Question

If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) is independent in \(\mathbb{R}^{n},\) and if \(\mathbf{y}\) is not in \(\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\},\) show that \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}, \mathbf{y}\right\\}\) is independent.

Short answer

\(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}, \mathbf{y}\}\) is independent as \(\mathbf{y}\) is outside the span of the other vectors.

Step-by-step solution

Definition of Linear Independence

A set of vectors \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m\}\) is called linearly independent if the only solution to the equation \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_m\mathbf{v}_m = \mathbf{0}\) is \(c_1 = c_2 = \ldots = c_m = 0\). We know that the set \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\}\) is linearly independent.

Constructing the Equation with \(\mathbf{y}\)

Add the vector \(\mathbf{y}\) to the set and consider the equation \(c_1\mathbf{x}_{1} + c_2\mathbf{x}_{2} + \ldots + c_k\mathbf{x}_{k} + c_{k+1}\mathbf{y} = \mathbf{0}\). We need to show that the only solution is \(c_1 = c_2 = \ldots = c_k = c_{k+1} = 0\).

Assume Solution and Isolate \(\mathbf{y}\)

Suppose there is a solution where \(c_1, c_2, \ldots, c_{k+1}\) are not all zero. Also assume, without loss of generality, that \(c_{k+1} eq 0\). Then we can rearrange the equation to \(\mathbf{y} = -\frac{c_1}{c_{k+1}}\mathbf{x}_{1} - \frac{c_2}{c_{k+1}}\mathbf{x}_{2} - \ldots - \frac{c_k}{c_{k+1}}\mathbf{x}_{k}\).

Contradiction with \(\mathbf{y}\) Outside Span

The expression in Step 3 shows that \(\mathbf{y}\) is a linear combination of \(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\), contradicting the given that \(\mathbf{y}\) is not in \(\operatorname{span}\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\}\).

Conclusion of Independence

Since assuming \(c_{k+1} eq 0\) leads to a contradiction, it must be that \(c_{k+1} = 0\). Consequently, all \(c_i = 0\) for \(i = 1, 2, \ldots, k\) which means the set \(\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}, \mathbf{y}\}\) is indeed independent.

Key concepts

Span of Vectors

The concept of the span of vectors is foundational in linear algebra. When we talk about the "span" of a set of vectors, we refer to all possible vectors that can be created through linear combinations of these vectors.

The span of vectors tells us everything about how these vectors can "cover" a vector space. Consider a set \(\{\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\}\). The span includes all vectors that you can get by multiplying each vector by a scalar and adding the results. If you take, say two vectors in 2D space, their span will form either a line or fill the entire 2D plane, depending on whether they are parallel or not.

Now, if a vector \( \mathbf{y} \) is outside this span, it means no linear combination of \( \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k \) can produce \( \mathbf{y} \). Hence, \( \mathbf{y} \) is a new direction not captured by the span of the current set, and it becomes vital in establishing the independence of an augmented set of vectors.

Vector Spaces

Vector spaces are a central object of study in linear algebra. They encapsulate the notion of vectors behaving in certain structured ways. A vector space over a field, typically the real numbers \(\mathbb{R}\), is constituted by a set of vectors that satisfy specific rules under addition and scalar multiplication.

Some rules include:

• The sum of two vectors in the space must also be in the space (closure under addition).
• Multiplying a vector by a scalar must result in another vector in the space (closure under scalar multiplication).
• Vectors can be added in any order (commutativity), and scalar multiplication is associative.

In terms of properties, vector spaces come equipped with a zero vector, and for each vector, there exists an inverse such that their sum produces the zero vector.

This framework is broad, meaning within any vector space; vectors can span subspaces or behave in structured ways that are useful in understanding linear systems, dimensions, and basis. Understanding vector spaces is key to grasping other abstract concepts in math.

Linear Combinations

A linear combination involves adding together scalar multiples of vectors to yield new vectors. Suppose you have vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_m \) and scalars \( c_1, c_2, \ldots, c_m \). A linear combination of these vectors can be expressed as \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \ldots + c_m\mathbf{v}_m \).

This concept marks the building block of linear algebra, as virtually all operations in vector spaces can be reduced to linear combinations.

• They describe vector spans, as in which vectors can form others.
• They help determine the rank of a matrix and solutions to linear equations.

In the context of this concept, when a set of vectors is linearly independent, no vector in the set can be written as a linear combination of the other vectors. This understanding directly levels up the solution of any exercise involving vector independence, as it delves into determining whether a vector can contribute a new vector combination that expands the span of a given set.