How many linearly independent vectors can exist in n-dimensions?

1) How many linearly independent vectors can exist in n-dimensions?
2) To define a n-dimensional space we need n unique unit vectors... Is it right??... Must these vectors be mutually orthogonal?

linear algebra

1) The Dimension is given by a number of "base" vectors, which is n in your case. The thing is that there is an infinite number of basis, there is an infinite number of n groups of linearly independent vectors, but there is one elementary, canonical base.

To understand take R2: ecanonical base is (1,0) and (0,1). But also (1,1) and (-1,2) are linearly independent, the idea of independence is not to find n constants , at l;east one of them non-zero, so that the linear combination of the n vectors is equal to 0. Also you can never find less than n vectors independent, that would mean that your dimension was wrong to begin with .

A base is the minimal number of linear vectors.

Any other vector in the space can be written in function of them, so it will be linearly dependent with the n base vectors.

So: bases have n vectors each, infinite number of basis, your independent vectors can be found infinite numbers, but always judged n at a time.

2)It is right, an n-dimensional space is defined by its base vectors, (other than its operations). And they are orthogonal because that is how they are defined , normally orthonormal ( ortho=orthogonal and normal = norme 1).

Orthogonality is the linear independence relation ( take the three geaometrical space canonical base vectors, which I know as i, j,k, no matter what you try to multiply them with, you will never get to the vector 0. They are orhtogonal=independent, normal because they module is 1, and so they form a base.

Re: linear algebra

can u make the idea a little brief and crisp... i'm not able to understand what you have said....

linear algebra

Brief and Crisp... hmmmm...

1) N-Dimensional space => the value of N is given by the minimum number of linearly independent vectors you can possibly find.

In R2( RxR) you can find maximum 2 independent vectors, such as (1,0) and (0,1) which are also by chance the ones with module/norme/...1.
But there are an infinity of pairs of linear independent vectors in this space.

In general, there are an infinity of n-tuples of linear independent vectors in an N-dim space.

Can't get any clearer than that, sorry

2) A space is defined by its base. A base can be chosen arbitrarily from the infinity of n-tuples of linear independent vectors. From any such n-tuple you can use an algorithm of Schmidt for example to extract the orthonomal vectors (they are already orthogonal, it;s the normalisation part that's more tricky without the algorithm...)

I hope I'm not mistaken, but for all Iremember and reason, the set of orthonomal vectors is not unique, there is one for every base you may choose to orthonormalise by the (Gram)Schmidt procedure

https://en.wikipedia.org/wiki/Gram-Schmidt_process

Re: linear algebra

1- to discribe n dimetioanal space you need n indipendent vectors

2- the vectors are not unique .
example: a=(2,0,0) , b=(0,2,0) , c= (0, 0,2)
a=(2,0,0) , b=(1,3,0) , c= (1, 2,3)
they both discribe 3 dimentioanl space => not unique


3- the vectors does not have to be unit vectors.
example: a=(2,0,0) , b=(0,3,0) , c= (1, 2,3)
a=(2,0,0) , b=(0,2,0) , c= (0, 0,2)


4- the vectors does not need to be orthogonal to each other. the only condition is independent.
example: a=(2,0,0) , b=(5,3,0) , c= (1, 2,3)