91Ó°ÊÓ

In the realm of Linear Algebra, a spanning set plays a crucial role in understanding vector spaces. Essentially, a set of vectors spans a vector space if linear combinations of these vectors can generate every possible vector within that space. For a set of vectors to span a space like \(\mathbb{R}^3\), you would need three linearly independent vectors. Each vector would occupy one degree of freedom in the three-dimensional space.

In our exercise, by examining the matrix composed of the set \(S\), we transformed it into row echelon form. From this form, we observed that the resulting matrix has two non-zero rows, meaning it has a rank of 2. Since the rank must equal the dimension of the space (3 in this case) for the set to span \(\mathbb{R}^3\), we know \(S\) does not span the space.

• The condition for spanning is met when the rank of the matrix matches the dimension of the space.
• A rank less than the dimension indicates the vectors do not cover the entire space.
• This leads us to conclude the set \(S\) doesn't span \(\mathbb{R}^3\).

When considering sets of vectors, a key concept is linear independence. A set of vectors is said to be linearly independent if no vector in the set can be written as a linear combination of the others. In simpler terms, each vector adds a new dimension to the combinations possible and none can be formed by scaling and adding the others.

For our given exercise, the method of transforming the matrix through Gaussian elimination reveals that the rank is 2. This is less than the number of vectors (5 in this case) in set \(S\). Since there are more vectors than the dimension offered by the rank, it indicates that there is some dependence among the vectors. In fact, the solution reveals that three vectors (or more precisely, since one is the zero vector which is intrinsically dependent) are linearly dependent.

• Linear dependence arises when vectors are not all contributing to the span of the vector space.
• This can often be seen when certain vectors reduce to zeros in row reduction.
• The presence of the zero vector confirms linear dependence, as it cannot exist independently.

Gaussian elimination is a powerful algorithm used to solve systems of linear equations, find ranks of matrices, and determine linear independence and spanning sets. The process involves manipulating a matrix to reach row echelon form, allowing easy interpretation of the matrix properties.

In our problem, Gaussian elimination helps tell us whether the vectors in set \(S\) span \(\mathbb{R}^3\) or are linearly independent. Starting with the matrix formed from the vectors, the elimination involves:

• Swapping rows to position larger pivot elements first, enhancing numerical stability and ease of computation.
• Using row operations (adding, subtracting, scaling) to zero out elements below or above pivots, systematically simplifying the matrix.
• Identifying non-zero rows in the reduced matrix to determine the rank.
  
  Subtle but important, Gaussian elimination transformed our initial matrix into one with two significant rows, telling us the rank was 2. This indicates both the vectors' inability to span \(\mathbb{R}^3\) and their linear dependence.