Let $V$ be a vector space over a field $F$ and let $B$ be a basis for $V$. In here $V$ can be infinite dimensional. Prove that every non zero vector $\mathbf{u} \in V$ can be expressed uniquely as a linear combination $$\mathbf{v} = c_1\mathbf{v_1}+c_2\mathbf{v_2}+...+c_m\mathbf{v_m}$$ for some $m \in \mathbb{N}$, some $c_1,c_2,...,c_m \in F$ and some $\mathbf{v_1,...v_m} \in B$ such that $c_i \neq 0$ for all $i$ and $\mathbf{v_i \neq v_j}$ whenever $i \neq j$.

Do note this question is absolutely not the same as the usual theorem we learned where if $B$ is a finite basis then all vectors are uniquely expressed by $B$. In this question, $B$ may not be finite, and furthermore, $\mathbf{v_1,...,v_m}$ are just some vectors in $B$ and may not be all the vectors in $B$.

I hope someone can verify my proof and let me know if there is any errors.

Since $V = \text{span}(B)$, every non zero vector $\mathbf{v} \in V$ can be expressed as a linear combination of vectors from $B$.

Assume a contradiction that $\mathbf{v}$ can be expressed in two ways.

$$\mathbf{v} = c_1\mathbf{v_1}+c_2\mathbf{v_2}+...+c_m\mathbf{v_m}$$ $$\mathbf{v} = d_1\mathbf{w_1}+d_2\mathbf{w_2}+...+d_n\mathbf{w_n}$$ where $m$ and $n$ may not be the same. But of course $\mathbf{w_1,w_2,...,w_n} \in B$. And do note our question said that $c_i \neq 0$, hence $d_i \neq 0$ as well.

Our first intuitive is to show that $m = n$, or else we cannot proceed. Here i present an error in my thinking at first in which i wrote show that $$(\mathbf{v_1,v_2,...,v_m}) = (\mathbf{w_1,w_2,...,w_n})$$ which is wrong as writing in this form suggest ordered vector and we are comparing $\mathbf{v_1}$ with $\mathbf{w_1}$, $\mathbf{v_2}$ with $\mathbf{w_2}$. But $\mathbf{v_1}$ can possibly be equals to $\mathbf{w_n}$ here. The correct way to write is as such: show that $$\{\mathbf{v_1,...,v_m}\} = \{\mathbf{w_1,...,w_n}\}$$ are the same set.

Take any $\mathbf{v_i} \in \{\mathbf{v_1,...,v_m}\}$, then assume a contradiction that $\mathbf{v_i} \not \in \{\mathbf{w_1,...,w_n}\}$. Then since $c_1\mathbf{v_1}+c_2\mathbf{v_2}+...+c_m\mathbf{v_m}- d_1\mathbf{w_1}-d_2\mathbf{w_2}-...-d_n\mathbf{w_n} = \mathbf{v-v} = 0$, we have $\mathbf{v_i} = \dfrac{1}{c_i}(d_1\mathbf{w_1}+...+d_n\mathbf{w_n}-c_1\mathbf{v_1}-...-c_{i-1}\mathbf{v_{i-1}}-c_{i+1}\mathbf{v_{i+1}}-...-c_m\mathbf{v_m}$, thus $$\mathbf{v_i} \in \text{span}(\{\mathbf{w_1,w_2,...,w_n,v_1,...v_{i-1},v_{i+1},...,v_m}\})$$

This is a contradiction as by the very definition of basis, the set $B$ is a linearly independent set and hence no element inside can be a linear combination of the other elements inside $B$. (side note: in this part of the proof, we only assumed no $\mathbf{v_i} \in \{\mathbf{w_1,...,w_n}\}$, but there can be such cases where $\mathbf{w_i} \in \{\mathbf{v_1,...v_{i-1},v_{i+1},...v_m}\}$ but it does not matter. Look at big picture that no element in $B$ can be expressed as LC as others.) Hence since it is a contradiction, then $$\mathbf{v_i} \in \{\mathbf{w_1,...,w_n}\}$$

WLOG, we can show that $$\mathbf{w_i} \in \{\mathbf{v_1,...,v_m}\}$$

Hence $$\{\mathbf{v_1,...,v_m}\} = \{\mathbf{w_1,...,w_n}\}$$ and again WLOG, we assume that $\mathbf{v_i} = \mathbf{w_i}$ for all $i$ and rewrite $$\mathbf{v} = c_1\mathbf{v_1}+c_2\mathbf{v_2}+...+c_m\mathbf{v_n}$$ $$\mathbf{v} = d_1\mathbf{v_1}+d_2\mathbf{v_2}+...+d_n\mathbf{v_n}$$

And the next part is very easy as $$(c_1-d_1)\mathbf{v_1}+(c_2-d_2)\mathbf{v_2}+...+(c_n-d_n)\mathbf{v_n} = \mathbf{v-v} = 0$$

and by linear independence $c_i - d_i = 0$ for all $i$ and thus $c_i = d_i$ for all $i$.

  • $\begingroup$ The argumentation gets simpler if one augments the sum and writes $v=\sum_{i\in I}c_iv_i$ (where $B=\{v_i\}_{i\in I}$) and demands $c_i=0$ for all but finitely many $i$ ... $\endgroup$ – Hagen von Eitzen Oct 6 '17 at 17:26
  • $\begingroup$ always wanted to do that but i feel safer when i write out the terms >.< $\endgroup$ – nan Oct 6 '17 at 17:28
  • $\begingroup$ For me this ‘property’ is the definition of a basis. So whar definition do you use? $\endgroup$ – Bernard Oct 6 '17 at 17:40
  • $\begingroup$ Let $V$ be a non zero vector space over a field $F$. A subset $B$ of $V$ is called a basis if the following two conditions are satisfied: i) $V$ is spanned by $B$. $$V =\text{Span}(B)$$ ii) $B$ is a linearly independent set in $V$. $\endgroup$ – nan Oct 6 '17 at 17:59

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