If a subspace of a vector space is the span of two independent sets of vectors, then the number of vectors of the two sets must be the same. This number is called the dimension of the space. This property is a consequence of the following result.
Let #m# and #n# be natural numbers, #V# a vector space, and #\vec{a}_1 ,\ldots ,\vec{a}_n#, #\vec{b}_1,\ldots ,\vec{b}_m# vectors of #V# such that
- #V= \linspan{\vec{a}_1 ,\ldots ,\vec{a}_n }# and
- #\vec{b}_1 ,\ldots ,\vec{b}_m# is an independent set of vectors.
Then:
- #m\leq n#.
- If #V# is also spanned by #\vec{b}_1 ,\ldots ,\vec{b}_m#, then #m=n#.
Proof of the first statement: The vector #\vec{b}_1# is a linear combination of #\vec{a}_1 ,\ldots ,\vec{a}_n #. Because #\vec{b}_1# is not the zero vector, one of the scalars in the linear combination must be distinct from zero, so, thanks to the Exchange theorem, we can exchange the vector #\vec{b}_1 # with one of the vectors #\vec{a}_1 ,\ldots ,\vec{a}_n # without changing the span. Since we can change the order of the vectors #\vec{a}_1 ,\ldots ,\vec{a}_n# (refer to the theory Standard operations on spanning sets), we may assume that we can exchange #\vec{b}_1# with #\vec{a}_1#. Thus we find
\[ V=\ \linspan{\vec{b}_1 ,\vec{a}_2 ,\ldots ,\vec{a}_n } \] The vector #\vec{b}_2# is a linear combination of these vectors. There is definitely a scalar in this linear combination that is not equal to zero, and associated with one of #\vec{a}_2 ,\ldots ,\vec{a}_n #, for otherwise #\vec{b}_2# would be dependent on #\vec{b}_1#. Thus, again thanks to the Exchange theorem, we can exchange the vector #\vec{b}_2 # with one of the vectors #\vec{a}_2 ,\ldots ,\vec{a}_n #. Again we may (possibly after renumbering) assume that we can exchange #\vec{b}_2# with #\vec{a}_2#, so \[ V=\ \linspan{\vec{b}_1 ,\vec{b}_2 ,\vec{a}_3 ,\ldots ,\vec{a}_n } \] We continue this way; thanks to the Exchange theorem, each of the vectors #\vec{b}_1 ,\ldots ,\vec{b}_m # can be exchanged in such a way that eventually \[V=\ \linspan{\vec{b}_1 ,\ldots,\vec{b}_m ,\vec{a}_{m+1} ,\ldots ,\vec{a}_n } \] We conclude that #m\leq n#.
Proof of the second statement: Applying the first statement to the independent system #\vec{b}_1 ,\ldots ,\vec{b}_m# in #\linspan{\vec{a}_1 ,\ldots ,\vec{a}_n }# we conclude that #m\leq n#. Next, applying the first statement to the independent sequence #\vec{a}_1 ,\ldots ,\vec{a}_n# in #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_m }#, we find #n\leq m#. We conclude that #m=n#.
This result justifies the following definition of dimension:
An independent spanning subset of a vector space #V# is called a basis of #V# (the plural is bases).
The number of elements of a basis is called the dimension of #V# and denoted as #\dim{V}#.
If #V# is the null space, then the dimension is #0#. If there is no finite number of spanning vectors of #V#, then the dimension is infinite, and we write #\dim{V}=\infty#.
For every natural number #n# the set of vectors in #\mathbb{R}^n# \[\vec{e}_1 =\rv{1,0,0,\ldots ,0},\phantom{x}
\vec{e}_2 =\rv{0,1,0,\ldots ,0},\phantom{x}\ldots,\phantom{x}\vec{e}_n =\rv{0,0,0,\ldots ,1}\] is a basis of #\mathbb{R}^n#; it is also called the standard basis of #\mathbb{R}^n#. Accordingly, #\dim{\mathbb{R}^n}=n#. Similarly, #\dim{\mathbb{C}^n}=n#. Here, too, we use the name standard basis.
