Prove Choice of Scalars Unique

Date: 03/13/2001 at 18:45:41
From: Mary
Subject: Linear Algebra

I am really stuck on this problem.  

Suppose B = {v1,v2, ... vn} is a basis for a vector space V.  Let u 
element of V be any vector.  Prove that the expression of u in terms 
of the basis B is unique.

I have no clue what to do.  Thanks for your help!

Date: 03/14/2001 at 07:08:35
From: Doctor Luis
Subject: Re: Linear Algebra

Hi Mary,

You could start by using the fact that, by definition, Span(B) = V

Span(B), if you recall, is just the set of all finite linear 
combinations of elements in B.

V = Span(B) just means that for any element u of V we can find scalars
c_1, c_2, ..., c_n such that

  u = c_1 v_1 + c_2 v_2 + ... + c_n v_n

Now, the problem is asking you to prove that this choice of scalars is
unique (i.e., you can't find another linear combination which also 
gives u).

Problems of this type (uniqueness), are usually attacked using the
powerful mathematical technique of contradiction, which works by
reducing the alternative choice to an absurdity. In our case, this
involves reducing the opposite scenario, u represented as TWO DISTINCT
linear combinations in Span(B), to an impossible situation.

So, let u be represented by two distinct linear combinations, like 
this:

 u = c_1 v_1 + c_2 v_2 + ... + c_n v_n
   = d_1 v_1 + d_2 v_2 + ... + d_n v_n

You can see that this implies the following equation,

  (c_1-d_1) v_1 + (c_2-d_2) v_2 + ... + (c_n-d_n) v_n = 0

Now, using the fact that B is a linearly independent set (since it is
a basis), we conclude that the coefficients in the equation above must 
all be zero - which contradicts what we supposed originally, that our 
two linear combinations were distinct.

That's pretty much it. Simple argument, no?

If you have any problems understanding the above, pay close attention
to the definitions of the terms 'linear independence', 'linear 
combination', 'span', and 'basis'; then come back and read the problem 
again and/or my outline above.

I hope this helps.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/