dimension formulae for vector spaces
In this entry we look at various formulae involving the dimension^{} of a vector space^{}.
Throughout this entry, $K$ will be a field, and $V$ and $W$ will be vector spaces over $K$. The dimension of a vector space $U$ over $K$ will be denoted by $dim(U)$, or by ${dim}_{K}(U)$ if the ground field needs to be emphasized.
All of these formulae potentially involve infinite^{} cardinals, so the reader should have a basic knowledge of cardinal arithmetic in order to understand them in full generality.
Subspaces
If $S$ and $T$ are subspaces^{} of $V$, then
$$dim(S)+dim(T)=dim(S\cap T)+dim(S+T).$$ |
Rank-nullity theorem
The rank-nullity theorem^{} states that if $\varphi :V\to W$ is a linear mapping, then the dimension of $V$ is the sum of the dimensions of the image and kernel of $\varphi $:
$$dim(V)=dim(\mathrm{Im}\varphi )+dim(\mathrm{Ker}\varphi ).$$ |
In particular, if $U$ is a subspace of $V$ then
$$dim(V)=dim(V/U)+dim(U).$$ |
The rank-nullity theorem can also be stated in terms of short exact sequences^{}: if
$$0\u27f6U\u27f6V\u27f6W\u27f60$$ |
is a short exact sequence of vector spaces over $K$, then
$$dim(V)=dim(U)+dim(W).$$ |
This can be generalized to infinite exact sequences^{}: if
$$\mathrm{\cdots}\u27f6{V}_{n+1}\u27f6{V}_{n}\u27f6{V}_{n-1}\u27f6\mathrm{\cdots}$$ |
is an exact sequence of vector spaces over $K$, then
$$\sum _{n\mathrm{even}}dim({V}_{n})=\sum _{n\mathrm{odd}}dim({V}_{n}).$$ |
(This is indeed a generalization^{}, because any finite exact sequence of vectors spaces terminating with $0$ at both ends can be considered as an infinite exact sequence in which all remaining terms are $0$.)
Direct sums
If ${({V}_{i})}_{i\in I}$ is a family of vector spaces over $K$, then
$$dim\left(\underset{i\in I}{\oplus}{V}_{i}\right)=\sum _{i\in I}dim({V}_{i}).$$ |
Cardinality of a vector space
The cardinality of a vector space is determined by its dimension and the cardinality of the ground field:
$$|V|=\{\begin{array}{cc}{|K|}^{dim(V)},\hfill & \text{if}dim(V)\text{is finite};\hfill \\ \mathrm{max}\{|K|,dim(V)\},\hfill & \text{if}dim(V)\text{is infinite}.\hfill \end{array}$$ |
The effect of the above formula^{} is somewhat different depending on whether $K$ is finite (http://planetmath.org/FiniteField) or infinite. If $K$ is finite, then it reduces to
$$|V|=\{\begin{array}{cc}{|K|}^{dim(V)},\hfill & \text{if}dim(V)\text{is finite};\hfill \\ dim(V),\hfill & \text{if}dim(V)\text{is infinite}.\hfill \end{array}$$ |
If $K$ is infinite, then it can be expressed as
$$ |
Change of ground field
If $F$ is a subfield of $K$, then $V$ can be considered as a vector space over $F$. The dimensions of $V$ over $K$ and $F$ are related by the formula
$${dim}_{F}(V)=[K:F]\cdot {dim}_{K}(V).$$ |
In this formula, $[K:F]$ is the degree of the field extension $K/F$, that is, the dimension of $K$ considered as a vector space over $F$.
Space of functions into a vector space
If $S$ is any set, then the set ${K}^{S}$ of all functions from $S$ into $K$ becomes a vector space over $K$ if we define the operations^{} pointwise, that is, $(f+g)(x)=f(x)+g(x)$ and $(\lambda f)(x)=\lambda f(x)$ for all $f,g\in {K}^{S}$, all $x\in S$, and all $\lambda \in K$. The dimension of this vector space is given by
$$dim({K}^{S})=\{\begin{array}{cc}|S|,\hfill & \text{if}S\text{is finite};\hfill \\ {|K|}^{|S|},\hfill & \text{if}S\text{is infinite}.\hfill \end{array}$$ |
The case where $S$ is infinite is not straightforward to prove. Proofs can be found in books by Baer[1] and Jacobson[2], among others.
