91Ó°ÊÓ

Linear independence is a critical concept in linear algebra, especially when determining if a set of vectors can form a basis. When we say vectors are linearly independent, it means no vector in the set can be written as a linear combination of the others.
More formally, a set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \} \) is linearly independent if the only solution to the equation \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{0} \) is \( c_1 = c_2 = c_3 = 0 \). This result shows that there are no redundancies among the vectors, meaning they are all essential components of the space they span.
As illustrated in the exercise above, solving the system of equations confirms the vectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \), \( \mathbf{v}_2 = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \), and \( \mathbf{v}_3 = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \) are indeed linearly independent. The process involved expressing each coefficient \( c_1, c_2, \) and \( c_3 \) with zero to satisfy the equation, showing that no vector can be discarded without losing the span.

A linear combination involves creating a new vector by adding up scaled versions of given vectors. In simpler terms, it is combining vectors by multiplying each by a scalar and then adding the results.
Imagine you have grocery bags of different weights, and by picking certain combinations of these bags, you can reach a desired total weight. That's like forming a linear combination.
In our exercise, the vector \( \begin{bmatrix} s \ t \ t \end{bmatrix} \) is expressed as a linear combination of the vectors \( \mathbf{v}_1, \mathbf{v}_2, \) and \( \mathbf{v}_3 \) by using coefficients \( s, \) \( (t-s), \) and \( s \), respectively. This means any vector where the second and third entries are equal in the set \( H \) can be perfectly described using the given vectors. Linear combinations not only help describe vectors but also illustrate how elements work within a vector space.

A vector space is a fundamental structure in linear algebra. It is like a container where vectors live and interact.
Vectors within this space can be added together or multiplied by scalars, and the space itself must follow certain rules, such as closure under addition and scalar multiplication, and the existence of a zero vector.
The exercise discusses a subset called \( H \) in \( \mathbb{R}^3 \), defined by vectors whose second and third elements are the same. This subset is also a vector space since:

• It is non-empty, containing at least the zero vector \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \).
• It is closed under addition, meaning if you take any two vectors from \( H \), their sum is also in \( H \).
• It is closed under scalar multiplication. Multiplying any vector in \( H \) by a scalar gives another vector in \( H \).

Understanding vector spaces helps to grasp how vectors can be manipulated and what new combinations or transformations are possible within a given space.