Let S = {v_1, v_2, ..., v_n} be a finite set of vectors in a vector space V. Show that S is a basis for V iff every member of V can be written uniquely as a linear combination of the vectors in S.

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- Nov 29th 2009, 10:25 PMCoda202Vector Spaces and Basis
Let S = {v_1, v_2, ..., v_n} be a finite set of vectors in a vector space V. Show that S is a basis for V iff every member of V can be written uniquely as a linear combination of the vectors in S.

- Nov 29th 2009, 10:31 PMaman_cc
- Nov 30th 2009, 04:39 AMHallsofIvy
In particular, to prove the vectors are independent, saying " every member of V can be written uniquely as a linear combination of the vectors in S" means that the

**zero**vector can be written**uniquely**as such a linear combination. There is one "obvious" linear combination and now you know it is the only one. - Nov 30th 2009, 09:34 AMRaoh
suppose $\displaystyle S$ is a basis of V.

in case $\displaystyle n=3$

$\displaystyle S=\left \{ v_{1},v_{2},v_{n} \right \}$ is the given basis.

let's take any $\displaystyle x$ vector of $\displaystyle V$

$\displaystyle x\in S\Rightarrow$ $\displaystyle x=\lambda_{1} v_{1}+\lambda_{2} v_{2}+\lambda_{3} v_{3} $

or $\displaystyle x=\lambda'_{1} v_{1}+\lambda'_{2} v_{2}+\lambda'_{3} v_{3}$

We have,

$\displaystyle \left ( \lambda'_{1}-\lambda_{1} \right ) v_{1}$+$\displaystyle \left (\lambda'_{2}-\lambda _{2} \right ) v_{2}$+$\displaystyle \left (\lambda'_{3}-\lambda _{3} \right ) v_{3}=0$

gives $\displaystyle \left ( \lambda'_{1}-\lambda_{1} \right ) =0$,$\displaystyle \left (\lambda'_{2}-\lambda _{2} \right )=0$ and $\displaystyle \left (\lambda'_{3}-\lambda _{3} \right ) =0$ since the system $\displaystyle \left \{ v_{1},v_{2},v_{3} \right \}$ is independent .

Therefore,$\displaystyle \lambda'_{1}=\lambda_{1}, \lambda'_{2}=\lambda _{2}$ and $\displaystyle \lambda'_{3}=\lambda _{3}$.