1. ## vector spaces

show that every set with more than n vectors is necessarily linearly dependent in a n-dimensional vector space.

The following shows how i give it a try. am i getting things right. is there any way of tackling the same problem?

proof:

suppose B = {e1,e2,...,en} is a basis for V. i.e dim(V)= n.

Let S = {s1,s2,...sN} n < N.
then 1 of the elements in S can be expressed as a linear combination of other elements. say s1.

s1
= k1*s2 + k2*s2 + ...+kN*sN

= k1(a1*e1 +... + an*en) + k2(b1*e1 +...+ bn*en) +...+ KN(x1*e1 +...+xN*en)

= ...(some steps)

= y1*e1 + y2*e2 + ... + yn*en

2. Originally Posted by mtmath
show that every set with more than n vectors is necessarily linearly dependent in a n-dimensional vector space.

The following shows how i give it a try. am i getting things right. is there any way of tackling the same problem?

proof:

suppose B = {e1,e2,...,en} is a basis for V. i.e dim(V)= n.

Let S = {s1,s2,...sN} n < N.
then 1 of the elements in S can be expressed as a linear combination of other elements. say s1.

s1
= k1*s2 + k2*s2 + ...+kN*sN

= k1(a1*e1 +... + an*en) + k2(b1*e1 +...+ bn*en) +...+ KN(x1*e1 +...+xN*en)

= ...(some steps)

= y1*e1 + y2*e2 + ... + yn*en
I have proven that if n vectors span $\mathbb{R}^n$ then they are lin ind. in the lin alg sticky proofs. Since n vectors in dim=n form a basis, they must be lin ind.

Now looking at your problem. If we have at least n+1 vectors in $\mathbb{R}^n$, then one of them has to be lin. dep. Since n vectors from a basis, then any additional vector beyond the basis must be a lin. combination of the basis.

3. Originally Posted by dwsmith
I have proven this in the lin alg sticky proofs
i've gone through the proofs.i can figure out which one is it.

4. Originally Posted by mtmath
i've gone through the proofs.i can figure out which one is it.
I thought that may be the case so I edited my post. There really isn't much to show since we have n+k vectors and n vectors in n dimension form a basis.

A basis is the minimum spanning set. If we have a minimum spanning set, then any other vector must be dependent.

5. Here is a general proof for a Vector Space that shows when $k+1$ will be lin. ind. and lin. dep.

Let $\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k$ be lin. ind. vectors in V. If we add a vector $\mathbf{x}_{k+1}$, do we still have a set of lin. ind. vectors?

(i) Assume $\mathbf{x}_{k+1}\in$ $Span (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k)$

$\mathbf{x}_{k+1}=c_1\mathbf{x}_{1}+...+c_k\mathbf{ x}_{k}$

$c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma thbf{x}_{k+1}=0$

$c_{k+1}=-1$

$-1\neq 0$; therefore, $(\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1})$ are lin. dep.

(ii) Assume $\mathbf{x}_{k+1}\notin$ $Span (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k)$

$c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma thbf{x}_{k+1}=0$

$c_{k+1}=0$ otherwise $\mathbf{x}_{k+1}=\frac{-c_1}{c_{k+1}}\mathbf{x}+....+\frac{-c_k}{c_{k+1}}\mathbf{x}_k$ which is a contradiction.

$c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma thbf{x}_{k+1}=0$ since $(\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1})$ are lin ind., $c_1=...c_{k+1}=0$.