# Linear Algebra proof

• Nov 13th 2007, 08:25 PM
beaveman
Linear Algebra proof
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?
• Nov 13th 2007, 08:35 PM
ThePerfectHacker
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.
• Nov 13th 2007, 08:40 PM
beaveman
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.
• Nov 13th 2007, 08:47 PM
ThePerfectHacker
Correct not everything can be expressed in linearly independent vectors.
• Nov 13th 2007, 09:20 PM
beaveman
A nice google search helped me find the answer to my question:

Elements of Operator Theory - Google Book Search

Thanks for your help, though! :)
• Nov 14th 2007, 09:49 AM
ThePerfectHacker
What is so hard? Consider $\{ (0,1,0),(1,0,0) \}$ this is linearly independent but $(0,0,1)$ cannot be obtained as a linear combination of those two elements.
• Nov 15th 2007, 05:25 AM
kalagota
just try to understand the proof.. Ü

Suppose v has a unique representation of linear combinations of vectors $v_1, v_2, ... ,v_n$, i.e.
$v = \sum_{i=1}^n a_iv_i = a_1v_1 + a_2v_2 + ... + a_nv_n$

and suppose the $S:= \{ v_1, v_2, ..., v_n \}$ is linearly dependent.

then, $\exists v_j \in S$ which is a linear combination of the other vectors in S (assuming you have proven this already) i.e.

$v_j = - (b_j^{-1}b_1v_1 + b_j^{-1}b_2v_2 + ... + b_j^{-1}b_{j-1}v_{j-1} + b_j^{-1}b_{j+1}v_{j+1} + ... + b_j^{-1}b_nv_n$ ---> (*)

from
$v = \sum_{i=1}^n a_iv_i = a_1v_1 + a_2v_2 + ... + a_jv_j + ... + a_nv_n$, we replace $v_j$ by (*), and since V is an abelian group, we can rearrange and combine like terms, so that,

$v = \sum_{i=1}^n a_iv_i = (a_1 - a_jb_j^{-1}b_1)v_1 + (a_2 - a_jb_j^{-1}b_2)v_2 + ... + (a_{j-1} - a_jb_j^{-1}b_{j-1})v_{j-1} +$
$(a_{j+1} - a_jb_j^{-1}b_{j+1})v_{j+1} + ... + (a_n - a_jb_j^{-1}b_n)v_n$

and clearly, $a_k - a_jb_j^{-1}b_k = a_k \Leftrightarrow a_jb_j^{-1}b_k = 0$, 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, $a_k - a_jb_j^{-1}b_k \neq a_k$ for all k.
which contradicts now that v has a unique representation of linear comb of the vectors in S.

therefore, S must be lin. indep set. QED