In the case #V=\{ \vec{0}\}# the vector space #V# is the span of zero vectors. Therefore, the dimension of this vector space is zero.
The vectors #\vec{e}_1 , \ldots , \vec{e}_n# form an independent set that spans #\mathbb{R}^n#: each vector #\rv{x_1, x_2,\ldots ,x_n}# can in fact be written as the linear combination #x_1 \vec{e}_1 + \cdots + x_n \vec{e}_n#. Previously, we established that this set is independent. Therefore #\vec{e}_1 , \ldots , \vec{e}_n# is a basis of #\mathbb{R}^n#. Because their number is #n#, we have #\dim{\mathbb{R}^n}=n#.
The vector space #P# of all polynomials in #x# has basis #1#, #x#, #x^2,\ldots# and has infinite dimension. After all, if there is a finite number of powers of #x# having a non-trivial linear combination that equals #\vec{0}#, then the coefficient of the highest power of #x# occurring in it would be equal to #0#, because it occurs at most once in the linear combination. But that means that highest power is not in the linear combination, a contradiction.
The following properties are useful for determining the dimension.
If #V# is a vector space with #\dim{V} =n\lt\infty#, then each basis of #V# consists of exactly #n# vectors. Let #\vec{b}_1 ,\ldots ,\vec{b}_m# be a set of vectors in #V#.
- If #m\lt n#, then #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_m }# is a proper subspace of #V#.
- If #m\gt n#, then the set #\vec{b}_1 ,\ldots ,\vec{b}_m# is dependent.
- If #m=n#, then the set #\vec{b}_1 ,\ldots ,\vec{b}_m# is a basis of #V# if and only if the sequence is independent.
1. If #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_m }# would be equal to #V#, then by thinning out the set #\vec{b}_1 ,\ldots ,\vec{b}_m# we would obtain a basis of at most #m# elements for #V#, so #n=\dim{V}\le m#, a contradiction with #m\lt n#.
2. This follows directly from the above statement on the number of spanning vectors.
3. If the sequence is dependent, then #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_m }# is spanned by less than #m=n# vectors and #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_m }# is a proper subspace of #V# because of part 1. If #\vec{b}_1 ,\ldots ,\vec{b}_n# are independent and not equal to #V#, then there is a vector #\vec{a} \in V# with #\vec{a}\not\in\linspan{\vec{b}_1 ,\ldots ,\vec{b}_n }#, so #\vec{b}_1 ,\ldots ,\vec{b}_n ,\vec{a}# would be an independent set of vectors in #V# with more than #n# elements. Therefore, #\linspan{\vec{b}_1 ,\ldots ,\vec{b}_n }=V# and #\vec{b}_1 ,\ldots ,\vec{b}_n# is a basis of #V#.
The third statement is a direct consequence of the statements 1 and 2.
If #n=\dim{V}=\infty# the first statement still holds, but the third does not.
The first statement is true because #m# is finite as #m\lt\infty# and, by definition, #V# is not spanned by a finite number of vectors.
The third statement is not true in general, as the following counterexample shows: If #V# is the vector space of polynomials in #x#, then #x#, #x^2#, #x^3,\ldots# is a sequence of #m=\infty# linearly independent vectors of the subspace of #V# whose span consists of all polynomials that take value #0# at #0#. Thus, this sequence has the same number of vectors as #n# (namely #\infty#) but does not span #V#.
What is the dimension of the plane #\mathbb{E}^2 # and of the space #\mathbb{E}^3#?
#\dim{\mathbb{E}^2} = 2# and #\dim{\mathbb{E}^3}=3#
Select two vectors in #\mathbb{E}^2# which are both distinct from the zero vector and which are not scalar multiples of each other. Then, any other vector in the plane can be written as a linear combination of these two (this is especially well known for the two vectors that have been given the coordinates #\rv{1,0}# and #\rv{0,1}#, respectively, in a coordinate system). Moreover, the two are linearly independent. From these two facts, we conclude that they form a basis.
The reasoning for #\mathbb{E}^3# is similar.