# Thread: n dimensional vectors -linear dependence

1. ## n dimensional vectors -linear dependence

i read a proof that there do no exist n+1 linearly independent vectors in a n-dimensional space but i dont quite get some bits
(1)Let $\bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n + 1}$ be linearly independent vectors in n dimensional space
(2)then $\bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n}$ must be linearly independent
(3)Hence the matrix A whose columns are $\bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n}$ is nonsingular ie the only solution to $A\bf {c} = \bf0$ is $\bf{c} = \bf0$ , where c is a column vector THIS I GET
(4)then the books says therefore the equation $A\bf c = \bf{v}_{n + 1}$ has unique soln for c ie. ${\bf{v}_{n + 1}} = {c}_{1}{\bf{v}_{1}} + {c}_{2}{\bf {v}_{2}} + {c}_{3}{\bf{v}_{3}} + ... + {c}_{n}{\bf {v}_{n}}$ CAN SOMEONE PLEASE EXPLAIN HOW THIS IS, as in how does $A\bf c = \bf{v}_{n + 1}$ have a soln and how it's unique??
(5)then it just follows that $\bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n + 1}$ would be linearly dependent and contary to assumption which i get...

2. c = A^-1*v_n+1

3. that means in a n-dimensional space
you can assume that v_1,...v_n are linearly independent
then v_1,...v_n-1 are linearly independent
then you can find a soln for $A{\bf c}={\bf {v}_{n}}$ as you said similiary ${\bf c}={A}^{-1}{\bf {v}_{n}}$
which means ${\bf{v}_{n}}= {c}_{1}{\bf{v}_{1}}+{c}_{2}{\bf{v}_{2}}+...+{c}_{n-1}{\bf{v}_{n-1}}$
and that means the original assumption is wrong as v_1,...v_n will not be linearly independent ???

4. that means in a n-dimensional space
you can assume that v_1,...v_n are linearly independent
then v_1,...v_n-1 are linearly independent
then you can find a soln for as you said similiary
which means
and that means the original assumption is wrong as v_1,...v_n will not be linearly independent ???
No. The matrix A will have only n-1 columns and c will have n elements so the product Ac is not defined.

c = A^-1*v_n+1
Skalkaz used the result, which you are expected to know, that A is invertible if and only if it's columns are linearly independent.