# vector spaces

#### 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

#### dwsmith

MHF Hall of Honor
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 $$\displaystyle \mathbb{R}^n$$ 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 $$\displaystyle \mathbb{R}^n$$, 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.

Last edited:

#### mtmath

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

#### dwsmith

MHF Hall of Honor
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.

#### dwsmith

MHF Hall of Honor
Here is a general proof for a Vector Space that shows when $$\displaystyle k+1$$ will be lin. ind. and lin. dep.

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

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

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

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

$$\displaystyle c_{k+1}=-1$$

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

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

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

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

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