30/09/2014

Dimension (Vector Space)

Dimension:


Definition 2.1
A vector space is finite-dimensional if it has a basis with only finitely many vectors.
(One reason for sticking to finite-dimensional spaces is so that the representation of a vector with respect to a basis is a finitely-tall vector, and so can be easily written.) From now on we study only finite-dimensional vector spaces. We shall take the term "vector space" to mean "finite-dimensional vector space". Other spaces are interesting and important, but they lie outside of our scope.
To prove the main theorem we shall use a technical result.
Lemma 2.2 (Exchange Lemma)
Assume that  B=\langle \vec{\beta}_1,\dots,\vec{\beta}_n \rangle  is a basis for a vector space, and that for the vector  \vec{v}  the relationship  \vec{v}=c_1\vec{\beta}_1+c_2\vec{\beta}_2+\cdots +c_n\vec{\beta}_n  has  c_i\neq 0 . Then exchanging  \vec{\beta}_i  for  \vec{v}  yields another basis for the space.
Proof
Call the outcome of the exchange  \hat{B}=\langle \vec{\beta}_1,\dots,\vec{\beta}_{i-1},\vec{v},\vec{\beta}_{i+1},\dots,\vec{\beta}_n \rangle .
We first show that \hat{B} is linearly independent. Any relationship  d_1\vec{\beta}_1+\dots+d_i\vec{v}+\dots+d_n\vec{\beta}_n=\vec{0} among the members of \hat{B}, after substitution for \vec{v},
d_1\vec{\beta}_1+\dots +d_i\cdot(c_1\vec{\beta}_1+\dots+c_i\vec{\beta}_i+\dots+c_n\vec{\beta}_n) +\dots+d_n\vec{\beta}_n =\vec{0} \qquad\qquad(*)
gives a linear relationship among the members of B. The basis B is linearly independent, so the coefficient d_ic_i of \vec{\beta}_i is zero. Because c_i is assumed to be nonzero, d_i=0. Using this in equation (*) above gives that all of the other d's are also zero. Therefore \hat{B} is linearly independent.
We finish by showing that \hat{B} has the same span as B. Half of this argument, that [{\hat{B}}]\subseteq[B], is easy; any member d_1\vec{\beta}_1+\dots+d_i\vec{v}+\dots+d_n\vec{\beta}_n of [{\hat{B}}] can be written d_1\vec{\beta}_1+\dots+d_i\cdot(c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_n)+\dots+d_n\vec{\beta}_n, which is a linear combination of linear combinations of members of B, and hence is in [B]. For the [B]\subseteq[{\hat{B}}] half of the argument, recall that when \vec{v}=c_1\vec{\beta}_1+\dots+c_n\vec{\beta}_nwith c_i\neq 0, then the equation can be rearranged to \vec{\beta}_i=(-c_1/c_i)\vec{\beta}_1+\dots+(1/c_i)\vec{v}+\dots+(-c_n/c_i)\vec{\beta}_n. Now, consider any member d_1\vec{\beta}_1+\dots+d_i\vec{\beta}_i+\dots+d_n\vec{\beta}_n of [B], substitute for \vec{\beta}_iits expression as a linear combination of the members of \hat{B}, and recognize (as in the first half of this argument) that the result is a linear combination of linear combinations, of members of \hat{B}, and hence is in [{\hat{B}}].
Theorem 2.3
In any finite-dimensional vector space, all of the bases have the same number of elements.
Proof
Fix a vector space with at least one finite basis. Choose, from among all of this space's bases, one  B=\langle \vec{\beta}_1,\dots,\vec{\beta}_n \rangle  of minimal size. We will show that any other basis  D={\langle \vec{\delta}_1,\vec{\delta}_2,\ldots \rangle }  also has the same number of members, n. Because  B  has minimal size,  D  has no fewer than  n  vectors. We will argue that it cannot have more than  n  vectors.
The basis  B  spans the space and  \vec{\delta}_1  is in the space, so  \vec{\delta}_1 is a nontrivial linear combination of elements of  B . By the Exchange Lemma,  \vec{\delta}_1  can be swapped for a vector from  B , resulting in a basis  B_1 , where one element is  \vec{\delta}  and all of the  n-1  other elements are  \vec{\beta} 's.
The prior paragraph forms the basis step for an induction argument. The inductive step starts with a basis  B_k  (for  1\leq k<n ) containing  k  members of  D  and  n-k members of  B . We know that  D  has at least  n  members so there is a  \vec{\delta}_{k+1} . Represent it as a linear combination of elements of  B_k . The key point: in that representation, at least one of the nonzero scalars must be associated with a \vec{\beta}_i  or else that representation would be a nontrivial linear relationship among elements of the linearly independent set  D . Exchange  \vec{\delta}_{k+1}  for  \vec{\beta}_i  to get a new basis  B_{k+1}  with one  \vec{\delta}  more and one  \vec{\beta}  fewer than the previous basis  B_k .
