On a characterization of finite-dimensional vector spaces

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


Introduction
In paper [5], 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 charachterizing complex one-dimensional Banach algebras (see, e.g., [1-3, 8, 10]).
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.

Preliminaries
As is well-known (see, e.g., [4,9]), 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 where I is the identity matrix of an appropriate size, in which case C is the inverse of A, i.e., Generally, for a linear operator on a (real or complex) vector space, the existence of a left inverse implies being invertible. 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., Key words and phrases. Weak solution, scalar type spectral operator, Gevrey classes. and hence, there exists an inverse A −1 : R(A) → X for the operator A, where R(A) is its range (see, e.g., [7]). Equality (2.1) also implies that the inverse operator A −1 is the restriction of C to R(A).
Further, as is easily seen, for a linear operator on a (real or complex) vector space, the existence of a right inverse immediately implies being surjective, which, provided the underlying vector space is finite-dimensional, by the rank-nullity theorem (see, e.g., [6,7]), is equivalent to being injective, and hence, being invertible.
With the underlying space being infinite-dimensional, the arithmetic of infinite cardinals does not allow to conclude, by the rank-nullity theorem, that the surjectivity of a linear operator on the space is equivalent to its injectivity. Hence, 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 (see, e.g., [6,7]) but has the right shift linear operator 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, the underlying space is necessarily finite-dimensional, but also serves as an inspiration for proving the "if " part of the subsequent characterization.

Theorem 3.1 (Characterization of Finite-Dimensional Vector Spaces).
A (real or complex) vector space X is finite-dimensional iff right-sided invertibility for linear operators on X implies invertibility. Proof.
"Only if " part. Suppose that the vector space X is finite-dimensional with dim X = n (n ∈ N) and let B := {x 1 , . . . , x n } be an ordered basis for X.
For an arbitrary linear operator A : X → X on X, which has a right inverse C, i.e., AC = I, where I is the identity operator on X, let [A] B and [C] B be the matrix representations of the operators A and C relative to the basis B, respectively (see, e.g., [7]).
where I n is the identity matrix of size n (see, e.g., [7]).
"If " part. Let us prove this part by contrapositive, assuming that the vector space X is infinite-dimensional. Suppose that B := {x i } i∈I is a basis for X, where I is an infinite indexing set, and that J := {i(n)} n∈N is a countably infinite subset of I.
Let us define a linear operator A : X → X on X as follows: is the unique basis representation of a vector x ∈ X relative to the basis B, in which all but a finite number of the coefficients c i , i ∈ I, called the coordinates of x relative to B, are zero (see, e.g., [6,7]).
As is easily seen, A is a linear operator on X, which is non-invertible since The linear operator C : X → X on X defined as follows: 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, a hence, of the entire statement.