Question

Let \(T: V \rightarrow W\) be a linear transformation and let \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\) denote vectors in \(V\). a. If \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{n}\right)\right\\}\) is linearly independent, show that \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) is also independent. b. Find \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) for which the converse of part (a) is false.

Short answer

A: If the image is independent, the original is independent. B: Use \(T(x, y) = (x, 0)\).

Step-by-step solution

Understand the Definitions

A set of vectors \(\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\}\) in a vector space is linearly independent if the only solution to \(a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + \ldots + a_{n}\mathbf{v}_{n} = \mathbf{0}\) is when all coefficients \(a_{i}\) are zero. A linear transformation \(T: V \rightarrow W\) maps vectors from one vector space to another.

Prove Part (a)

Assume that \(\{T(\mathbf{v}_{1}), \ldots, T(\mathbf{v}_{n})\}\) is linearly independent. This means if \(a_{1}T(\mathbf{v}_{1}) + a_{2}T(\mathbf{v}_{2}) + \ldots + a_{n}T(\mathbf{v}_{n}) = \mathbf{0}\), then \(a_{1} = a_{2} = \ldots = a_{n} = 0\). Since \(T\) is a linear transformation, \(T(a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + \ldots + a_{n}\mathbf{v}_{n}) = a_{1}T(\mathbf{v}_{1}) + a_{2}T(\mathbf{v}_{2}) + \ldots + a_{n}T(\mathbf{v}_{n})\). Thus, \(T(a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + \ldots + a_{n}\mathbf{v}_{n}) = \mathbf{0}\) implies \(a_{1} = a_{2} = \ldots = a_{n} = 0\). Hence, \(\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\}\) is linearly independent.

Approach Part (b)

To show that the converse is false, we need an example where \(\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\}\) is linearly independent, but \(\{T(\mathbf{v}_{1}), \ldots, T(\mathbf{v}_{n})\}\) is not. Consider any linear transformation where the image dimension is lower than the domain.

Choose a Transformation for Part (b)

Take \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) defined by \(T(x, y) = (x, 0)\). This projects any vector onto the x-axis.

Analyze the Chosen Transformation

The vectors \((1, 0)\) and \((0, 1)\) in \(\mathbb{R}^{2}\) are linearly independent. Under \(T\), they map to \((1, 0)\) and \((0, 0)\), which are not independent. Thus, the converse is false for this transformation.

Key concepts

Linearly Independent Vectors

When talking about vector spaces, understanding the concept of linear independence is crucial. A set of vectors is said to be linearly independent if no vector in the set can be written as a combination of the others. This means that the only solution to the equation

• \(a_{1}\mathbf{v}_{1} + a_{2}\mathbf{v}_{2} + \ldots + a_{n}\mathbf{v}_{n} = \mathbf{0}\)

is when all the coefficients \(a_{i}\) are zero.

Linear independence is a fundamental concept because it helps in determining the span and dimension of vector spaces. If the transformation of vectors under consideration is linearly independent, it means these vectors do not lie on the same plane or line. They retain their individual "direction" or "information."

For instance, in a 2-dimensional plane, two vectors like

• \((1,0)\) and \((0,1)\)

are linearly independent because they form a basis and cover the entire plane without overlapping each other. In practical terms, checking the linear independence of a set of vectors involves solving a system of simultaneous linear equations to check if the only solution is the trivial one (all zero coefficients).

Vector Spaces

A vector space is an essential core concept in mathematics and supports vector operations like addition and scalar multiplication. When we talk about vector spaces, we're describing a collection of vectors that can be neatly combined and manipulated through these operations while remaining within the same space. In technical terms, a vector space over a field \(F\) is a set \(V\) along with two operations that allow for vector addition and scalar multiplication.

Each vector space has independent basis vectors that span the space. The dimension of a vector space is determined by these basis vectors, which are minimal independent sets that can construct every other vector in the space through linear combinations.

• To visualize a vector space, think of a flat, infinite plane in three-dimensional space. Any vector in this sub-plane can be represented by combining basis vectors of the plane.
• In contrast, three basis vectors in full three-dimensional space can span the entire volume.

A thorough grasp of vector spaces enables us to better understand projections, transformations, and many other operations that vectors are subject to.

Projection Mapping

Projection mapping is a specific type of linear transformation. It reduces the dimension of a vector while preserving some of its original information. Think about projecting an object’s shadow. The shadow retains the object's outermost boundary but loses information about its depth.

Mathematically, a projection is a function \(P\) from a vector space to itself such that \(P^2 = P\). This means applying the projection twice leaves the vector unchanged from the first application. For example, the transformation \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) defined by \(T(x, y) = (x, 0)\) is a projection.

• This example "flattens" any vector onto the x-axis, losing the y-component.
• The transformation is useful in deriving conclusions and simplifying problems by focusing on the most critical dimensions.

Projections are not always linearly independent. Hence, as interpreted from the problem where \((1,0)\) remained independent whereas \((0,1)\) mapped to zero, we can observe that projections can sometimes lead to a loss of independent dimensions.