Research Article | Open Access

Volume 2021 |Article ID 8895949 | https://doi.org/10.1155/2021/8895949

Marat V. Markin, "On a Characterization of Finite-Dimensional Vector Spaces", International Journal of Mathematics and Mathematical Sciences, vol. 2021, Article ID 8895949, 3 pages, 2021. https://doi.org/10.1155/2021/8895949

# On a Characterization of Finite-Dimensional Vector Spaces

Revised08 Jan 2021
Accepted13 Jan 2021
Published12 Feb 2021

#### Abstract

We provide a characterization of the finite dimensionality of vector spaces in terms of the right-sided invertibility of linear operators on them.

#### 1. Introduction

In [1], found is a characterization of one-dimensional (real or complex) normed algebras in terms of the bounded linear operators on them, echoing the celebrated Gelfand–Mazur theorem characterizing complex one-dimensional Banach algebras (see, e.g., [26]).

Here, continuing along this path, we provide a simple characterization of the finite dimensionality of vector spaces in terms of the right-sided invertibility of linear operators on them.

#### 2. Preliminaries

As is well-known (see, e.g., [7, 8]), a square matrix with complex entries is invertible iff it is one-sided invertible, i.e., there exists a square matrix of the same order as such thatwhere is the identity matrix of an appropriate size, in which case is the (two-sided) inverse of , i.e.,

Generally, for a linear operator on a (real or complex) vector space, the existence of a left inverse implies is invertible, i.e., injective. Indeed, let be a linear operator on a (real or complex) vector space and a linear operator be its left inverse, i.e.,where is the identity operator on . Equality (3), obviously, implies thatand hence, there exists an inverse for operator , where is its range (see, e.g., [9]). Equality (3) also implies that the inverse operator is the restriction of to .

Furthermore, as is easily seen, for a linear operator on a (real or complex) vector space, the existence of a right inverse, i.e., a linear operator such thatimmediately implies being surjective, which, provided the underlying vector space is finite dimensional, by the rank-nullity theorem (see, e.g., [9, 10]), is equivalent to being injective, i.e., being invertible.

With the underlying space being infinite-dimensional, the arithmetic of infinite cardinals does not allow to directly infer by the rank-nullity theorem that the surjectivity of a linear operator on the space is equivalent to its injectivity. In this case, the right-sided invertibility for linear operators need not imply invertibility. For instance, on the (real or complex) infinite-dimensional vector space of bounded sequences, the left shift linear operatoris noninvertible since(see, e.g., [9, 10]), but the right shift linear operatoris its right inverse, i.e.,where is the identity operator on .

Not only does the above example give rise to the natural question of whether, when the right-sided invertibility for linear operators on a (real or complex) vector space implies their invertibility, i.e., injectivity, the underlying space is necessarily finite dimensional but also serve as an inspiration for proving the “if” part of the subsequent characterization.

#### 3. Characterization

Theorem 1 (characterization of finite-dimensional vector spaces). A (real or complex) vector space is finite-dimensional iff, for linear operators on , right-sided invertibility implies invertibility.

Proof. “Only if” part. Suppose that the vector space is finite-dimensional with and let be an ordered basis for .
For an arbitrary linear operator on , which has a right inverse, i.e., a linear operator such thatwhere is the identity operator on . Let and be the matrix representations of the operators and relative to the basis , respectively (see, e.g., [7, 8]), thenwhere is the identity matrix of size (see, e.g., [7, 8]).
By the multiplicativity of determinant (see, e.g., [7, 8]), equality (11) implies thatWhence, we conclude thatwhich, by the determinant characterization of invertibility, it implies that matrix is invertible, and hence, so is the operator (see, e.g., [7, 8]).
“If” part. Let us prove this part by contrapositive, assuming that the vector space is infinite-dimensional. Suppose that is a (Hamel) basis for (see, e.g., [9, 10]), where is an infinite indexing set and that is a countably infinite subset of .
Let us define a linear operator as follows:whereis the basis representation of a vector relative to , in which all but a finite number of the coefficients , , called the coordinates of relative to , are zero (see, e.g., [9, 10]).
As is easily seen, is a linear operator on , which is noninvertible, i.e., noninjective, sinceThe linear operator on defined as follows:which is a right inverse for sinceThus, on a (real or complex) infinite-dimensional vector space, there exists a noninvertible linear operator with a right inverse, which completes the proof of the “if” part, and hence, of the entire statement.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright © 2021 Marat V. Markin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.