Hey n22.
Which specific parts are you having trouble with?
Hi I need some help understanding this theorem,thanks
I dont understand this fully ..especiallly the bit about a contradiction ..what is the implication of this theorem??
Let w be a subspace of a vector space V and dim(V) =n
then i)dim(W)≤n;
and
ii)dim(W)=n iff W=V
Proof) i) any basis B of W must contain at most n vectors ,since otherwise B is Linearly dependendent by theorem(*). Hence dim(W)≤n.
ii)If W=V then dim(W)=dim(V)=n
conversely suppose that dim(W)=n and let B={w,,w2,...wn} be a basis of W.
Suppose for a contradiction that W≠V.Then there is v∈V such that v does not exist in W ..
a0v+ a1w1+a2w2+...anWn=0
for some scalars a0,a1,....an
then a0=0 since otherwise..
v=[-1/(a0)][ ]∈Span(B)=W
so a1w1+a2w2+...anWn=0
a1w1+a2w2+...anwn=0
a1=a2-...=an=0
since B is linear independent
this shows X={v1,w1,....,wn} is linear independent,which is impossible since X contains n vectors ,so X is Linearly depenedent by theorem(*)
ehence W=V.
(*)theorem
Let X={v1,v2,....Vn} be a basis of a vector space V and Y={w1,w2,...Wm} a subset of V.
i)If m>n then Y is linear dependent..
ii)if m<n then Y does not span V.
Basically if two vectors are independent then it means av + bw = 0 implies that a = 0 and b = 0.
You are assuming independence (since both sets are meant to be disjoint) and coming to a contradiction. In other words, you assume the opposite of what you are trying to prove (W = V) and arrive at a contradiction so that you can show that the initial premise (W = V) is actually true.
If you can prove that av + bw = 0 implies that a != 0 or b != 0 then you have shown that the initial assumption (W != V) is false and that the opposite (W = V) is actually true.
This idea known as proof by contradiction is probably the most used (and important) ways of proving things in all of mathematics.