Math, asked by targetexamcurrentaff, 7 months ago

The dimensions of a linear space does not depend on a choice of .......​

Answers

Answered by somavarapuneelima
0

Answer: READ THE BELOW ONE HPOE U GRT THE ANSWER

In mathematics, the dimension theorem for vector spaces states that all bases of a vector space have equally many elements. This number of elements may be finite or infinite (in the latter case, it is a cardinal number), and defines the dimension of the vector space.

Formally, the dimension theorem for vector spaces states that

Given a vector space V, any two bases have the same cardinality.

As a basis is a generating set that is linearly independent, the theorem is a consequence of the following theorem, which is also useful:

In a vector space V, if G is a generating set, and I is a linearly independent set, then the cardinality of I is not larger than the cardinality of G.

In particular if V is finitely generated, then all its bases are finite and have the same number of elements.

While the proof of the existence of a basis for any vector space in the general case requires Zorn's lemma and is in fact equivalent to the axiom of choice, the uniqueness of the cardinality of the basis requires only the ultrafilter lemma,[1] which is strictly weaker (the proof given below, however, assumes trichotomy, i.e., that all cardinal numbers are comparable, a statement which is also equivalent to the axiom of choice). The theorem can be generalized to arbitrary R-modules for rings R having invariant basis number.

In the finitely generated case the proof uses only elementary arguments of algebra, and does not require the axiom of choice nor its weaker variants.

Similar questions