Problem 24 - Differential Equations and Linear Algebra

Question

Let \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) be a basis for the vector space \(V,\) and suppose that \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\) are the linear transformations satisfying $$\begin{array}{ll} T_{1}\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}+\mathbf{v}_{2}, & T_{1}\left(\mathbf{v}_{2}\right)=\mathbf{v}_{1}-\mathbf{v}_{2} \\ T_{2}\left(\mathbf{v}_{1}\right)=\frac{1}{2}\left(\mathbf{v}_{1}+\mathbf{v}_{2}\right), & T_{2}\left(\mathbf{v}_{2}\right)=\frac{1}{2}\left(\mathbf{v}_{1}-\mathbf{v}_{2}\right) \end{array}$$ Find \(\left(T_{1} T_{2}\right)(\mathbf{v})\) and \(\left(T_{2} T_{1}\right)(\mathbf{v})\) for an arbitrary vector in \(V\) and show that \(T_{2}=T_{1}^{-1}\)

Step-by-Step Solution

Answer

We found that both \( T_2T_1(\mathbf{v}) \) and \( T_1T_2(\mathbf{v}) \) yield the original vector, \( \mathbf{v} \), indicating that operating \( T_2 \) after \( T_1 \) or vice versa yields the identity transformation. Thus, we can conclude that \( T_2 \) is the inverse of \( T_1\), i.e., \( T_1^{-1} = T_2 \).

Step 1: Express the Arbritrary Vector

Any vector \( \mathbf{v} \) in the vector space V can be expressed as a linear combination of the basis vectors as follows: \[ \mathbf{v} = a\mathbf{v}_1 + b\mathbf{v}_2 \]

Step 2: Compute \( T_1T_2(\mathbf{v})\)

First, let's compute \( T_2(\mathbf{v})\): \[ T_2(\mathbf{v}) = aT_2(\mathbf{v}_1) + bT_2(\mathbf{v}_2) = \frac{a}{2}(\mathbf{v}_1+\mathbf{v}_2) + \frac{b}{2}(\mathbf{v}_1 - \mathbf{v}_2) = \frac{a+b}{2}\mathbf{v}_1 + \frac{a-b}{2}\mathbf{v}_2\] Then compute \(T_1(T_2(\mathbf{v}))\) which is equivalent to \( T_1T_2(\mathbf{v})\): \[ T_1T_2(\mathbf{v}) = T_1\left(\frac{a+b}{2}\mathbf{v}_1 + \frac{a-b}{2}\mathbf{v}_2\right) = \frac{a+b}{2}T_1(\mathbf{v}_1) + \frac{a-b}{2}T_1(\mathbf{v}_2) = \frac{a+b}{2}(\mathbf{v}_1 + \mathbf{v}_2) + \frac{a-b}{2}(\mathbf{v}_1 - \mathbf{v}_2) = a\mathbf{v}_1 + b\mathbf{v}_2 = \mathbf{v} \]

Step 3: Compute \( T_2T_1(\mathbf{v}) \)

Similarly, let's compute \( T_1(\mathbf{v}) \): \[ T_1(\mathbf{v}) = aT_1(\mathbf{v}_1) + bT_1(\mathbf{v}_2) = a(\mathbf{v}_1 + \mathbf{v}_2) + b(\mathbf{v}_1 - \mathbf{v}_2) = (a+b)\mathbf{v}_1 + (a-b)\mathbf{v}_2 \] Then compute \(T_2(T_1(\mathbf{v}))\), which is equivalent to \(T_2T_1(\mathbf{v}) \): \[ T_2T_1(\mathbf{v}) = T_2((a+b)\mathbf{v}_1 + (a-b)\mathbf{v}_2) = \frac{a+b}{2}T_2(\mathbf{v}_1) + \frac{a-b}{2}T_2(\mathbf{v}_2) = \frac{a+b}{2}\frac{1}{2}(\mathbf{v}_1 + \mathbf{v}_2) + \frac{a-b}{2}\frac{1}{2}(\mathbf{v}_1 - \mathbf{v}_2) = a\mathbf{v}_1 + b\mathbf{v}_2 = \mathbf{v} \]

