided the group in the wring way

i made some mistakes in latex
but its only about one formula for the dimention of the sum of spans

here is the edited:
ok so we need to find

dim(sp{v1,..vk,u,w})=dim(sp{v1,..,vk,u}+sp{w})

dim(sp{v1,..,vk,u}+sp{w})=dim(sp{v1,..,vk,u})+dim( sp{w})- $dim(sp{v1,..,vk,u}\cap sp{w})$=k+1

beacuse $dim(sp({v1,..,vk,u}}) \cap sp{w})=0$

so

dim(sp{v1,..vk,u,w})=k+1

correct?

1. How do you get $dim(sp({v1,..,vk,u}}) \cap sp{w})=0$?

2. If (1) were true, then

dim(sp{v1,..,vk,u}+sp{w})=dim(sp{v1,..,vk,u})+dim( sp{w})- $dim(sp{v1,..,vk,u}\cap sp{w})$=dim(sp{v1,..,vk,u})+dim(sp{w})-0=(k+1)+1-0=k+2, NOT k+1.

I'll say again: hatsoff has already given the COMPLETE proof.
• Jun 22nd 2011, 02:01 PM
transgalactic
Re: dimention of group question
oohh yes i got it
soryy i am a little slow on the comprehetion side in this hour

thanks :)