Purchase Solution

Linearly Independent Subsets : Partially Ordered by Inclusion

Not what you're looking for?

Ask Custom Question

Let X be any vector space over the field F, let L be a linearly independent subset
of X, and A be the set of linearly independent subsets of X containing L.
Then A is partially ordered by inclusion - why does it follow?

Purchase this Solution
Solution Summary

Partially Ordered by Inclusion is investigated. Linearly independent subsets for partially ordered by inclusions are examined.

Solution Preview

Because inclusion is a partial order on ANY family of subsets.
You just have to check A subset A for all A, A ...

Purchase this Solution
Free BrainMass Quizzes
Exponential Expressions

In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

Graphs and Functions

This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.

Probability Quiz

Some questions on probability

Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

View More Free Quizzes
Related BrainMass Solutions

  • Vector space

    Decide for each of the following subsets if it is a subspace. Justify your answer by giving a proof or a counterexample: {see attachment} b) choose [a,b]=[ ]. Show that f(x) = sinx and g(x)=cosx are linearly independent. ?

  • Vecor Spaces and Linear Combinations

    d) Subsets of linearly independent sets are linearly independent. True.

  • Spanning sets and vectors

    If it is i) then by definition, x(k+1) is not a member of the spanning set of V (all members must be linearly independent of each other).

  • Zorn's Lemma

    Zorn's lemma states that: every non-empty partially ordered set in which every chain (i.e. totally ordered subset) has an upper bound contains at least one maximal element. Consider the set P of all subrings of R, containing 1 but not ½.

  • Minimal linearly dependent sets of columns

    Since all the subsets of 3 column vectors of B are linearly independent, it follows that the minimal linearly dependent set of columns of B is the set {c_1,c_2,c_3,c_4 }.

  • Lebesque Number and Connectivity

    Restate the lemma for the following special cases: (1) A covering by two open sets (2) The sets {Ui} are linearly ordered by inclusion Using the restated lemma for special case (1), prove that the unit n-sphere Sn, n ≥ 2, is simply connected.

  • Cyclic Vector

    Since , then and are linearly independent. Therefore, is a basis for . We are done. A cyclic vector is proven. The solution is detailed and well presented.

  • Discrete Math: Subsets and Elements

    Determine: (a) The number of subsets of S (b) The number of subsets of S with at most four elements (c) The number of ordered lists with elements chosen form S (with possible repetitions) (d) The number of ordered lists with nine elements

  • Sets and Functions : Sets and Set Inclusion

    95732 Sets and Functions : Sets and Set Inclusion Let U be the single-element set { 1 }. There are two subsets, the empty set φ and { 1 } itself.

View More Related Solutions