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Let \(S=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) be a linearly independent subset of a vector space \(\mathrm{V}\) over the field \(Z_{2}\). How many vectors are there in \(\operatorname{span}(S)\) ? Justify your answer.

Short Answer

Expert verified
There are \(2^n - 1\) vectors in the span of S.

Step by step solution

01

Understand the Concepts

In this problem, we are working with a linearly independent subset within a vector space over the field Z_2. Linearly independent means that no vector in the set S can be expressed as a linear combination of the others, and Z_2 is a field with only two elements: 0 and 1.
02

Linear Combinations

Since the set S is linearly independent, every vector in the span of S can be written as a unique linear combination of the vectors in S. A linear combination of the vectors in S is of the form: \(c_1u_1 + c_2u_2 + \cdots + c_nu_n\) where \(c_1, c_2, \ldots , c_n\) are scalar coefficients from the field Z_2.
03

Determine the Possible Coefficient Values

Since we are working with the field Z_2, each coefficient can only have two possible values: 0 or 1. That means for each of the n vectors in S, there are 2 possible values for the scalar coefficients.
04

Calculate the Total Number of Vectors in the Span of S

To find the total number of vectors in the span of S, we can multiply the number of possible values for each coefficient: Number of Vectors = \(2^n\) However, one of these combinations corresponds to the zero vector (when all coefficients are 0), which we need to subtract from the total count: Number of Vectors in the span of S = \(2^n - 1\) So, there are \(2^n - 1\) vectors in the span of S.

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