On a Characterization of Finite-Dimensional Vector Spaces

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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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$$\begin{array}{c}\text{(1)}& \text{AC}=I\text{right\hspace{0.17em}inverse}\text{or\hspace{0.17em}CA}=I\text{left\hspace{0.17em}inverse},\end{array}$$where $I$ is the identity matrix of an appropriate size, in which case $C$ is the (two-sided) inverse of $A$, i.e.,$$\begin{array}{}\text{(2)}& \text{AC}=\text{CA}=I.\end{array}$$

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 $A:X⟶X$ 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.,$$\begin{array}{}\text{(3)}& \text{CA}=I,\end{array}$$where $I$ is the identity operator on $X$. Equality (3), obviously, implies that$$\begin{array}{}\text{(4)}& \ker A=0,\end{array}$$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 a right inverse, i.e., a linear operator $C:X⟶X$ such that$$\begin{array}{}\text{(5)}& \text{AC}=I,\end{array}$$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 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 $l_{\infty }$ of bounded sequences, the left shift linear operator$$\begin{array}{}\text{(6)}& l_{\infty }\mathrm{\ni }x≔x_1,x_2,x_3,\dots ⟼Lx≔x_2,x_3,x_4,\dots \in l_{\infty }\end{array}$$is noninvertible since$$\begin{array}{}\text{(7)}& \ker L=x_1,\mathrm{0,0},\dots \ne 0\end{array}$$(see, e.g., [9, 10]), but the right shift linear operator$$\begin{array}{}\text{(8)}& l_{\infty }\mathrm{\ni }x≔x_1,x_2,x_3,\dots \mapsto Rx≔0,x_1,x_2,\dots \in l_{\infty }\end{array}$$is its right inverse, i.e.,$$\begin{array}{}\text{(9)}& \text{LR}=I,\end{array}$$where $I$ is the identity operator on $l_{\infty }$.

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 $X$ 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 $\dim X=nn\in \mathbb{N}$ and let $B≔x_1,\dots ,x_n$ be an ordered basis for $X$.

For an arbitrary linear operator $A:X⟶X$ on $X$, which has a right inverse, i.e., a linear operator $C:X⟶X$ such that$$\begin{array}{}\text{(10)}& \text{AC}=I,\end{array}$$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, 8]), then$$\begin{array}{}\text{(11)}& A_B{C}_B=I_n,\end{array}$$where $I_n$ is the identity matrix of size $n$ (see, e.g., [7, 8]).

By the multiplicativity of determinant (see, e.g., [7, 8]), equality (11) implies that$$\begin{array}{}\text{(12)}& \det A_B\det C_B=\det A_B{C}_B=\det I_n=1.\end{array}$$

Whence, we conclude that$$\begin{array}{}\text{(13)}& \det A_B\ne 0,\end{array}$$which, by the determinant characterization of invertibility, it implies that matrix $A_B$ 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 $B≔{x_i}_{i\in I}$ is a (Hamel) basis for $X$ (see, e.g., [9, 10]), where $I$ is an infinite indexing set and that $J≔{in}_{n\in \mathbb{N}}$ is a countably infinite subset of $I$.

Let us define a linear operator $A:X⟶X$ as follows:$$\begin{array}{c}\text{(14)}& A{x}_{i1}≔0,A{x}_{in}≔x_{in-1},\hspace{1em}n\ge 2,\\ A{x}_i≔x_i,\hspace{1em}i\in I\backslash J,\\ X\mathrm{\ni }x={\displaystyle \sum _{i\in I}{c}_i{x}_i}\mapsto Ax≔{\displaystyle \sum _{i\in I}{c}_{i}A{x}_i},\end{array}$$where$$\begin{array}{}\text{(15)}& {\displaystyle \sum _{i\in I}{c}_i{x}_i}\end{array}$$is the basis representation of a vector $x\in X$ relative to $B$, in which all but a finite number of the coefficients $c_i$, $i\in I$, 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$$\begin{array}{}\text{(16)}& \ker A=\text{span}{x}_{i1}\ne 0.\end{array}$$

The linear operator $C:X⟶X$ on $X$ defined as follows:$$\begin{array}{c}\text{(17)}& C{x}_{in}≔x_{in+1},\hspace{1em}n\in \mathbb{N},C{x}_i≔x_i,\hspace{1em}i\in I\backslash J,\\ X\mathrm{\ni }x={\displaystyle \sum _{i\in I}{c}_i{x}_i}⟼Cx≔{\displaystyle \sum _{i\in I}{c}_{i}C{x}_i},\end{array}$$which is a right inverse for $A$ since$$\begin{array}{c}\text{(18)}& \text{AC}{x}_{in}=A{x}_{in+1}=x_{in},\hspace{1em}n\in \mathbb{N},\text{AC}{x}_i=A{x}_i=x_i,\hspace{1em}i\in I\backslash J.\end{array}$$

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.

References

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