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 aloows 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 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 ∈ 𝔽ⁿ, where 𝔽 is the ground field (either ℚ or ℝ or ℂ). We used this correspondence between V and 𝔽ⁿ 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 𝔽ⁿ. 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.

We can think of isomorphic vector spaces as having the same structure and the same vectors as each other, but different labels on those vectors. Note that if 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.

Take any isomorphism T : V ⇾ W. Since T is injective, kerT = {0}, dim kerT = 0. Since T is surjective, imT = V, dim imT = dimV. 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.

Example 1: Let is start with the vector space 𝔽^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 ℝⁿ = ℝ × ℝ × ⋯ × ℝ that is built of ordered n-tuples (x₁, … , x_n) ∈ ℝⁿ. We can organize these n-tuples as row vectors or as column vectors; which shows that these three vectors spaces are isomorphic: ℝⁿ ≌ 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 𝔽ⁿ.

The fact that 𝔽ⁿ, 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 𝔽ⁿ (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, 𝔽⁴ of 4-tuples and 𝔽^2,2 of 2×2 matrices. They are isomorphic vector spaces because there is a natural isomorphism between them: $\left[ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right] \, \mapsto \, \begin{bmatrix} a & b \\ c & d \end{bmatrix} .$

End of Example 1

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When we speak about isomorphisms and vector spaces being “the same”, we only mean with respect to the 8 defining properties of vector spaces (i.e., properties based on vector addition and scalar multiplication). We can add column vectors in the exact same way that we add row vectors, or we can add diagonal matrices, and similarly scalar multiplication works the exact same for those three types of vectors. However, other operations like matrix multiplication may behave differently on these three sets (e.g., if 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^Tu = v · u for all v, u ∈ ℝⁿ. Indeed, the quantity v · u is a scalar in ℝ, whereas v^Tu 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 𝔽ⁿ.

Let us choose a basis { 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 𝔽ⁿ ⇾ V is

$\left\{ x_i \right\}_{1 \le i \le n} \,\mapsto \, \sum_{i=1}^n x_i {\bf e}_i .$

Example 2: For any real number 𝑎, the vector space V = span{ e^𝑎x, xe^𝑎x, x²e^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, xe^𝑎x, x²e^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 x² are linearly independent.

Hence, three functions e^𝑎x, xe^𝑎x, x²e^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.

$\begin{align*} [\,T\,]_{\beta \leftarrow \alpha} &= \left[ \left[ T(1, 0, 0) \right]_{\beta} \vert \left[ T(0, 1, 0) \right]_{\beta} \vert \left[ T(0, 0, 1) \right]_{\beta} \right] \\ &= \left[ \left[ e^{ax} \right]_{\beta} \vert \left[ x\,e^{ax} \right]_{\beta} \vert \left[ x^2 e^{ax} \right]_{\beta} \right] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} . \end{align*}$

Since [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 ℝⁿ⁺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 : ℝⁿ⁺¹ ℝ_≤n[x], so $T^{-1} \left( a_0 , a_1 , a_2 , \ldots , a_n \right) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n .$

End of Example 2

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While it’s true that there’s nothing really “mathematically” new about isomorphisms, the important thing is the new perspective that it gives us. It is very useful to be able to think of vector spaces as being the same as each other, as it can provide us with new intuition or cut down the amount of work that we have to do. More generally, isomorphisms are used throughout all of mathematics, not just in linear algebra. In general, they are defined to be invertible maps that preserve whatever the relevant structures or operations are. In our setting, the relevant operations are scalar multiplication and vector addition, and those operations being preserved is exactly equivalent to the invertible map being a linear transformation.

Introduction to Linear Algebra

Systems of Linear Equations

Matrix Algebra

Vector Spaces

Eigenvalues, Eigenvectors

Euclidean Spaces

Matrix Decompositions

Applications

Functions of Matrices

Miscellany

Preliminaries

Glossary

Reference

Isomorphism