# Vector Spaces and Basis

• Nov 29th 2009, 10:25 PM
Coda202
Vector 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 PM
aman_cc
Quote:

Originally Posted by Coda202
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.

This follows directly for the defintion. A set S is basis IFF
1. It is linearly independent
2. It spans the entire vector space

Can you show both the above hold under the condition given in the question? It is rather straight fwd.
• Nov 30th 2009, 04:39 AM
HallsofIvy
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 AM
Raoh
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 $\displaystyle \lambda'_{3}=\lambda _{3}$.