If $X=(x_1,x_2,\dots)$ is an infinite real row vector and $A=(a_{ij}),0<i,j< \infty$ is an infinite real matrix, one may or may not be able to define the matrix product $XA$. For which A can one define right multiplication on the space $Z$ defined as $Z=\{(a) \in \mathbb{R}^{\infty}|\mbox{ }a_n=0 \text{ for all but finitely many n} \}$

  • $\begingroup$ Hint: You mention that one cannot always define $XA$ in $\mathbb{R}^\omega$, what specifically goes wrong with a naive approach? $\endgroup$ – Eric Stucky Oct 11 '13 at 15:50
  • $\begingroup$ @EricStucky That is what is written in my book .. $\endgroup$ – user43418 Oct 11 '13 at 15:53

Hint: You mention that one cannot always define $XA$ in $\mathbb{R}^\omega$; in particular, the product that we want to use $$(XA)_i =\sum_{j\in\mathbb{N}}x_ia_{ij}$$ has a problem. What is that problem, and to what extent does having finitely many nonzero $x_i$ alleviate it?

  • $\begingroup$ Why does this have 3 unexplained downvotes? This is a solid hint. $\endgroup$ – Jack M Oct 11 '13 at 18:17
  • $\begingroup$ In retrospect, perhaps it's because I messed up my indices: the sum should be $(XA)_j=\sum_i x_ia_{ij}$. $\endgroup$ – Eric Stucky Dec 11 '16 at 3:42

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