Basis Theorem

## Theorem 1

Let $\beta =${$\vec{b}_1,\vec{b}_2,...,\vec{b}_n$} be a vector space V. Then for each $\vec{x} \in V$, there is a unique set of coefficients $c_1,c_2,...,c_n$ such that $\vec{x} = c_1\vec{b}_1+...+c_n\vec{b}_n$

Proof

Existence
Let $\beta =${$\vec{b}_1,...,\vec{b}_2$} be a basis for the vector space V.
Let $\vec{x} \in V$
Since $\beta$ spans V, there exists $c_1,...,c_n$ such that $\vec{x} = c_1\vec{b}_1+...+c_n\vec{b}_n$

Suppose there is a vector $\vec{x}$ such that
$\vec{x} = c_1\vec{b}_1+...+c_n\vec{b}_n$
and
$\vec{x} = d_1\vec{b}_1+...+d_n\vec{b}_n$
where $(c_i-d_i) \neq 0$ for some i

Then,
$\vec{0} = \vec{x} - \vec{x}$
$\vec{0} = (c_1\vec{b}_1 +...+ c_n\vec{b}_n) - (d_1\vec{b}_1 +...+ d_n\vec{b}_n)$
$\vec{0} = (c_1-d_1)(\vec{b}_1)+(c_2-d_2)(\vec{b}_2)+...+(c_n-d_n)(\vec{b}_n)$

Since $\beta$ is linearly independent, $c_i-d_i = 0$ for all i. There can therefore be no second set of coefficients to make the same $\vec{x}$

page revision: 0, last edited: 28 Mar 2015 20:36