^{n}, where 𝔽 is either ℚ (rational numbers) or ℝ (real numbers) or ℂ (complex numbers). This aloows us to extend vector operations from 𝔽

^{n}to a vector space. Therefore, any finite dimensional vector space has the same algebraic structure as 𝔽

^{n}even though its vectors may not be expressed as

*n*-tuples.

# Isomorphism

We have already studied invertible square matrices. Let us generalize the results obtained in the general context of vector spaces. Recall that every finite-dimensional vector space*V*has a basis β, and we can use that basis to represent a vector

**v**∈

*V*as a coordinate vector [

**v**]

_{β}∈ 𝔽

^{n}, where 𝔽 is the ground field (either ℚ or ℝ or ℂ). We used this correspondence between

*V*and 𝔽

^{n}to motivate the idea that these vector spaces are “the same” in the sense that, in order to do a linear algebraic calculation in

*V*, we can instead do the corresponding calculation on coordinate vectors in 𝔽

^{n}. We now make this idea of vector spaces being “the same” a bit more precise and clarify under exactly which conditions this “sameness” happens.

A vector transformation *T* : *V* ⇾ *W* 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*.

*T*:

*V*⇾

*W*is an isomorphism, then the inverse

*T*

^{−1}:

*W*⇾

*V*is linear, hence also an isomorphism.

**Theorem 1:**Two finite-dimensional vector spaces

*V*and

*W*are isomorphic precisely when they have the same dimension.

*T*:

*V*⇾

*W*. Since

*T*is injective, ker

*T*= {

**0**}, dim ker

*T*= 0. Since

*T*is surjective, im

*T*=

*V*, dim im

*T*= dim

*V*. Hence \[ \dim V = \dim\,\mbox{im}\,T + \dim\,\ker\,T = \dim W. \] Conversely, if

*V*and

*W*have the same dimension

*n*, take bases {

**e**

_{i}} of

*V*and {

**u**

_{i}} of

*W*. Since they have the same number of elements by assumption, we may index them by the same index set 1 ≤

*i*≤

*n*. The mapping \[ {\bf x} = \sum_p x_i {\bf e}_i \ \mapsto \ {\bf y} = \sum_p x_i {\bf u}_i \] defines an isomorphism between

*V*and

*W*.

^{m,n}of rectangular

*m*×

*n*matrices with entries from ground field 𝔽. Our first example starts with the simplest 1×1 matrices [𝑎]. Since we have an obvious mapping [𝑎] ↦ 𝑎, which establishes an isomorphism 𝔽

^{1,1}⇾ 𝔽.

Consider the direct product of real axis ℝ^{n} = ℝ × ℝ × ⋯ × ℝ that is built of ordered *n*-tuples (*x*_{1}, … , *x*_{n}) ∈ ℝ^{n}. We can organize these *n*-tuples as row vectors or as column vectors; which shows that these three vectors spaces are isomorphic: ℝ^{n} ≌ ℝ^{1,n} ≌ ℝ^{n,1}.

The fact that we write the entries of vectors in 𝔽^{1,n} in a row whereas we
write those from 𝔽^{n,1} in a column is often just as irrelevant as if we used a
different font when writing the entries of the vectors in one of these vector
spaces. Indeed, vector addition and scalar multiplication in these spaces are
both performed entrywise, so it does not matter how we arrange or order those
entries. Moreover, we can identify these vectors with diagonal square matrices, or write them in come other fancy way (say in circle)m the resulting objects will be again isomorpic to 𝔽^{n}.

The fact that 𝔽^{n}, 𝔽^{1,n}, and 𝔽^{n,1} are isomorphic justifies something that
is typically done right from the beginning in linear algebra—treating members of 𝔽^{n} (vectors), members of 𝔽^{n,1} (column vectors), and members of 𝔽^{1,n} (row vectors) as the same thing.

Furthermore, the set 𝔽^{m,n} of *m* × *n* matrices is isomorphic to the set 𝔽^{n,m} of *n* × *m* matrices because transposing
twice gets us back to where we started.

