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Let $\mathrm{CFM}_\mathbb{N}(\mathbb{R})$ the infinite matrices $(a_{i,j})_{i,j \in \mathbb{N}}$ with real coefficients such that $\forall j \in \mathbb{N} $,$a_{ij}=0$ except for finite values of i, this is a monoid with the usual product in matrices , i.e, we have that for the matrices $A=(A_{ij})$ and $B=(B_{ij})$ the product is defined by $$(AB)_{ij}=\sum_{k=1}^{\infty}{A_{ik}B_{kj}}$$

Could someone explain me an element of this monoid? I can understand the condition given.

And what is the difference of the usual multiplication in matrices with this new condition in the product in $\mathrm{CFM}_\mathbb{N}(\mathbb{R})$ .

Thanks for your time and help.

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    I am trying to prove that has left inverse but not a right inverse. – 2017-02-02
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    the book said that the matrix $(\delta_{i,2j})_{i,j\in N}$ is left invertible but not right – 2017-02-02
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    I edited exactly as the book say. – 2017-02-02

2 Answers 2

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An element is just an infinite matrix in which each column has only finitely many nonzero entries (CFM stands for Column-Finite Matrices). That condition ensures that when you come to multiply two infinite matrices you don't have to worry about any infinite sums; the sums $\sum_{k=1}^{\infty}A_{ik}B_{kj}$ are all finite sums, since for fixed $j$ all but finitely many of the $B_{kj}$ are zero.

EDIT: The matrix you ask about in the comments is $$A=\pmatrix{0&0&\dots\cr1&0&\dots\cr0&0&\dots\cr0&1&\dots\cr\vdots&\vdots&\dots\cr}$$ It has the left inverse $$B=\pmatrix{0&1&0&0&\dots\cr0&0&0&1&\dots\cr\vdots&\vdots&\vdots&\vdots&\dots\cr}$$ $A$ has no right inverse because the 1st row of $A$ is all zeros, so no matter what $C$ is the 1st row of $AC$ will be all zeros.

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As Gerry Myerson already said, $\newcommand{\cfm}{\operatorname{CFM}_{\Bbb N}(\Bbb R)}\cfm$ can be seen as matrices where the index ranges over the whole $\Bbb N$, instead of just $\{1,\cdots, n\}$, with the request that each succession $A_{1j},A_{2j},\cdots, A_{kj},\cdots$ is eventually $0$. This allows each sum $\sum\limits_{k=1}^\infty A_{ik}A_{kj}$ to be actually finite.

A convenient way to see $\cfm$ is the following: recall that the map \begin{align}M(n,\Bbb R)&\to \operatorname{end}_{\Bbb R}(\Bbb R^n)\\A&\mapsto \left[x\mapsto A\cdot x\right]\end{align} is an isomorphism of unital $\Bbb R$-algebras (hence, of monoids)

Analogously, let's call $\Bbb R^{(\Bbb N)}:=\bigoplus\limits_{n\in\Bbb N} \Bbb R$ the set of real sequences $x=(x_n\,:\,n\in\Bbb N)$ such that $x_n=0$ eventually.

Given $A\in\cfm$ and $x\in \Bbb R^{(\Bbb N)}$, we can consider the vector $\Bbb R^{(\Bbb N)}\ni y=A\cdot x$ given by $$y_i:=\sum_{k=1}^\infty A_{ik}x_k$$

In analogy with the finite case, the map \begin{align}\cfm&\to\operatorname{end}_{\Bbb R}\left(\Bbb R^{(\Bbb N)}\right)\\ A&\mapsto[x\mapsto A\cdot x]\end{align} is an isomorphism. Therefore, you can without loss of generality see $\cfm$ as the algebra of linear maps $V\to V$, where $V$ is a $\Bbb R$-vector space with $\dim V=\aleph_0$, with the identification $\Phi_{A\cdot B}=\Phi_{A}\circ\Phi_{B}$.