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Questions tagged [hamel-basis]

A Hamel basis (often simply called basis) of a vector space $V$ over a field $F$ is a linearly independent subset of $V$ that spans it.

A Hamel basis (often simply called basis) of a vector space $V$ over a field $F$ is a linearly independent subset of $V$ that spans it. In other words, a subset $B$ of $V$ is a basis when every element of $V$ can be expressed in one and only one way as a (finite) linear combinations of elements of $B$. It can be proved that all bases of a vector space have the same cardinal, which is (by definition) the dimension of $V$.

The term "Hamel basis" is used mostly in reference to vector spaces of infinite dimension, where other notions of bases are used.

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What is the difference between a Hamel basis and a Schauder basis?

Let $V$ be a vector space with infinite dimensions. A Hamel basis for $V$ is an ordered set of linearly independent vectors $\{ v_i \ | \ i \in I\}$ such that any $v \in V$ can be expressed as a finite linear combination of the $v_i$'s; so $\{ v_i \…
Lor
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Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of $X$ is uncountable.

Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of $X$ is uncountable. Can anyone help how can I solve the above problem?
mintu
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What is a basis for the vector space of continuous functions?

A natural vector space is the set of continuous functions on $\mathbb{R}$. Is there a nice basis for this vector space? Or is this one of those situations where we're guaranteed a basis by invoking the Axiom of Choice, but are left rather…
I. J. Kennedy
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Vector Spaces and AC

I know that the proof that every vector space has a basis uses the Axiom of Choice, or Zorn's Lemma. If we consider an axiom system without the Axiom of Choice, are there vector spaces that provably have no basis?
user40170
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Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$?

Assuming the Axiom of Choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$, the vector space of real sequences?
Fernando Martin
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Finding a basis of an infinite-dimensional vector space?

The other day, my teacher was talking infinite-dimensional vector spaces and complications that arise when trying to find a basis for those. He mentioned that it's been proven that some (or all, do not quite remember) infinite-dimensional vector…
InterestedGuest
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Does $\mathbb{R}^\mathbb{R}$ have a basis?

I'm studying linear algebra on my own and saw basis for $\mathbb{R}, \mathbb{R}[X], ...$ but there is no example of $\mathbb{R}^\mathbb{R}$ (even though it is used for many examples). What is a basis for it? Thank you
user12205
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Is every vector space basis for $\mathbb{R}$ over the field $\mathbb{Q}$ a nonmeasurable set?

The existence of subsets of the real line which are not Lebesgue measurable can be argued using the Axiom of Choice. For example, define an equivalence relation on $[0, 1]$ by $a \thicksim b$ if and only if $a - b \in \mathbb{Q}$ and let $S \subset…
Ben Passer
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A Hamel basis for $\ell^p$?

I am looking for an explicit example for a Hamel basis for $\ell^{p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one can express any element of the vector space as a…
Abhishek Gupta
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What's an example of a vector space that doesn't have a basis if we don't accept Choice?

I've read that the fact that all vector spaces have a basis is dependent on the axiom of choice, I'd like to see an example of a vector space that doesn't have a basis if we don't accept AoC. I'm also interested in knowing why this happens. Thanks!
YoTengoUnLCD
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Why isn't this a counterexample of Banach-Steinhaus theorem?

The theorem from Wikipedia is as follows Let $X$ be Banach space, $Y$ be a normed space and $F$ be family of linear bounded operators $f:X \to Y$ such that $\forall x \in X \sup_{f \in F} \|f(x)\|_Y < \infty$. Then $$ \sup_{f \in F ,\ \|x\| = 1} \|…
Physor
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Confusion about the Hamel Basis

Alright, so I'm reading a book on Hilbert spaces and functional analysis, and here it defines a "Hamel basis" to be a "maximal linearly independent set". I take this to mean that $S$ is a Hamel basis if $S$ is linearly independent, and there is no…
user3002473
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At most finitely many (Hamel) coordinate functionals are continuous - different proof

If $X$ is a vector space over $\mathbb R$ and $B=\{x_i; i\in I\}$ is a Hamel basis for $X$, then for each $i\in I$ we have a linear functional $a_i(x)$ which assigns to $x$ the $i$-th coordinate, i.e., the functions $a_i$ are uniquely determined by…
Martin Sleziak
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What is a Hamel basis?

According to Mathworld, a Hamel basis is a basis for $\mathbb R$ considered as a vector space over $\mathbb Q$. According to Wikipedia, the term is used in the context of infinite-dimensional vector spaces over $\mathbb R$ or $\mathbb C$. According…
J. W. Tanner
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Every vector space has a basis proof using Zorn's lemma in linear algebra

Every vector space has a basis proof by using Zorn's lemma. How will I prove this statement by using Zorn's lemma?
user39206
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