1
$\begingroup$

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 $Z=\{(a) \in \mathbb{R}^{\infty}|\mbox{ }a_n=0 \text{ for all but finitely many n} \}$

$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.