Understanding the Basis and Dimension of a Vector Space Generated by {-1 0, 2 5, 23 3 -2 1}
In the realm of linear algebra and vector spaces, the concepts of basis and dimension are fundamental to understanding the structure and properties of a given set of vectors. In this article, we will explore these concepts through a specific example involving the set of vectors {(-1, 0), (2, 5, 23, 3), (3, -2, 1)}. This article will guide you step-by-step through the process of determining whether this set generates a basis for a vector space and what its dimension is.
Approach 1: Using Zero Components
To start, we will use the presence of zero components to our advantage. The first vector, (-1, 0), has a zero in the second component, implying that we have at least a two-dimensional subspace. Here is the step-by-step process:
Select the vector with a zero component, which is (-1, 0). This ensures we have a two-dimensional subspace. Choose a second vector arbitrarily. For simplicity, let’s select the vector (2, 5, 23, 3). Identify the zero component of the first vector (which is the second component in this case) and the corresponding component of the second vector (which is 5). Multiply the second vector by -1 to make the corresponding component of the second vector equal to the component of the first vector (0). This results in the vector (-2, -5, -23, -3). Calculate the difference between -3 and 1, which is 4. Multiply the first vector by 2 to make the third component of the first vector equal to 4. This transforms the vector to (-2, 0, 4, 0). Add the first transformed vector (-2, 0, 4, 0) to the second vector (-2, -5, -23, -3) to get (-4, -5, -19, -3). Check if this new vector is equal to the original third vector (3, -2, 1). Since it is not, the first two vectors are linearly independent with the third vector.Since the three vectors are linearly independent, they form a basis for the vector space, and the dimension of this space is 3.
Approach 2: Using Matrix Row Reduction
A second method involves constructing a matrix with the given vectors as columns, row reducing it, and identifying the vectors with pivotal 1’s. Here are the steps:
Create a matrix with the vectors as columns:( -1 2 3 0 5 -2 23 3 -2 3 1 1 )Row reduce the matrix:
( 1 0 0 0 1 0 0 0 1 0 0 0 )Identify the pivot columns. In this case, all columns are pivot columns because the matrix is already in reduced row echelon form. Therefore, all three original vectors form a basis for the vector space, and the dimension is 3.
This method confirms that the three vectors are linearly independent and span the vector space, making them a basis for the space.
Why This Works
The reason both methods yield the same result is that row reduction in matrices allows us to identify linearly independent vectors. Specifically, the vectors corresponding to the pivot columns in the reduced row echelon form are linearly independent and span the same space as the original vectors. Thus, the number of pivot columns (or the number of linearly independent vectors) is the dimension of the vector space.
Conclusion
In summary, for the given set of vectors {(-1, 0), (2, 5, 23, 3), (3, -2, 1)}, the dimension of the vector space they generate is 3. This can be determined by either using the presence of zero components or row reduction of the matrix formed by the vectors.