# On a Characterization of Finite-Dimensional Vector Spaces

Authors: Marat V. Markin
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

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## Abstract

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


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## Body

## 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., [2–6]).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 A with complex entries is invertible iff it is one-sided invertible, i.e., there exists a square matrix C of the same order as A such that(1)AC=Iright inverseor CA=Ileft inverse,where I is the identity matrix of an appropriate size, in which case C is the (two-sided) inverse of A, i.e.,(2)AC=CA=I.Generally, for a linear operator on a (real or complex) vector space, the existence of aleft inverse implies is invertible, i.e., injective. Indeed, let A:X⟶X be a linear operator on a (real or complex) vector space X and a linear operator C:X⟶X be its left inverse, i.e.,(3)CA=I,where I is the identity operator on X. Equality (3), obviously, implies that(4)kerA=0,and hence, there exists an inverse A−1:RA⟶X for operator A, where RA is its range (see, e.g., [9]). Equality (3) also implies that the inverse operator A−1 is the restriction of C to RA.Furthermore, as is easily seen, for a linear operator on a (real or complex) vector space, the existence of aright inverse, i.e., a linear operator C:X⟶X such that(5)AC=I,immediately 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 beinginfinite-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 l∞ of bounded sequences, the left shift linear operator(6)l∞∋x≔x1,x2,x3,…⟼Lx≔x2,x3,x4,…∈l∞is noninvertible since(7)kerL=x1,0,0,…≠0(see, e.g., [9, 10]), but the right shift linear operator(8)l∞∋x≔x1,x2,x3,…↦Rx≔0,x1,x2,…∈l∞is its right inverse, i.e.,(9)LR=I,where I is the identity operator on l∞.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 spaceX is finite-dimensional iff, for linear operators on X, right-sided invertibility implies invertibility.Proof.
“Only if” part. Suppose that the vector space X is finite-dimensional with dimX=nn∈ℕ and let B≔x1,…,xn be an ordered basis for X.
For an arbitrary linear operatorA:X⟶X on X, which has a right inverse, i.e., a linear operator C:X⟶X such that(10)AC=I,where I is the identity operator on X. Let AB and CB be the matrix representations of the operators A and C relative to the basis B, respectively (see, e.g., [7, 8]), then(11)ABCB=In,where In is the identity matrix of size n (see, e.g., [7, 8]).
By themultiplicativity of determinant (see, e.g., [7, 8]), equality (11) implies that(12)detABdetCB=detABCB=detIn=1.
Whence, we conclude that(13)detAB≠0,which, by the determinant characterization of invertibility, it implies that matrix AB is invertible, and hence, so is the operator A (see, e.g., [7, 8]).
“If” part. Let us prove this part by contrapositive, assuming that the vector space X is infinite-dimensional. Suppose that B≔xii∈I is a (Hamel) basis for X (see, e.g., [9, 10]), where I is an infinite indexing set and that J≔inn∈ℕ is a countably infinite subset of I.
Let us define a linear operatorA:X⟶X as follows:(14)Axi1≔0,Axin≔xin−1,n≥2,Axi≔xi,i∈I\J,X∋x=∑i∈Icixi↦Ax≔∑i∈IciAxi,where(15)∑i∈Icixiis the basis representation of a vector x∈X relative to B, in which all but a finite number of the coefficients ci, i∈I, called the coordinates of x relative to B, are zero (see, e.g., [9, 10]).
As is easily seen,A is a linear operator on X, which is noninvertible, i.e., noninjective, since(16)kerA=spanxi1≠0.
The linear operatorC:X⟶X on X defined as follows:(17)Cxin≔xin+1,n∈ℕ,Cxi≔xi,i∈I\J,X∋x=∑i∈Icixi⟼Cx≔∑i∈IciCxi,which is a right inverse for A since(18)ACxin=Axin+1=xin,n∈ℕ,ACxi=Axi=xi,i∈I\J.
Thus, 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.



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*Source: 8895949-2021-02-12.xml*