TheoremBase

Properties of Dimension

In a finite-dimensional vector space independent families extend to bases, independent and spanning families are compared with the dimension, subspaces have at most the dimension of the space, and Euclidean space RnR^n has dimension n.

Statement

In the setting of Linear Algebra: Standing Notation, let KK, with ++, ⋅\cdot, 00 and 11, be a field, let VV, with its addition and scalar multiplication, be a finite-dimensional vector space over KK with dimension dim⁡V\dim V, let m∈N0m\in\mathbb{N}_{0}, and let u=(u1,…,um)u=(u_{1},\dots,u_{m}) be an mm-tuple in VV. Linear independence is as in Linearly Independent and Linearly Dependent Finite Families (i), spanning as in Span of a Finite Family; Spanning Families (iii), and bases as in Finite Basis of a Vector Space (i).

Extension to a basis. If uu is linearly independent, then there are p∈N0p\in\mathbb{N}_{0} and a pp-tuple (w1,…,wp)(w_{1},\dots,w_{p}) in VV such that

(u1,…,um,w1,…,wp),(u_{1},\dots,u_{m},w_{1},\dots,w_{p}),

the (m+p)(m+p)-tuple whose ii-th component is uiu_{i} for i∈[m]i\in[m] and wi−mw_{i-m} for i∈{m+1,…,m+p}i\in\{m+1,\dots,m+p\}, is a basis of VV.

Independent families. If uu is linearly independent, then m≤dim⁡Vm\leq\dim V, and m=dim⁡Vm=\dim V if and only if uu is a basis of VV.

Spanning families. If uu spans VV, then m≥dim⁡Vm\geq\dim V, and m=dim⁡Vm=\dim V if and only if uu is a basis of VV.

Dimension zero. dim⁡V=0\dim V=0 if and only if V={0}V=\{0\}.

Subspaces. Let WW be a linear subspace of VV, a vector space over KK with the operations of VV by Linear Subspaces Are Vector Spaces; Intersections (i). Then WW is finite-dimensional, dim⁡W≤dim⁡V\dim W\leq\dim V, and dim⁡W=dim⁡V\dim W=\dim V if and only if W=VW=V.

Euclidean space. Let n∈Nn\in\mathbb{N}. In the real vector space Rn\mathbb{R}^{n} of ℝⁿ Is a Real Vector Space (i), the standard basis vectors (e1,…,en)(e_{1},\dots,e_{n}) form a basis, and

dim⁡Rn=n.\dim\mathbb{R}^{n}=n.

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