Dimension of a vector space

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The dimension of a vector space (also called the Hamel dimension of a vector space, to distinguish it from the Hilbert dimension in the case of Hilbert spaces) is the number of vectors that form a [Hamel] basis of the vector space.

Introduction

Given a vector space, we can consider sets of vectors S of a vector space V and it can be examined if they possess some of these properties:

Linear independence it is said that a set of vectors ${displaystyle Ssubset V}$ is linearly independent if for any finite number of vectors it is fulfilled that:

${displaystyle lambda _{1}v_{1}+lambda _{2}v_{2}+dots +lambda _{n}v_{n}=0Rightarrow lambda _{1}=lambda _{2}=lambda _{n}=dots =0,quad v_{i}in S,lambda _{i}in mathbf {K} }$

Note that in a finite dimension vector space n, the maximum number of linearly independent vectors is n.

Generator unitgiven a linear subspace L it is said that a set S is generator L Yes:

${displaystyle xin LLeftrightarrow exists mu _{1}dots mu _{n}:x=mu _{1}v_{1}+dots mu _{n}v_{n},quad v_{i}in S,mu _{i}in mathbf {K} }$

Note that a finite set of m elements can generate at most a subspace L of dimension to the highest m.

A set that is linearly independent (1) and generator of the vector space (2) is said to be a vector basis. It can be shown that all the bases of a vector space are sets with the same number of elements (that is, sets that have the same cardinal number). And the common number of elements of any basis is precisely the dimension of the vector space.

Note an important fact, if you change the body of the climbers, $mathbb{R}$ a $mathbb{C}$ , then the same M point will be determined by the complex z_M =x + andi, that's for a single parameter.

The P dimension is 1 above $mathbb{C}$ and two over $mathbb{R}$ :

${displaystyle {begin{matrix}dim _{mathbb {C} }P&=&1\dim _{mathbb {R} }P&=&2end{matrix}}}$

A real plane is therefore a complex line. The name complex plane to designate a real plane with complex writing of the coordinates (x + yi instead of (x; y)) It's wrong, but very common.

The environment is three-dimensional and therefore requires three reals (x, and, z) to define a point. You cannot be considered as a space on $mathbb{C}$ .

In the theory of relativity, a fourth variable is added: time, and a point (x, y, z, t) of this four-dimensional space corresponds to an event or happening (the coordinates tell us where and when it happened).

In some current theories, physicists work on a model of space with eleven dimensions, but on the whole, and not the real ones. Like the whole $mathbb{Z}$ of the integers is not a body but a ring, the space is not vectorial (it is said to be a module). However, the definition of the dimension is valid in such spaces. In this example, most dimensions are rolled over themselves, like a snake that bites the tail. Its curvature is enormous, because its radio is microscopic, less than a core. Vector spaces do not have curvature.

Formal definition

More formally the dimension of a vector space is defined as the cardinal of a vector base for that space. By the axiom of choice all space has a base (even the space {0}, since the void is a base), and since it can be shown that all the vector bases have the same cardinal, then it turns out the concept of dimension is well defined. It should be noted that there are both finite and infinite dimension vector spaces (the vector space of the polynomials of a variable, for example it has dimension $aleph _{0}$ .

The dimension of a space also coincides with the following two cardinals:

The maximum number of linearly independent vectors of that space.
The minimum number of vectors that form a generator set for all space.

Dimension of a subspace

The definition remains the same in the case of a subspace, but there is a particular method of calculating it when the subspace is defined as space generated by the vector system. Let's see it in an example. In space ${mathbb {R}}^{3}$ , be it vectors:

${displaystyle {vec {u}}{begin{pmatrix}1\3\-2end{pmatrix}},{vec {v}}{begin{pmatrix}-2\1\3end{pmatrix}},{vec {w}}{begin{pmatrix}-3\5\4end{pmatrix}}{mbox{ y }}{vec {n}}{begin{pmatrix}4\5\-7end{pmatrix}}}$

Four vectors cannot be independent in ${mathbb {R}}^{3}$ , therefore there must be dependency relationships:

{displaystyle x{begin{pmatrix}1\3\-2end{pmatrix}}+y{begin{pmatrix}-2\1\3end{pmatrix}}+z{begin{pmatrix}-3\5\4end{pmatrix}}+t{begin{pmatrix}4\5\-7end{pmatrix}}={vec {0}}}

which can be written in matrix form:

{displaystyle {begin{pmatrix}1&-2&-3&4\3&1&5&5\-2&3&4&-7end{pmatrix}}{begin{pmatrix}x\y\z\tend{pmatrix}}={vec {0}}}

Let's call the previous matrix A, and X the column vector. This relationship means that vector X belongs to the kernel of A, denoted Ker A (from German Kern, kernel). The generated space is the set of

{displaystyle x{vec {u}}+y{vec {v}}+z{vec {w}}+t{vec {n}}}

that is, of the A·X: it is the image of A.

