dimension formulae for vector spaces

In this entry we look at various formulaeinvolving 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 arithmeticin 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

$$0⟶U⟶V⟶W⟶0$$

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

$$\mathrm{\cdots }⟶V_{n+1}⟶V_n⟶V_{n-1}⟶\mathrm{\cdots }$$

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 spacesterminating with $0$ at both endscan be considered as an infinite exact sequencein 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 spaceis determined by its dimension and the cardinality of the ground field:

The effect of the above formula is somewhat differentdepending on whether $K$ is finite (http://planetmath.org/FiniteField) or infinite.If $K$ is finite, then it reduces to

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

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

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 sectioncan be applied to give a formula for the dimension of $V^{*}$:

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 thatan infinite-dimensional Banach space can be isomorphic to its dualare sometimes surprised to learn that an infinite-dimensional vector spacecannot 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 spacesone is usually only interested in the continuous linear functionals,and the resulting ‘continuous’ dualis 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 sectionis 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

In the special case $W=V$ this can be simplified to

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 formulais to show that the dimension is alwaysat 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 dimensionis 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 tothe 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