91Ó°ÊÓ

In the context of vector spaces, linear independence is a foundational concept. Consider a set of vectors \( \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} \). The set is linearly independent if no vector in the set can be represented as a linear combination of the others. This means the only solution to the equation:\[ b_1 \mathbf{v}_1 + b_2 \mathbf{v}_2 + \ldots + b_n \mathbf{v}_n = \mathbf{0} \]is when all coefficients are zero, that is, \( b_1 = b_2 = \cdots = b_n = 0 \).
This property is crucial; it ensures that each vector contributes uniquely to the vector space.
In our exercise, the original basis \( \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} \) is already known to be linearly independent.When each vector is multiplied by a nonzero scalar, such as \( \{c_1 \mathbf{v}_1, \ldots, c_n \mathbf{v}_n\} \), the set remains linearly independent because multiplying by a nonzero scalar does not introduce any dependencies among the vectors.

The concept of a spanning set is about covering the entire space. A set of vectors \( \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} \) spans a vector space \( V \) if every vector in \( V \) can be built or expressed as a linear combination of these vectors.
For instance, if you look at a spanning set in a 2D space, you need enough vectors to reach any point on the plane.
In our scenario, the original basis \( \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} \) is a spanning set for \( V \). When we consider the set \( \{c_1 \mathbf{v}_1, \ldots, c_n \mathbf{v}_n\} \), since each \( c_i \) is nonzero, we can rearrange any vector in \( V \) using these new vectors by solving \( \mathbf{v} = \frac{a_1}{c_1}(c_1 \mathbf{v}_1) + \cdots + \frac{a_n}{c_n}(c_n \mathbf{v}_n) \).
This confirms that our new set also spans \( V \), maintaining the completeness of the vector coverage across the space.

Scalar multiplication involves taking a vector and multiplying it by a number, known as a scalar. This operation produces a new vector that points in the same direction as the original, but its magnitude is scaled by the scalar factor.
For example, multiplying a vector \( \mathbf{v} \) by a scalar \( c \) results in \( c \mathbf{v} \). If \( c \) is greater than one, the vector becomes longer; if \( c \) is between zero and one, the vector shrinks. Crucially, multiplying a vector by a non-zero scalar does not change its direction in the vector space.
In problem-solving, scalar multiplication is instrumental in modifying bases of vector spaces. Our exercise used scalar multiplication to adjust an existing basis \( \{\mathbf{v}_1, \ldots, \mathbf{v}_n\} \) by scalars \( c_1, \ldots, c_n \), forming \( \{c_1 \mathbf{v}_1, \ldots, c_n \mathbf{v}_n\} \). This transformation does not affect linear independence or the span of the basis, meaning the new set continues to act as a basis, providing both independence and full coverage of the vector space.