Step 4: Conclude \(T_1^{-1} = T_2\)

As you can see, both \( T_2T_1(\mathbf{v}) \) and \( T_1T_2(\mathbf{v}) \) yield the original vector, \( \mathbf{v} \). This shows that operation of \( T_2 \) after \( T_1 \) or it's reverse yields the identity transformation. Thus, we can conclude that \( T_2 \) is the inverse of \( T_1\), i.e., \( T_1^{-1} = T_2 \).

Key Concepts

Basis Vectors

In the realm of linear algebra and vector spaces, basis vectors are fundamental building blocks. A set of basis vectors for a vector space is a group of vectors that are linearly independent and span the space. This means that any vector in the space can be expressed as a unique linear combination of the basis vectors. For example, in our exercise, \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) form a basis for the vector space \( V \). This implies that any vector \( \mathbf{v} \) in \( V \) can be written as \( a\mathbf{v}_1 + b\mathbf{v}_2 \), where \( a \) and \( b \) are scalars. This concept is crucial because it allows us to understand and describe the entire space using just a few vectors. By knowing the basis, we gain the ability to easily perform transformations and other operations on the vector space.

Inverse Transformations

An inverse transformation is a key concept in linear algebra that refers to reversing the effect of a given transformation. If \( T \) is a linear transformation from a vector space \( V \) to itself, its inverse \( T^{-1} \) is a transformation such that when you apply \( T \) and then \( T^{-1} \), you get back to your starting point; mathematically, \( T(T^{-1}(\mathbf{v})) = T^{-1}(T(\mathbf{v})) = \mathbf{v} \) for any vector \( \mathbf{v} \) in \( V \).

• In the exercise, it was calculated that applying \( T_2 \) after \( T_1 \) or vice versa resulted in the identity transformation, meaning the original vector \( \mathbf{v} \) was retrieved.
• This confirms that \( T_2 \) is indeed the inverse of \( T_1 \), noted as \( T_2 = T_1^{-1} \).

Inverse transformations are essential in solving systems of linear equations and understanding the structure of the vector spaces and transformations involved.

Vector Space

A vector space is a fundamental concept in linear algebra that defines a collection of vectors. This collection is closed under vector addition and scalar multiplication. In simpler terms, if you have a vector space and you add two vectors in this space, the result is still a vector in the space. Similarly, if you multiply a vector by a scalar (a real or complex number), the result is also within the vector space.

• In our exercise, the vector space \( V \) is the environment where transformations \( T_1 \) and \( T_2 \) take place.
• This space is defined by its basis \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), ensuring that any vector in \( V \) can be expressed as a linear combination of these basis vectors.

The concept of vector spaces is important because it forms the backdrop against which all other linear algebra operations and transformations occur. Understanding the properties and structures of vector spaces is crucial for deeper insights into how vectors interact through transformations.

Linear Combination

A linear combination refers to the sum of scalar multiples of vectors. For vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \), and scalars \( a_1, a_2, \ldots, a_n \), a linear combination can be written as \( a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \ldots + a_n\mathbf{v}_n \). This concept demonstrates how vectors in a space can be constructed and manipulated through weighted sums.

• In the exercise, any vector \( \mathbf{v} \) in the vector space \( V \) is expressed as a linear combination of the basis vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) such that \( \mathbf{v} = a\mathbf{v}_1 + b\mathbf{v}_2 \).
• This expression is crucial for transforming vectors in \( V \) because it allows us to calculate transformations \( T_1 \) and \( T_2 \) in terms of their action on the basis vectors.

Understanding linear combinations is essential because it provides insight into how different vectors can span a space and how transformations affect these vectors, as demonstrated in the exercise.