Let be a nonzero vector in
.
Then there are constants and such that and hence
.
So which means .
there is
v1,..,vk,u,w vectors on space V
the group {v1,..,vk,u} is independant
and
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
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})- =k+1
beacuse
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.
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})- =k+1
beacuse
so
dim(sp{v1,..vk,u,w})=k+1
correct?