How to write this vector in terms of Hamel basis?

Question

Let $(c,c,c,...)$ be a vector in $l_\infty$, where $c \ne 0$. How to write this vector in terms of Hamel basis?

Depends which basis. For instance, there is a basis that has $(c,c,c,\ldots)$ as an element, so it's pretty easy in that one. — spaceisdarkgreen, Jul 29 '18 at 05:44
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma. — max_zorn, Jul 29 '18 at 06:02
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $\Bbb{R^3}$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $\ell_\infty$. They only exist by virtue of the Axiom of Choice, you know. — Jyrki Lahtonen, Jul 29 '18 at 06:03
Possible duplicate of What is the basis of the vector space $l^\infty$? — Jyrki Lahtonen, Jul 29 '18 at 06:09

spaceisdarkgreen · Answer 1 · 2018-07-29T06:22:37.370

It's not clear exactly which misconceptions you have here, so I'll just say a few things and hopefully it will clarify.

The set $\{(1,0,0,\ldots),(0,1,0,0,\ldots), \ldots\}$ is not a Hamel basis for $l_\infty$ as is evidenced by the fact that $(c,c,\ldots)$ cannot be written as a finite linear combination of those vectors.
There is no such thing as the Hamel basis. There are generally many of them. There is also not generally a canonical basis, so saying "the" doesn't even work implicitly.
It is generally impossible to explicitly exhibit a basis (and $l_\infty$ is not an exception), so there's no way to answer your question in any kind of generality. However, from the usual general proof that a basis exists, it's reasonably clear that, for some given element $v$ of your vector space $V,$ that there is a basis that contains $v$ as an element. More generally, for any linearly independent subset $L\subseteq V,$ there is a basis $B$ with $L\subseteq B.$ (The usual argument applies Zorn's lemma to the set of all linearly independent subsets of $V,$ and we can just modify it to instead consider the set of all linearly independent subsets of $V$ that contain $L$ as a subset.) So there are certainly bases in which your vector $(c,c,\ldots)$ has a simple expression.

1 Answers1