I have shown that if a set of vectors (v1,...,vn) is linearly independant, then every vector v that is an element of <v1,...,vn> has a unique representation as a linear combination of vectors v1,...,vn.
Now I need to show that the converse statement is true, that is if a vector v has a unique representation as a linear combination of vectors v1,...,vn then the set of vectors (v1,...,vn) are linearly independant. Does anyone have suggestions on you show that?
November 13th 2007, 08:35 PM
Suppose we can write a vector x as a_1v_1+...+a_nv_n and as b_1v_1+...+b_nv_n. We want to show a_1=b_1,...,a_n=b_n for uniqueness. So that means, a_1v_1+...+a_nv_n = b_1v_1+...+b_nv_n. Subtract them, (a_1-b_1)v_1+...+(a_n-b_n)v_n = 0. Since these vector are linearly independent it means there is only the trivial representation. And so that means, a_1 - b_1 = 0 ... a_n - b_n =0 so a_1 = b_1 ... a_n=b_n. Q.E.D.
November 13th 2007, 08:40 PM
That is how I showed the first part - that there is a unique representation. I don't think that works to show the converse that if there is a unique representation, then the set of vectors (v1,...,vn) are independant.
November 13th 2007, 08:47 PM
Correct not everything can be expressed in linearly independent vectors.
November 13th 2007, 09:20 PM
A nice google search helped me find the answer to my question:
What is so hard? Consider this is linearly independent but cannot be obtained as a linear combination of those two elements.
November 15th 2007, 05:25 AM
just try to understand the proof.. Ü
Suppose v has a unique representation of linear combinations of vectors , i.e.
and suppose the is linearly dependent.
then, which is a linear combination of the other vectors in S (assuming you have proven this already) i.e.
from , we replace by (*), and since V is an abelian group, we can rearrange and combine like terms, so that,
and clearly, , for k=1,2,...,n
but we have to note that S must not contain the zero vector since v has a unique representation of linear comb of vectors in S. hence, for all k.
which contradicts now that v has a unique representation of linear comb of the vectors in S.