Let us consider two four dimensional vector spaces, 𝔽^{4} of 4-tuples and 𝔽^{2,2} of 2×2 matrices.
They are isomorphic vector spaces because there is a natural isomorphism
between them:
\[
\left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right] \, \mapsto \, \begin{bmatrix} a & b \\ c & d \end{bmatrix} .
\]

**A**∈ 𝔽

^{m,n}, then

**A x**makes sense when

**x**is a column vector, but not when it is a row vector).

As an even simpler example of an isomorphism, we have implicitly been
using one when we say things like **v**^{T}**u** = **v** · **u** for all **v**, **u** ∈ ℝ^{n}. Indeed, the
quantity **v** · **u** is a scalar in ℝ, whereas **v**^{T}**u** is actually a 1 × 1 matrix (after
all, it is obtained by multiplying a 1 × *n* matrix by an *n* × 1 matrix), so it does
not quite make sense to say that they are “equal” to each other. However, the
spaces ℝ and ℝ^{1,1} are trivially isomorphic, so we typically sweep this technicality under the rug.

**Theorem 2:**Every finite-dimensional vector space

*V*of dimension

*n*≥ 1 is isomorphic to 𝔽

^{n}.

**e**

_{i}}, 1 ≤

*i*≤

*n*, of the vector space

*V*. Hence each element

**x**∈

*V*has a unique representation as a linear combination of the basis elements, say \( \displaystyle {\bf x} = \sum_i x_i {\bf e}_i . \) Let

*f*(

**x**) denote the

*n*-tuple formed by the components of

**x**: \[ f({\bf x}) = \left( x_1 , x_2 , \ldots , x_n \right) \in \mathbb{F}^n \] This map is bijective by definition. It is linear and its inverse 𝔽

^{n}⇾

*V*is \[ \left\{ x_i \right\}_{1 \le i \le n} \,\mapsto \, \sum_{i=1}^n x_i {\bf e}_i . \]

*V*= span{

*e*

^{𝑎x},

*x*

*e*

^{𝑎x},

*x*²

*e*

^{𝑎x}} and ℝ³ are isomorphic. The standard way to show that two spaces are isomorphic is to construct an isomorphism between them. To this end, consider the linear transformation

*T*: ℝ³ ⇾

*V*, defined by \[ T \left( c_1 , c_2 , c_3 \right) = c_1 e^{ax} + c_2 x\,e^{ax} + c_3 x^2 e^{ax} . \] It is straightforward to show that this function is a linear transformation, so we just need to convince ourselves that it is invertible. To this end, we need to show that vectors (functions)

*e*

^{𝑎x},

*x*

*e*

^{𝑎x},

*x*²

*e*

^{𝑎x}are linearly independent. Indeed, suppose we have the relation \[ c_1 e^{ax} + c_2 x\,e^{ax} + c_3 x^2 e^{ax} = 0 \qquad \iff \qquad c_1 + c_2 x + c_3 x^2 = 0 \tag{A} \] because the function

*e*

^{𝑎x}≠ 0 for any

*x*. Therefore, this function can be canceled out and we reduce Eq.(A) is reduced to \[ c_1 + c_2 x + c_3 x^2 = 0 , \] which is equivalent to show that monomials 1,

*x*, and

*x*² are linearly independent.

Hence, three functions *e*^{𝑎x}, *x**e*^{𝑎x}, *x*²*e*^{𝑎x} form a basis for *V*. Therefore, we can construct the standard matrix
[*T*]_{β←α}, where α = {**i**, **j**, **k**} is the standard basis of ℝ³.

*T*]

_{β←α}is clearly invertible (the identity matrix is its own inverse),

*T*is invertible too and is thus an isomorphism.

We can generalize this example and consider the set ℝ_{≤n}[*x*] of polynomials with real coefficients of degree up to *n*. This set is isomorphic to ℝ^{n+1}:
\[
T \left( a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \right) = \left( a_0 , a_1 , a_2 , \ldots , a_n \right) .
\]
It is straightforward to show that this function is a linear transformation,
so we just need to convince ourselves that it is invertible. To this end, we
just explicitly construct its inverse *T*^{−1} : ℝ^{n+1} ℝ_{≤n}[*x*], so
\[
T^{-1} \left( a_0 , a_1 , a_2 , \ldots , a_n \right) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n .
\]