Understanding the Dimension of a Given Vector Space

In this article, we will explore the concept of dimension in the context of a vector space defined by a set of vectors. Specifically, we will analyze the given vector space ( L {111-1 010-1 1010 0-101} ) and determine its dimension. We will use a systematic approach involving linear algebra principles to achieve this.

Initial Setup and Redundancy Check

Let's begin by noting the given vectors:

a 111 b 010-1 c 1010 d 0-101

By examining the vectors, we can simplify this set. Let's start with the equations:

a c

b -d

This implies that vectors c and d can be expressed in terms of a and b. Therefore, the vectors c and d are redundant and can be discarded. This simplifies our problem to the following set:

[a c] [111 1010]

Analyzing the Dimension

The next step is to determine the maximum dimension of the vector space ( L ).

Given the set of vectors, we have:

a 111 b 010-1 c 1010 d 0-101

Notably, the relationship a c indicates that if vector c is proportional to vector a, then the dimension is 1. However, if c is not proportional to a, then the dimension is 2. This leads us to conclude that the maximum dimension of ( L ) is 2.

Typical Element Form

For a typical element in ( L ), consider the following form:

x a b a - b 1010 010-1

This form helps us to explore the relationship between the vectors in a more detailed manner. Specifically, since a c and b -d, we can express the set ( L ) as:

L {a b a - b : a, b in mathbb{R}} {1010 010-1 : a, b in mathbb{R}} text{span} {1010 010-1}

This indicates that a basis for ( L ) could be {1010 010-1}}.

Conclusion and Verification

The vectors 1010 and 010-1 are linearly independent (not multiples of each other). Thus, they can form a basis for ( L ). Therefore, the dimension of ( L ) is 2.

By summarizing, the basis for ( L ) is {1010 010-1}} and the dimension of ( L ) is 2. This conclusion is supported by the process of solving the system of equations:

a c

b -d

And the general form of a vector in ( L ) being expressed as:

x abcd c -d c d c 0 0 -d 0 d 1010 c 0-101 d

Thus, the basis vectors i 1010 and j 010-1 form a basis, and we conclude that

dim L 2