All Questions
Tagged with axiom-of-choice linear-algebra
71
questions
3
votes
0
answers
43
views
Discontinuous linear map and AC
The question arises when I am constructing an elementary proof for the following claim:
Given a normed vector space $V$, the following are equivalent:
$V$ is finite dimensional
Every linear map $T:V\...
- 73
5
votes
2
answers
125
views
No 3 vectors independent over $\mathbb{Z}$ in $\mathbb{Z}^2$, without AoC
Q: Are there three $\mathbb{Z}^2$ vectors independent over $\mathbb{Z}$ ?
Context: This problem arise naturally when I'm characterizing possible sub-"latice" in $\mathbb{Z}^2$. Formally let ...
- 635
4
votes
1
answer
151
views
Without choice, what can be the (finite) automorphism groups of $\mathbb F_2$-vector spaces?
My motivation for this question is similar to the one in this question. However, that question only asks about the possibility of an infinite $\mathbb F_2$-vector space having trivial automorphism ...
- 41
1
vote
0
answers
49
views
The natural map $V^*\otimes W\rightarrow \mathrm{Hom}(V, W)$ in infinite-dimensional case.
Let $V$ and $W$ be vector spaces over a field $K$. Define a function $\varphi:V^*\times W\to V^*\otimes W$ as $\varphi(f, w)= f\otimes w$.
Define a function $h:V^*\times W\to \mathrm{Hom}(V, W)$ as $h(...
- 399
5
votes
1
answer
124
views
$\bigcap\limits_{\varphi\in E^*}\ker(\varphi)$ and the Axiom of Choise
Context.
Give a nonzero $K$-vector space $E$, it is known that $\displaystyle \bigcap_{\varphi\in E^*}\ker(\varphi)=0$ under AC.
It is also known that, without AC, there are models of ZF in which some ...
- 15.3k
0
votes
0
answers
36
views
Do we need the axiom of choice when extracting a linearly independent spanning set from any spanning set of a separable Hilbert space? [duplicate]
Let $H$ be a separable Hilbert space and $\{v_n\}$ be some countable subset of $H$ such that its linear span is dense in $H$. $\{v_n\}$ need NOT be linearly independent.
(Here, linear independence of ...
- 7,829
1
vote
0
answers
18
views
Embedding a vector space in its bidual and the axiom of choice [duplicate]
Let $V$ be a vector space over some field $\mathbb F$. The dual space is defined as the set of all linear maps $f : V \to \mathbb F$ which is again a vector space over $\mathbb F$ (with pointwise ...
- 4,624
2
votes
0
answers
69
views
In ZF, can it be proven that a non-trivial real vector space has a non-zero linear functional?
I've been through a number of older similar threads and it does not seem clear that this question is answered anywhere. I know this can be proven without the full strength of the Axiom of Choice but ...
- 59
6
votes
2
answers
289
views
Characterization of basis in terms of universal property: axiom of choice
I wonder if the proof of the following statement requires the axiom of choice:
(Characterization of basis in terms of universal property) Let $V$ be a vector space, and let $S$ be a non-empty subset ...
- 361
0
votes
1
answer
147
views
Is the Axiom of Choice needed in this kind of reasoning?
I do not understand if the Axiom of Choice is needed in a certain kind of reasoning.
Example 1. Let $X,Y$ be two sets and $f:X\to Y$ be a function. Let $A,B$ be two subsets of $X$ such that $A\subset ...
- 234
1
vote
1
answer
254
views
Automorphisms of finite-dimensional vector spaces without the Axiom of Choice
Let $k$ be an infinite field (for example, $\mathbb{C}$), and let $V$ be a finite-dimensional vector space over $k$ in a ZF-model without the Axiom of Choice (AOC). (If I am not mistaken, a finitely ...
- 799
6
votes
1
answer
168
views
Classifying Vector Spaces without AC
Using the axiom of choice we can give a simple classification of all vector spaces over a given field $K$ up to isomorphism: Any $K$-vector-space $V$ is just isomorphic to $\bigoplus_{i\in B}K$ where $...
- 183
3
votes
1
answer
146
views
Characterizing Vector Spaces without (provable) Basis in ZF
Without the axiom of choice certain vector spaces cannot be proven to have a (hamel) basis in ZF alone and I am wondering whether there exists some criteria characterizing such spaces.
Here is what I ...
- 183
2
votes
1
answer
156
views
(Non)necessity of the axiom of choice in proving associativity of the tensor product
I am trying to prove the associativity of the tensor product $\otimes$ of vector spaces, along the line with the sketch discussed on a previously-asked question, but I got stuck on the last point.
...
- 406
8
votes
1
answer
266
views
Basis for $\mathbb{R}^\mathbb{N}$ implies axiom of choice?
Let $\mathbb{R}^\mathbb{N}$ denote the vector space over $\mathbb{R}$ of sequences of real numbers, with multiplication and addition defined by component. It's well-known that though the subspace $\...
- 7,080