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Claim: For any singly infinite non-invertible matrix $A$, let $A$ to be injective and $A=BC$, where $B$ is invertible, and $C$ is a product of shifting matrices.

Is this claim true? Any reference would be nice.

By shifting matrices, I mean ones only on the superdiagonal or subdiagonal, and zeroes elsewhere.

Chengpei
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  • When you say invertible do you mean (a) "invertible as a bounded linear operator" or (b) "invertible solely when considering it as a matrix"? As an example, the compact linear operator $K=\operatorname{diag}(\frac1n)_{n=1}^\infty$ does not have bounded inverse but when considered as an infinite matrix formally has inverse $\operatorname{diag}(1,2,3,\ldots)$ which is not defined everywhere anymore (i.e. there exist square-summable infinite vectors which are mapped to something non-square-summable). I am asking because choosing $A=K$ yields a counterexample if you go by definition (a). – Frederik vom Ende Mar 15 '20 at 09:25
  • @FrederikvomEnde Sorry that I did not make it clear. Let us say that it is (b). – Chengpei Mar 16 '20 at 23:32

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