Question

Suppose that \(V\) and \(W\) are vector spaces over \(F, V\) is finite dimensional, and \(\left\\{\alpha_{i}\right\\}_{i=1}^{k}\) is a linearly independent family of elements of \(V\). In addition, let \(\beta_{1}, \ldots, \beta_{k}\) be arbitrary elements of \(W\). Show that there exists an \(F\) linear map \(\rho: V \rightarrow W\) such that \(\rho\left(\alpha_{i}\right)=\beta_{i}\) for \(i=1, \ldots, k\).

Short answer

Question: Given two vector spaces \(V\) and \(W\) over a field \(F\), and a linearly independent family of elements \(\left\\{ \alpha_{i} \right\\}_{i=1}^{k}\) in \(V\), and arbitrary elements \(\beta_{1}, ..., \beta_{k}\) of \(W\), show that there exists a linear map \(\rho: V \rightarrow W\) such that \(\rho\left( \alpha_{i} \right) = \beta_{i}\) for \(i = 1, ..., k\). Answer: The linear map \(\rho\) exists because we can extend the linearly independent family \(\left\\{ \alpha_{i} \right\\}_{i=1}^{k}\) to a basis of \(V\), and then define \(\rho\) such that \(\rho\left( \alpha_{i} \right) = \beta_{i}\) for \(i = 1, ..., k\), and \(\rho\left( \alpha_{i} \right) = 0_W\) for \(i = k+1, ..., n\), where \(n\) is the dimension of \(V\) and \(0_W\) is the zero vector in \(W\).

Step-by-step solution

Extend the linearly independent family to a basis of V

As \(V\) is a finite-dimensional vector space with a linearly independent family \(\left\\{ \alpha_{i} \right\\}_{i=1}^{k}\), we can extend this family to a basis of \(V\). Let \(\left\\{ \alpha_{i} \right\\}_{i=1}^{n}\) be a basis of \(V\) with \(n \geq k\), where the first \(k\) elements are the given linearly independent family.

Define the linear map \(\rho: V \rightarrow W\)

Now, define the linear map \(\rho: V \rightarrow W\) by assigning the values \(\rho\left( \alpha_{i} \right) = \beta_{i}\) for \(i = 1, ..., k\), and for \(i = k+1, ..., n\), let \(\rho\left( \alpha_{i} \right) = 0_W\), where \(0_W\) is the zero vector in \(W\). Since the map is well-defined on a basis of \(V\), it can be uniquely extended to the entire vector space \(V\).

Show that \(\rho\) satisfies the conditions of the problem

To show that the linear map \(\rho\) satisfies the conditions of the problem, we need to prove that \(\rho\left( \alpha_{i} \right) = \beta_{i}\) for \(i = 1, ..., k\). By the definition of \(\rho\), this condition is true for the first \(k\) elements. Therefore, there exists a linear map \(\rho: V \rightarrow W\) such that \(\rho\left( \alpha_{i} \right) = \beta_{i}\) for \(i = 1, ..., k\), completing the proof.

Key concepts

Vector Spaces

In the world of linear algebra, a vector space is one of the fundamental structures we encounter. Simply put, a vector space is a collection of objects called vectors, which can be added together and multiplied by scalars (numbers from a given field, like real numbers, \( \mathbb{R} \) ). Think of it as a playground where these vectors live and interact according to specific rules.

For example, we require that vector addition is commutative and associative, and that scalar multiplication is compatible with field multiplication. Also, there must exist a zero vector in the space, which when added to any vector, does not change it. This makes the subject very user-friendly and easy to manipulate for various applications, ranging from physics to computer science. In the exercise provided, the vector spaces \( V \) and \( W \) over a field \( F \) behave under these rules, allowing for the linear mapping concept to be explored.

Linear Independence

The concept of linear independence is central to understanding the structure of vector spaces and it comes up often when we discuss the strength and capabilities of a set of vectors. A set of vectors in a vector space is considered 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 brings something unique to the table, adding new dimensions or directions that we could explore. This quality makes them extremely valuable, as they can form a sturdy foundation for the vector space. In the context of our exercise, the set \( \{ \alpha_i \}_{i=1}^{k} \) is given as linearly independent, ensuring that each \( \alpha_i \) contributes to the space without redundancy. This unique property is fundamental when extending this set to a full basis of the vector space \( V \) and constructing our desired linear map.

Basis Extension

Building off the concept of linear independence is the notion of a basis extension. What if we start with a set of linearly independent vectors, but want to cover the entire vector space? We perform a basis extension. A basis is a set of linearly independent vectors that spans the whole vector space, meaning that any vector in the space can be expressed as a combination of the basis vectors.

To extend to a basis, we add more vectors to our set, being careful not to disturb its linear independence. The extension process maintains the independent and spanning properties—like completing a puzzle with just the right pieces. In our example, the linearly independent family \( \{ \alpha_i \}_{i=1}^{k} \) is extended to form a complete basis for \( V \) by adding additional vectors until the space is fully spanned. This sets the stage to defining our function \( \rho \) with precision across the whole space.

Linear Algebra

Finally, wrapping up all these concepts together is the overarching field of linear algebra. It's the branch of mathematics dealing with vectors, vector spaces, linear mappings, and systems of linear equations. It's a powerful area of study because it provides the tools and language to describe lines, planes, and higher dimensional analogs.

Not only is linear algebra essential for understanding theoretical perspectives, but it's also practical. Its principles are instrumental in areas like computer graphics, engineering, and data science. In tackling the exercise, we engage with the essence of linear algebra by using its principles to construct a specific linear map \( \rho \) that meets certain criteria. We do this within the framework of vector spaces, harnessing the power of linear independence and basis extension to find our solution.