Kenshin
  • Kenshin
Hi I need some help remembering the definition and conditions of a basis: In which way can the following implications go? 1) => 2), 2) => 1) or both? 1) A basis for a subspace or any space has to satisfy the following 2 conditions: a) be linearly independent, b) span the subspace; 2) The "Basis theorem 15" in David C. Lay's 3rd edition of Linear Algebra and its applications says: Let H be a p-dimensional subspace of R^n. Any linearly independent set of exactly p elements in H is automatically a basis for H. Also, any set of p elements of H that spans H is automatically a basis for H.
Mathematics
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chestercat
  • chestercat
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Kenshin
  • Kenshin
I'm confused because: 2) seems to suggest that condition a) implies condition b) for whichever it satifies, the other must also be true. but in 1) it seems to suggest that both a) and b) need to be satisfied before we can even conclude it being a basis to begin with.
Lurker
  • Lurker
ya what about it
Lurker
  • Lurker
its say if u have p dimensional subspace, then having exactly p elements, whose lin combination is able to spam all of H, means they must be lin independant

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Lurker
  • Lurker
it's both
Kenshin
  • Kenshin
so does 2) mean p elements will always span all of p dimension so only L.I. is enough? and having p elements is also equivalent of saying it is L.I. so it just has to span in order to be basis? 2) just seems so confusing sorry
Lurker
  • Lurker
a) be linearly independent, b) span the subspace; p dim space u can only have p lin independant vectors at most
Lurker
  • Lurker
actually hmm
Lurker
  • Lurker
nvm this isnt really right
Lurker
  • Lurker
1 doesnt really have to imply 2
Lurker
  • Lurker
because for example if u have z=0 plane in r^3, your vectors will be dimension which is p3
Lurker
  • Lurker
but ud only need 2 independant vectors to define your whole basis
Lurker
  • Lurker
and 2 independant vectors in the xy plane
Lurker
  • Lurker
so 1 doesnt imply 2
Lurker
  • Lurker
but 2 implies 1
Kenshin
  • Kenshin
I still don't get what i wrote in my first reply to my own post sorry.
Lurker
  • Lurker
meh dont think too hard about it
Lurker
  • Lurker
theyre trying to make a big deal out of something trivial
Lurker
  • Lurker
there are 3 things u need to know really
Kenshin
  • Kenshin
"u can only have p lin independant vectors at most" but you can still have less and they still form a basis or not?
Lurker
  • Lurker
1) A vector space contains the 0 vector 2) k*U, a scalar times any vector in the vector space is still in the vector space 3)k1*U+k2*V , any linear combinations of vectors in a vector space stay in the vector space
Lurker
  • Lurker
yes that is right
Lurker
  • Lurker
like for example u can be in r^1000000000000, but still have only the plane in xy
Lurker
  • Lurker
Lurker
  • Lurker
this is a dimension 100 vector however, only 2 lin independant vectors are needed to define this plane
Lurker
  • Lurker
oh oops i read the question wrong this is what they are saying
Lurker
  • Lurker
they say u just need p elements but the condition is your p element vectors are lin independant
Lurker
  • Lurker
so ya its still both!
Lurker
  • Lurker
1 implies 2 and 2 implies 1
Kenshin
  • Kenshin
ok i get that part, but the "Also, any set of p elements of H that spans H is automatically a basis for H. " does that statement mean the Linear independence is assumed, or the previous sentence's L.I. is already taken into account?
Lurker
  • Lurker
no not true
Lurker
  • Lurker
they have to be lin independant
Lurker
  • Lurker
u can have 10 p element vectors and still span only r3
Lurker
  • Lurker
where as u can span that r^3 with only 3 independant p element vectors
Kenshin
  • Kenshin
yea that's what I thought too, thanks.
Lurker
  • Lurker
pls pay dan815
Lurker
  • Lurker
that is my QH accnt :)
Lurker
  • Lurker
byee
Kenshin
  • Kenshin
Is dan815 going to make a post here for me to 'best response' it? sorry I'm such a newbie here
Lurker
  • Lurker
oh u hae to click rate a quahlified helper orange button
Lurker
  • Lurker
the best response thing u can give u anyone
Kenshin
  • Kenshin
I clicked it, nothing shows up
Lurker
  • Lurker
ah okay dont worry about it, i guess its because no QH responded
Lurker
  • Lurker
xD well cheers!
Kenshin
  • Kenshin
thanks!
Error1603
  • Error1603
k

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