All Questions
Tagged with axiom-of-choice linear-algebra
22
questions
7
votes
1
answer
873
views
Logical strength of a statement about vector spaces
[Apologies if this is a really trivial question, I know virtually nothing about set theory, and the following came up while preparing undergraduate linear algebra lectures.]
I'm asking about the ...
- 36.5k
16
votes
1
answer
876
views
Examples of vector spaces with bases of different cardinalities
In this question Sizes of bases of vector spaces without the axiom of choice it is said that "It is consistent [with ZF] that there are vector spaces that have two bases with completely different ...
- 163
0
votes
0
answers
180
views
Eigenbases without the Axiom of Choice
I understand that in ZF set theory without the Axiom of Choice (AC), it is consistent to have models in which there exist vector spaces over some (unspecified) field $k$ without a basis.
So in ...
- 4,503
8
votes
1
answer
1k
views
If a vector space has a basis then its dual vector space has a basis
Consider the following statement:
If a vector space has a basis then its dual vector space also has a basis.
It is not an axiom of ZF. It clearly follows from the Axiom of Choice. But it is also ...
- 4,096
43
votes
2
answers
2k
views
Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?
If $V \hookrightarrow W$ and $W \hookrightarrow V$ are injective linear maps, then is there an isomorphism $V \cong W$?
If we assume the axiom of choice, the answer is yes: use the fact that every ...
- 62.6k
7
votes
1
answer
476
views
Exterior powers and choice
Under the assumption that any vector space has a basis (so under the assumption of the axiom of choice), we can prove the following algebraic statements :
1) If $\varphi:V\to W$ is an injective ...
- 995
6
votes
1
answer
394
views
Measure of rational hyperplanes of $\mathbb{R}$
Let's view $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and pick some basis $(v_\alpha)_{0 \leq \alpha < \mathfrak{c}}$ of it. We can then consider the subspace $L$ spanned by $(v_\alpha)_{0 &...
- 4,501
11
votes
1
answer
761
views
Axiom of choice and algebraic tensor product
The first part of the question was asked on Math-stackexchange.
Let $V$, and $W$ be vector spaces. By the universal property of the tensor product,
there is a canonical map from $V^*\otimes W^*$ ...
- 995
10
votes
1
answer
957
views
Relation between the Axiom of Choice and a the existence of a hyperplane not containing a vector
In a lot of problems in linear algebra one uses the existence, for each $E$ vector space over a field $k$, and each $x\in E$, of a Hyperplane $H$ such that $E=k\cdot x \oplus H$ (Let us denote $\...
- 179
14
votes
0
answers
762
views
Cardinality vs. isomorphism type of vector spaces without choice
One of the classical uses of the existence of bases of vector spaces (which is equivalent to the axiom of choice) is the following theorem:
If $V$ is an infinite vector space over a field $F$, and $...
- 38.5k
13
votes
1
answer
938
views
Axiom(s) of choice and bases of vector spaces
I'm not sure this question is more suitable for MO or for MSE, so feel free to move it to MSE if necessary.
I work here in ZF theory. Consider the following statements:
$(C)$ Axiom of choice: for ...
- 1,661
16
votes
1
answer
3k
views
How to construct a basis for the dual space of an infinite dimensional vector space?
Let $V$ be an infinite-dimensional vector space over a field $K$. Then it is known that $\dim V < \dim V^*$. More precisely, by a result attributed to Kaplansky and Erdos, we have $\dim V^* = |K|^{\...
- 2,791
12
votes
5
answers
1k
views
Does k(X) have a k-basis for every set X, without AC?
This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?.
For any field $k$, the field $k(x)$ of rational functions in one variable has an ...
- 34.4k
7
votes
1
answer
353
views
Sets of cardinalities of bases without choice
For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it ...
- 20.7k
19
votes
1
answer
474
views
Linear maps between arbitrarily chosen vectors of vector spaces $V$ and $W$
I recently came across this question:
Is the axiom of choice needed to prove the following statement:
Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. ...
- 191