### Basis

##### 4. Dimension

We used coordinates to construct isomorphisms between vector spaces V and euclidean spaces Rn. This produces a dictionary that translates the general linear algebra in V to the more concrete linear algebra in Rn.

The number n in Rn is the dimension of the euclidean space. It is also the number of vectors in the basis of V used for constructing the coordinates.

The dimension of a vector space is the number of vectors in a basis.

The dimension is infinite if there is no finite basis. From this exercise, we see that a vector space is finite dimensional if and only if it can be spanned by finitely many vectors.

Recall that the columns of any n by n invertible matrix is a basis of Rn. Thus the basis is not unique, and the definition of dimension must be justified by showing that different choices of bases will always have the same number of vectors. Since this is a linear algebra problem, we may translate the problem (by fixing any one basis and considering the corresponding coordinates) into the euclidean space. Therefore it is sufficient to show that any basis of Rn must contain n vectors. This special case has been proved by the characterization of bases of Rn.

Example The standard basis of Rn contains n vectors. Therefore dimRn = n.

The basis {1, t, t2, ..., tn} of Pn consists of n + 1 monomials. Therefore dimPn = n + 1.

The basis {Eij: 1 ≤ im, 1 ≤ jn} of M(m, n) contains mn matrices. Therefore dimM(m, n) = mn.

Example All the 2 by 2 symmetric matrices form a vector space Sym(2) (because it is a subspace of M(2, 2)). By

 A = [ a b ] = a[ 1 0 ] + b[ 0 1 ] + c[ 0 0 ] = aE11 + b(E12 + E21) + cE22, b c 0 0 1 0 0 1

and an argument similar to this example, we conclude that {E11, E12 + E21, E22} is a basis of Sym(2), and dimSym(2) = 3.

In general, the vector space Sym(n) of symmetric n by n matrices has a basis consisting of the following vectors

1. Eii: 1 ≤ in
2. Eij + Eji: 1 ≤ i < jn

From this it is easy to conclude that dimSym(n) = n(n + 1)/2.