Repeat the inductive step until no  \vec{\beta} 's remain, so that  B_n contains \vec{\delta}_1,\dots,\vec{\delta}_n. Now,  D  cannot have more than these n  vectors because any  \vec{\delta}_{n+1}  that remains would be in the span of  B_n  (since it is a basis) and hence would be a linear combination of the other \vec{\delta}'s, contradicting that D is linearly independent.
Definition 2.4
The dimension of a vector space is the number of vectors in any of its bases.
Example 2.5
Any basis for  \mathbb{R}^n  has  n  vectors since the standard basis  \mathcal{E}_n  has  n  vectors. Thus, this definition generalizes the most familiar use of term, that \mathbb{R}^n is n-dimensional.
Example 2.6
The space  \mathcal{P}_n  of polynomials of degree at most n has dimension  n+1 . We can show this by exhibiting any basis— \langle 1,x,\dots,x^n \rangle  comes to mind— and counting its members.
Example 2.7
A trivial space is zero-dimensional since its basis is empty.
Again, although we sometimes say "finite-dimensional" as a reminder, in the rest of this book all vector spaces are assumed to be finite-dimensional. An instance of this is that in the next result the word "space" should be taken to mean "finite-dimensional vector space".
Corollary 2.8
No linearly independent set can have a size greater than the dimension of the enclosing space.
Proof
Inspection of the above proof shows that it never uses that D  spans the space, only that  D  is linearly independent.
Example 2.9
Recall the subspace diagram from the prior section showing the subspaces of  \mathbb{R}^3 . Each subspace shown is described with a minimal spanning set, for which we now have the term "basis". The whole space has a basis with three members, the plane subspaces have bases with two members, the line subspaces have bases with one member, and the trivial subspace has a basis with zero members. When we saw that diagram we could not show that these are the only subspaces that this space has. We can show it now. The prior corollary proves that the only subspaces of  \mathbb{R}^3  are either three-, two-, one-, or zero-dimensional. Therefore, the diagram indicates all of the subspaces. There are no subspaces somehow, say, between lines and planes.
Corollary 2.10
Any linearly independent set can be expanded to make a basis.
Proof
If a linearly independent set is not already a basis then it must not span the space. Adding to it a vector that is not in the span preserves linear independence. Keep adding, until the resulting set does span the space, which the prior corollary shows will happen after only a finite number of steps.
Corollary 2.11
Any spanning set can be shrunk to a basis.
Proof
Call the spanning set  S . If  S  is empty then it is already a basis (the space must be a trivial space). If  S=\{\vec{0}\}  then it can be shrunk to the empty basis, thereby making it linearly independent, without changing its span.
Otherwise, S contains a vector \vec{s}_1 with \vec{s}_1\neq\vec{0} and we can form a basis  B_1=\langle \vec{s}_1 \rangle . If  [B_1]=[S]  then we are done.
If not then there is a  \vec{s}_2\in[S]  such that  \vec{s}_2\not\in[B_1] . Let  B_2=\langle \vec{s}_1,\vec{s_2} \rangle ; if  [B_2]=[S]  then we are done.
We can repeat this process until the spans are equal, which must happen in at most finitely many steps.
Corollary 2.12
In an  n -dimensional space, a set of  n  vectors is linearly independent if and only if it spans the space.
Proof
First we will show that a subset with  n  vectors is linearly independent if and only if it is a basis. "If" is trivially true— bases are linearly independent. "Only if" holds because a linearly independent set can be expanded to a basis, but a basis has  n  elements, so this expansion is actually the set that we began with.
To finish, we will show that any subset with  n  vectors spans the space if and only if it is a basis. Again, "if" is trivial. "Only if" holds because any spanning set can be shrunk to a basis, but a basis has  n  elements and so this shrunken set is just the one we started with.
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