More generally, we can consider the space ${V}^{S}$, which is really just the direct product^{} (http://planetmath.org/DirectProduct) of copies of $V$ indexed by $S$. We get
$$dim({V}^{S})=\{\begin{array}{cc}0,\hfill & \text{if}dim(V)=0;\hfill \\ |S|\cdot dim(V),\hfill & \text{if}S\text{is finite};\hfill \\ {|V|}^{|S|},\hfill & \text{otherwise}.\hfill \end{array}$$ |
Dual space
Given any basis $B$ of $V$, the dual space^{} ${V}^{*}$ is isomorphic^{} to ${K}^{B}$ via the mapping $f\mapsto {f|}_{B}$. So the formula of the previous section^{} can be applied to give a formula for the dimension of ${V}^{*}$:
$$dim({V}^{*})=\{\begin{array}{cc}dim(V),\hfill & \text{if}dim(V)\text{is finite};\hfill \\ {|K|}^{dim(V)},\hfill & \text{if}dim(V)\text{is infinite}.\hfill \end{array}$$ |
In particular, this formula implies that $V$ is isomorphic to ${V}^{*}$ if and only if $V$ is finite-dimensional. (Students who are familiar with the fact that an infinite-dimensional Banach space can be isomorphic to its dual are sometimes surprised to learn that an infinite-dimensional vector space cannot be isomorphic to its dual, for a Banach space is surely a vector space. But the term dual is used in different senses in these two statements, so there is no contradiction^{}. In the theory of Banach spaces one is usually only interested in the continuous^{} linear functionals^{}, and the resulting ‘continuous’ dual is a subspace of the full dual used in the above formula.)
Space of linear mappings
The set ${\mathrm{Hom}}_{K}(V,W)$ of all linear mappings from $V$ into $W$ is itself a vector space over $K$, with the operations defined in the obvious way, namely $(f+g)(x)=f(x)+g(x)$ and $(\lambda f)(x)=\lambda f(x)$ for all $f,g\in {\mathrm{Hom}}_{K}(V,W)$, all $x\in V$, and all $\lambda \in K$. The dual space ${V}^{*}={\mathrm{Hom}}_{K}(V,K)$ considered in the previous section is a special case of this. For any basis $B$ of $V$, the mapping $f\mapsto {f|}_{B}$ defines an isomorphism^{} between ${\mathrm{Hom}}_{K}(V,W)$ and ${W}^{B}$, so that from an earlier section we get
$$dim({\mathrm{Hom}}_{K}(V,W))=\{\begin{array}{cc}0,\hfill & \text{if}dim(W)=0;\hfill \\ dim(V)\cdot dim(W),\hfill & \text{if}dim(V)\text{is finite};\hfill \\ {|W|}^{dim(V)},\hfill & \text{otherwise}.\hfill \end{array}$$ |
In the special case $W=V$ this can be simplified to
$$dim({\mathrm{End}}_{K}(V))=\{\begin{array}{cc}dim{(V)}^{2},\hfill & \text{if}dim(V)\text{is finite};\hfill \\ {|K|}^{dim(V)},\hfill & \text{otherwise}.\hfill \end{array}$$ |
Tensor products
The dimension of the tensor product^{} (http://planetmath.org/TensorProduct) of $V$ and $W$ is given by
$$dim(V\otimes W)=dim(V)\cdot dim(W).$$ |
Banach spaces
The dimension of a Banach space, considered as a vector space, is sometimes called the Hamel dimension, in order to distinguish it from other concepts of dimension. For an infinite-dimensional Banach space $B$ we have
$$dim(B)=|B|.$$ |
The tricky part of establishing this formula is to show that the dimension is always at least the cardinality of the continuum^{}. A short proof of this is given in a paper by Lacey[3].
The above formula suggests that Hamel dimension is not a very useful concept for infinite-dimensional Banach spaces, which is indeed the case. Nonetheless, it is interesting to see how Hamel dimension relates to the usual concept of dimension in Hilbert spaces^{}. If $H$ is a Hilbert space, and $d$ is its dimension (meaning the cardinality of an orthonormal basis), then the Hamel dimension $dim(H)$ is given by
$$dim(H)=\{\begin{array}{cc}d,\hfill & \text{if}d\text{is finite};\hfill \\ {d}^{{\mathrm{\aleph}}_{0}},\hfill & \text{if}d\text{is infinite}.\hfill \end{array}$$ |
References
- 1 Reinhold Baer, Linear Algebra^{} and Projective Geometry, Academic Press, 1952.
- 2 Nathan Jacobson, Lectures in Abstract Algebra, Volume II: Linear Algebra, D. Van Nostrand Company Inc., 1953.
- 3 H. Elton Lacey, The Hamel Dimension of any Infinite Dimensional Separable Banach Space is c, Amer. Math. Mon. 80 (1973), 298.
Title | dimension formulae for vector spaces |
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Canonical name | DimensionFormulaeForVectorSpaces |
Date of creation | 2013-03-22 16:31:19 |
Last modified on | 2013-03-22 16:31:19 |
Owner | yark (2760) |
Last modified by | yark (2760) |
Numerical id | 25 |
Author | yark (2760) |
Entry type | Feature |
Classification | msc 15A03 |