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 basis of $X$ is uncountable. Can anyone help how can I solve the above problem?
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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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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} \|…
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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…
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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…
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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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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!
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Why isn't every Hamel basis a Schauder basis?

I seem to have tripped on the common Hamel/Schauder confusion. If $X$ is any vector space (not necessarily finite dimension) and $B$ is a linearly independent subset that spans $X$, then $B$ is a Hamel basis for $X$. If there exists a sequence…
Fequish
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What if the basis not countable, then what?

I'm second year physics & mathematics student, and self-studying Abstract linear algebra from the book Linear Algebra by Werner Greub. In the mean time I have come across several times to the notion of countable basis.I know what can or can not do…
Our
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What is the basis of the vector space $l^\infty$?

We know that every vector space has a Hamel basis and also every normed space need not have a Schauder basis. As the normed space $l^\infty$ is not Separable so can't have the Schauder basis, but on the other side $l^\infty$ is also a vector space…
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Vector space bases without axiom of choice

I want to find an example of a vector space with no base if we assume that axiom of choice is incorrect. This question might be duplicate so please alert me. Thanks.
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We have $n$ real numbers around the circle and among any consecutive 3 one is AM of the other two. Then all the numbers are the same or $3\mid n$.

There are $n$ real numbers around the circle and among any consecutive 3 one is arithmetic mean of the other two. Prove that all the numbers are the same or $3\mid n$. Hint was to use a linear algebra. It is obviously that if among some three…
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