We used coordinates to construct isomorphisms between vector spaces ** V** and euclidean spaces

The number `n` in **R**` ^{n}` is the dimension of the euclidean space. It is also the number of vectors in the basis of

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 **R**` ^{n}`. 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

Example The standard basis of **R**` ^{n}` contains

The basis {1, `t`, `t`^{2}, ..., `t ^{n}`} of

The basis {` E_{ij}`: 1 ≤

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 | ] = aE_{11} + b(E_{12} + E_{21}) + cE_{22}, |

b |
c |
0 | 0 | 1 | 0 | 0 | 1 |

and an argument similar to this example, we conclude that {**E**_{11}, **E**_{12} + **E**_{21}, **E**_{22}} is a basis of `Sym`(2), and `dim``Sym`(2) = 3.

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

: 1 ≤**E**_{ii}`i`≤`n`+**E**_{ij}: 1 ≤**E**_{ji}`i`<`j`≤`n`

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