It is intuitive that the larger the kernel, the smaller the image, in terms of dimensions. Specifically, if we call the dimension of its image range of A: rg A = dim (Im A), we have the relation:

rg A + dim (Ker A) = dim E (E: space of Entry from A, here E =

{displaystyle mathbb {R} ^{4}}

Let's find dim (Ker A):

${displaystyle left{{begin{matrix}(I)&x-2y-3z+4t=0\(II)&3x+y+5z+5t=0\(III)&-2x+3y+4z-7t=0end{matrix}}right.Leftrightarrow left{{begin{matrix}(I)&x-2y-3z+4t=0\(IV)=(II)-3(I)&7y+14z-7t=0\(V)=(III)+2(I)&-y-2z+t=0end{matrix}}right.}$

${displaystyle Leftrightarrow left{{begin{matrix}(I)&x-2y-3z+4t=0\(IV)/7&y+2z-t=0\-(V)&y+2z-t=0end{matrix}}right.}$

There remain two non-proportional equations, therefore independent, and each one subtracts 1 from the dimension, which is initially equal to 4. It turns out that dim (Ker A) = 2. It can be verified in another way: The two equations allow us to express y, then x as a function of z and t, therefore there are only two free variables left, and the dimension is 2.

Applying the formula: rg A = 4 - 2 = 2. The subspace is a plane.

Grassmann dimensions formula

Main article: Grassmann Formula

If U₁ and U₂ are subspaces of a finite-dimensional vector space, the following holds:

${displaystyle {mbox{dim}}(U_{1}+U_{2})={mbox{dim}} U_{1}+{mbox{dim}} U_{2}-{mbox{dim}}(U_{1}cap U_{2})}$

Examples

All euclid space has finite dimension on ${displaystyle scriptstyle mathbb {R} }$ .
The set of complex numbers $scriptstyle {mathbb {C}}$ is dimension 1 over $scriptstyle {mathbb {C}}$ I mean, ${displaystyle scriptstyle mathrm {dim} _{mathbb {C} }(mathbb {C})=1}$ , however, about ${displaystyle scriptstyle mathbb {R} }$ It's dimension 2, ${displaystyle scriptstyle mathrm {dim} _{mathbb {R} }(mathbb {C})=2}$ .
The whole of Hamilton's quaternions $scriptstyle {mathbb {H}}$ , satisfies ${displaystyle scriptstyle mathrm {dim} _{mathbb {R} }(mathbb {H})=4}$ and ${displaystyle scriptstyle mathrm {dim} _{mathbb {C} }(mathbb {H})=2}$ .
Given a body $scriptstyle {mathbb {K}}$ the whole:

{displaystyle mathbb {K} ^{n}=underbrace {mathbb {K} times dots times mathbb {K} } _{n}}

sat ${displaystyle scriptstyle mathrm {dim} _{mathbb {K} }left(mathbb {K} ^{n}right)=n}$ .

All set with body structure $scriptstyle {mathbb {K}}$ is a vectorial space on itself of dimension 1. If considered a subbody ${displaystyle scriptstyle mathbb {K} _{1}}$ then the original body is a vector space on the subbody. Yeah. $scriptstyle {mathbb {K}}$ is an algebraic extension ${displaystyle scriptstyle mathbb {K} _{1}}$ then the dimension of the original body on the subbody is finite.
The set of matrices ${displaystyle scriptstyle M_{ntimes n}(mathbb {K})}$ is a dimension vector space n².
The vector space of polynomials ${displaystyle scriptstyle mathbb {K} [X]}$ has infinite dimension over $scriptstyle {mathbb {K}}$ specifically, ${displaystyle scriptstyle mathrm {dim} _{mathbb {K} }(mathbb {K} [X])=aleph _{0}}$ .
The set of real numbers has dimension 1 when considered as the body of scales ${displaystyle scriptstyle mathbb {R} }$ , but if considered a numberable subbody of the real, such as the rational number ${displaystyle scriptstyle mathbb {Q} }$ then the dimension is infinite [not numberable], in particular, ${displaystyle scriptstyle mathrm {dim} _{mathbb {Q} }(mathbb {R})=aleph _{1}}$ .

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