Problem 24 - Differential Equations and Linear Algebra

Question

Let \(V\) be a vector space with basis \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\}\) and suppose \(T: V \rightarrow W\) is a linear transformation such that \(T\left(\mathbf{v}_{i}\right)=\mathbf{0}\) for each \(i=1,2, \ldots, k .\) Prove that \(T\) is the zero transformation; that is, \(T(\mathbf{v})=\mathbf{0}\) for each \(\mathbf{v} \in V.\)

Step-by-Step Solution

Answer

Let any arbitrary vector \(\mathbf{v} \in V\) be given with the basis \(\left\\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\right\\}\) and linear transformation T. Write \(\mathbf{v}\) as a linear combination of the basis vectors, apply the linear transformation using the linearity property, and use the given information \(T(\mathbf{v}_i) = \mathbf{0}\) for each \(i = 1, 2, \ldots, k\) to simplify the equation. The result is that \(T(\mathbf{v}) = \mathbf{0}\), proving that T is the zero transformation.

Step 1: Write an arbitrary vector in terms of the basis

Let's consider any arbitrary vector \(\mathbf{v} \in V\). Since \(\left\\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\right\\}\) is a basis for V, we can express any vector \(\mathbf{v}\) as a linear combination of the basis vectors. That is, \[\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k\] where \(c_1, c_2, \ldots, c_k\) are scalars.

Step 2: Apply the linear transformation T to the arbitrary vector

Now let's apply the linear transformation T to the vector \(\mathbf{v}\). By the definition of linearity, we have \[T(\mathbf{v})=T(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k)\] We can now use the linearity property of T (additivity and homogeneity) to distribute it over the linear combination. This gives us \[T(\mathbf{v})= c_1T(\mathbf{v}_1) + c_2T(\mathbf{v}_2) + \cdots + c_kT(\mathbf{v}_k)\]

Step 3: Use the given information to simplify

According to the problem statement, we are given that \(T(\mathbf{v}_i) = \mathbf{0}\) for each \(i = 1, 2, \ldots, k\). We can substitute this information into the equation we derived in Step 2 to get \[T(\mathbf{v}) = c_1\mathbf{0} + c_2\mathbf{0} + \cdots + c_k\mathbf{0}\]

Step 4: Simplify the equation and conclude

Finally, we can simplify the equation above as sum of zero vectors is also a zero vector. Thus, \[T(\mathbf{v}) = \mathbf{0}\] Since the vector v was arbitrary, this proves that the linear transformation T applied to any vector in V will result in the zero vector. Hence, T is the zero transformation.

Key Concepts

Linear Transformation

Linear transformation is a concept that plays a central role in understanding how vectors can be manipulated and changed within vector spaces. Informally, think of it as a function that takes vectors from one space and 'transforms' them into vectors in another space, while preserving the operations of vector addition and scalar multiplication. This means that if you have two vectors and you add them together, then transform the result, it's the same as transforming each vector first and then adding them together in the new space.

Similarly, if you multiply a vector by a scalar (a number) and then transform it, it's the same as transforming the vector first and then multiplying by the scalar in the new space. This preservation is formally expressed through two main properties: the additivity and the homogeneity of scalars. In our exercise example, we rely on these properties to show that a particular linear transformation is the zero transformation, which means it sends every vector to the zero vector of the target space, effectively collapsing the entire space into a single point.

Vector Space

A vector space can be thought of as the stage on which vectors can perform; it's a mathematical structure that allows for the addition of vectors and the multiplication of vectors by scalars. Imagine a vast field where you can move in any direction and by any distance—that's what a vector space is like for vectors. In this space, there are a few rules that must be followed, like being able to add any two vectors together to get another vector in the same space, and being able to multiply any vector by a number (scalar) and still have a vector in the same space.

In the exercise, the vector space is denoted by 'V', and it's the domain of our linear transformation 'T'. The vectors in space 'V' can be stretched, flipped, rotated, or even squished to a point, as is the case with our zero transformation. To demonstrate that 'T' is indeed the zero transformation, we use the fact that no matter what operations we perform within the vector space, the transformation respects these operations but sends every vector to the zero vector, showing the power of vector space rules at play.

Basis of Vector Space

The basis of a vector space is like a set of building blocks from which you can construct any vector in that space. It's the smallest set of vectors that, when combined in various ways (added together and multiplied by scalars), can describe every vector in the space. You can think of it as a recipe where you have a list of ingredients (the basis vectors), and by mixing these ingredients in different amounts, you can cook up any vector in the space.

In our exercise, the basis vectors are denoted by \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\), and every other vector in the space is a unique combination of these. When showing that our linear transformation 'T' is the zero transformation, we only need to check what 'T' does to these basis vectors because, by the rules of linear transformations, this determines what 'T' does to every vector in the space. This pivotal role of the basis simplifies the proof since, if each basis vector is mapped to zero, so is any linear combination of them, and thus every vector in 'V' under 'T' leads us to the zero vector in 'W'.