es
This section establishes a connection between vector spaces and show that all finite-dimensional spaces are equivalent to 𝔽n, where 𝔽 is either ℚ (rational numbers) or ℝ (real numbers) or ℂ (complex numbers). This allows us to extend vector operations from 𝔽n to a vector space. Therefore, any finite dimensional vector space has the same algebraic structure as 𝔽n even though its vectors may not be expressed as n-tuples.

Isomorphism

We often find two vector spaces related to each other in such a way that a particular isomorphism between them is singled out. Duality is one source of such "natural" isomorphisms.

A linear transformation T : V ⇾ W between two vector spaces that is both one-to-one and onto is said to be an isomorphism, and W is said to be isomorphic to V, which is abbreviated as V ≌ W.

Observe that if T : X ⇾ Y is an isomorphism, then the inverse T−1 : Y ⇾ X is a linear map, hence also an isomorphism.
Theorem 1: Two finite-dimensional vector spaces are isomorphic precisely when they have the same dimension.
Let T : X ⇾ Y be an isomorphism. Since T is injective, kerT = {0}, and dim ker(T) = 0. Also since T is surjective, Im(T) = Y, so dim Im(T) = dim(Y). Hence \[ \dim\,X = dim\mbox{Im}T + \dim\,\mbox{ker}(T) = \dim\,Y. \] Conversely, if X and Y have the same dimension n, take bases {ei} of X and {εj} of Y. Since they have the same number of elements by assumption, we may index them by the same index set 1 ≤ i ≤ n. The mapping \[ {\bf x} = \sum_i x_i {\bf e}_i \ \mapsto \ \sum_i x_i \varepsilon_i \] defines a n isomorphism between X and Y.
Example 1: Every polynomial is uniquely identified by its coefficients. Therefore, we get the isomorphism \[ \mathbb{R}_{\le 2} [x] \ni a + b\, x + c\, x^2 \mapsto \left[ a, b, c \right] \in \mathbb{R}^3 . \]
End of Example 1
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Theorem 2: Every finite-dimensional vector space of dimension n ≥ 1 is isomorphic to 𝔽n.
Let us choose a basis (ei), 1 ≤ i ≤ n, of the vector space X. Hence, each element x ∈ X has a unique representation as a linear combination of the ei's, say, \( \displaystyle {\bf x} = \sum_i x_i {\bf e}_i . \) Let f(x) denote the n-tuple formed by the the components of x: \[ f({\bf x}) = \left( x_1 , x_2 , \ldots , x_n \right) \in \mathbb{R}^n . \] This map is bijective by definition. It is linear and its inverse 𝔽n ⇾ X is \[ \left( x_i \right)_{1 \leqslant i \leqslant n} \mapsto \sum_{1 \leqslant i \leqslant n} x_i {\bf e}_i . \]
Example 5: The infinite set of monomials V₁ has the basis β = {[1, 0, 0], [0, 1, −1]} because every vector from V₁ is uniquely expanded as \[ \left[ a, b, -b \right] = a \left[ 1, 0, 0 \right] + b \left[ 0, 1, -1 \right] . \] This space is isomorphic to V₂ according to linear transformation \[ \phi \,V_1 \,\mapsto \, V_2 , \qquad \phi ([a, b, -b]) = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \] that preserves vector operations (vector addition and scalar multiplication). Indeed, \[ \phi \left( \left[ a_1 + a_2 , b_1 + _2 , - b_1 - b_2 \right] \right) = \begin{bmatrix} a_1 + a_2 & b_1 b_2 \\ -b_1 - b_2 & a_1 + a_2 \end{bmatrix} = \begin{bmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{bmatrix} = \phi ([ a_1 , b_1 , -b_1 ]) + \phi (a_2 , b_2 , -b_2 ]). \] This mapping also preserves scalar multiplication: \[ \phi \left( [\lambda a, \lambda b, -\lambda b]\right) = \begin{bmatrix} \lambda a & \lambda b \\ -\lambda b & a \end{bmatrix} = \lambda \begin{bmatrix} a & b \\ -b & a \end{bmatrix} = \lambda\,\phi \left( [a, b, -b] \right) . \] Any vector from V₂ can be uniquely expanded through two basis elements: \[ \begin{bmatrix} a & b \\ -b & a \end{bmatrix} = a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix} \phantom{-}0 & 1 \\ -1 & 0 \end{bmatrix} . \] We can also establish a linear transformation ψ V₂ ⇾ V₃ by formula \[ \psi \, : \,V_2 \mapsto V_2 , \qquad \psi \left( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \right) = a + b\, x - b\,x^2 + a\,x^3 . \]
End of Example 5
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The matrix description of a linear map T : X ⇾ Y between finite-dimensional vector spaces, is a consequence of choices of bases in both domain and target spaces. These choices are reflected by vertical isomorphisms

   

Example 7: An arbitrary m-by-n matrix A = [𝑎i,j] is often considered as a real or complex-valued function having the domain that is the Cartesian product [1..m] × [1..n], where Maple's notation of [1..m] = {1, 2, … , m} indicates the set of all integers within the interval [1, m]. It is customary to write a functions of discrete variables in indexed form as Ai, j instead of standard function form A(i, j).

Strictly speaking, our notation 𝔽m×n for the set of all 𝔽-valued m-by-n matrices is not accurate from math prospective and should be replaced by 𝔽[1..m]×[1..n]. However, the latter is not used in science literature mostly because it looks cumbersome for human eyes. We can also view [𝑎i,j] as a sequence of n column vectors in 𝔽m×1. This is the dual point of view where we hold j fixed and obtain a function of i for each j. From this point of view [𝑎i,j] is an element of (𝔽[1..m])[1..n]. This correspondence between 𝔽[1..m]×[1..n] and (𝔽[1..m])[1..n] is clearly an isomorphism, and is an example of a natural isomorphism.    ■

End of Example 7

We review next the various ways of looking at Cartesian n-space itself. One standard way of defining an ordered n-tuple is by induction. The ordered triplet (x, y, z) is defined as the ordered pair ((x, y), z), and the ordered n-tuple (x₁, x₂, … , xn) is defined as ((x₁, x₂, … , xn-1), xn). Thus, we define 𝔽n inductively by setting 𝔽¹ = 𝔽 and 𝔽n = 𝔽n-1 × 𝔽. .

The ordered n-tuple can also be defined as the function on [1..n] = {1, 2, … , n}, which assigns xi to i. Then

\[ \left( x_1 , x_2 , \ldots , x_n \right) = \left( 1, x_1 \right) , \left( 2, x_2 \right) , \ldots , \left( n, x_n \right) , \]
and Cartesian n-space is 𝔽[1..n] = 𝔽[1, 2, … , n].

Finally, we often wish to view Cartesian (n + m)-space as the Cartesian product of Cartesian n-space with Cartesian m-space, so we now take

\[ \left( x_1 , x_2 , \ldots , x_{n+m} \right) = \left( \left( x_1 , x_2 , \ldots , x_{n} \right) , \left( x_{n+1} , \ldots , x_{n+m} \right) \right) \]
and 𝔽n+m = 𝔽n × 𝔽m.

 

 

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