Consider the vector equation

`c`_{1}`b`_{1} + `c`_{2}`b`_{2} + ... + `c _{n}b_{n}` =

in a vector space ** V**, where

Vectors `b`_{1}, `b`_{2}, ..., ` b_{n}` ∈

`b`_{1},`b`_{2}, ...,span**b**_{n}.`V``b`_{1},`b`_{2}, ...,are linearly independent.**b**_{n}

The discussion at the beginning immediately gives us the following meaning of basis.

`B` = {`b`_{1}, `b`_{2}, ..., ` b_{n}`} is a basis ⇔

Example The standard basis vectors **e**_{1} = (1, 0, 0), **e**_{2} = (0, 1, 0), **e**_{3} = (0, 0, 1) span **R**^{3} and are linearly independent. Therefore the standard basis {**e**_{1}, **e**_{2}, **e**_{3}} is indeed a basis of **R**^{3}. The fact clearly extends to the standard basis of general **R**` ^{n}`.

The monomials 1, `t`, `t`^{2}, ..., `t ^{n}` span

= [A |
a |
b |
] = a[ |
1 | 0 | ] + b[ |
0 | 1 | ] + c[ |
0 | 0 | ] + d[ |
0 | 0 | ]
= aE_{11} + bE_{12}
+ cE_{21} + dE_{22} |

c |
d |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |

has the following two consequences

- Any 2 by 2 matrix
is a linear combination of`A`**E**_{11},**E**_{12},**E**_{21},**E**_{22}. - If a linear combination
of`A`**E**_{11},**E**_{12},**E**_{21},**E**_{22}(with`a`,`b`,`c`,`d`as coefficients) is the zero matrix, then all the entries of(which are`A``a`,`b`,`c`,`d`) must be zero.

The first consequence means exactly that `M`(2, 2) is spanned by
**E**_{11}, **E**_{12},
**E**_{21}, **E**_{22}. The second consequence means exactly that
**E**_{11}, **E**_{12},
**E**_{21}, **E**_{22} are linearly independent. Thus
{**E**_{11}, **E**_{12},
**E**_{21}, **E**_{22}} is a basis of `M`(2, 2).

In general, let ` E_{ij}` be the

Example In an earlier example, the system

3x_{1} |
+ x_{2} |
- x_{3} |
= | 2 |

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 6 |

was shown to have a unique solution. Since the number of rows is the same as the number of columns, by the basic principal of linear algebra, the general system

3x_{1} |
+ x_{2} |
- x_{3} |
= | b_{1} |

x_{1} |
- x_{2} |
+ x_{3} |
= | b_{2} |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | b_{3} |

has a unique solution for any right side. Since the general system is the same as the vector equation

`x`_{1}(3, 1, 2) + `x`_{2}(1, -1, 2) + `x`_{3}(-1, 1, 1)
= (`b`_{1}, `b`_{2}, `b`_{3}),

the existence for the general system means that (3, 1, 2), (1, -1, 2), (-1, 1, 1) span
**R**^{3}, and the uniqueness means the three vectors are linearly independent.
Thus we conclude that the three vectors form a basis of **R**^{3}.

The argument in the second example applies to general collections of euclidean vectors.
After all, being a basis of **R**^{n} means always existence (span **R**^{n})
plus uniqueness (independence). According to the
study on the basic
principal of linear algebra, the exact condition is described as follows.

If the columns of a matrix ** A** is a basis of

- Columns of
form a basis of`A`**R**^{n}. =`Ax`has a unique solution for any`b`.`b`=`Ax`has solutions for any`b`.`b`- The solution of
=`Ax`is unique.`b` is invertible.`A`

For the equivalence to invertibility, see here. For the criterion of euclidean basis from the computational viewpoint, see here.