Question

Exercise 7.3 .4 In each case, compute the action of \(S T\) and \(T S,\) and show that \(S T \neq T S\). a. \(S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) with \(S(x, y)=(y, x) ; T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) with \(T(x, y)=(x, 0)\) b. \(S: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) with \(S(x, y, z)=(x, 0, z)\) \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) with \(T(x, y, z)=(x+y, 0, y+z)\) c. \(S: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) with \(S(p)=p(0)+p(1) x+p(2) x^{2}\); \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) with \(T\left(a+b x+c x^{2}\right)=b+c x+a x^{2}\) d. \(S: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) with \(S\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a & 0 \\ 0 & d\end{array}\right]\); \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) with \(T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}c & a \\ d & b\end{array}\right]\)

Short answer

In all cases (a to d), the compositions ST and TS are not equal.

Step-by-step solution

Determine S(T(x, y)) for Part (a)

We begin by finding the result of applying transformations in the order S followed by T. Given functions \(S(x, y) = (y, x)\) and \(T(x, y) = (x, 0)\),1. Apply \(T\) first: \(T(x, y) = (x, 0)\).2. Apply \(S\) next: \(S(T(x, y)) = S(x, 0) = (0, x)\).So, \(ST(x, y) = (0, x)\).

Determine T(S(x, y)) for Part (a)

Now we'll apply transformations in the order T followed by S.1. Apply \(S\) first: \(S(x, y) = (y, x)\).2. Apply \(T\) next: \(T(y, x) = (y, 0)\).So, \(TS(x, y) = (y, 0)\).

Show ST ≠ TS for Part (a)

We found that \(ST(x, y) = (0, x)\) and \(TS(x, y) = (y, 0)\). Since these outputs are different, \(ST eq TS\).

Determine S(T(x, y, z)) for Part (b)

For functions \(S(x, y, z) = (x, 0, z)\) and \(T(x, y, z) = (x+y, 0, y+z)\):1. Apply \(T\) first: \(T(x, y, z) = (x+y, 0, y+z)\).2. Apply \(S\) next: \(S(x+y, 0, y+z) = (x+y, 0, y+z)\).

Determine T(S(x, y, z)) for Part (b)

Now apply transformations in the order T followed by S:1. Apply \(S\) first: \(S(x, y, z) = (x, 0, z)\).2. Apply \(T\) next: \(T(x, 0, z) = (x+0, 0, 0+z) = (x, 0, z)\).

Show ST ≠ TS for Part (b)

We found \(ST(x, y, z) = (x+y, 0, y+z)\) and \(TS(x, y, z) = (x, 0, z)\). Since these are different, \(ST eq TS\).

Determine S(T(p(x))) for Part (c)

For functions \(S(p) = p(0) + p(1)x + p(2)x^2\) and \(T(a + bx + cx^2) = b + cx + ax^2\):1. Apply \(T\) first: \(T(a + bx + cx^2) = b + cx + ax^2\).2. Apply \(S\) next: \(S(b + cx + ax^2) = b + ax + cx^2\).

Determine T(S(p(x))) for Part (c)

Apply transformations in the order T followed by S:1. Apply \(S\) first: \(S(p(x)) = p(0) + p(1)x + p(2)x^2\).2. Apply \(T\) next: \(T(p(0) + p(1)x + p(2)x^2) = p(1) + p(2)x + p(0)x^2\).

Show ST ≠ TS for Part (c)

We have \(ST(p(x)) = b + ax + cx^2\) and \(TS(p(x)) = p(1) + p(2)x + p(0)x^2\). Since these are different, \(ST eq TS\).

Determine S(T(A)) for Part (d)

Given transformations \(S\) and \(T\) using matrices:1. Apply \(T\) first to matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\): \(T(A) = \begin{bmatrix} c & a \ d & b \end{bmatrix}\).2. Apply \(S\) next: \(S(T(A)) = \begin{bmatrix} c & 0 \ 0 & b \end{bmatrix}\).

Determine T(S(A)) for Part (d)

Apply transformations in the order T followed by S:1. Apply \(S\) first: \(S(A) = \begin{bmatrix} a & 0 \ 0 & d \end{bmatrix}\).2. Apply \(T\) next: \(T(S(A)) = \begin{bmatrix} 0 & a \ d & 0 \end{bmatrix}\).

Show ST ≠ TS for Part (d)

We find \(ST(A) = \begin{bmatrix} c & 0 \ 0 & b \end{bmatrix}\) and \(TS(A) = \begin{bmatrix} 0 & a \ d & 0 \end{bmatrix}\). These differ, so \(ST eq TS\).

Key concepts

Non-commutative Operations

In mathematics, an operation is said to be non-commutative when the order in which the operations are performed affects the outcome. In the case of matrix transformations or functions, this often means that applying one transformation followed by another may produce a different result than applying them in the reverse order. Consider the exercise you've been working on, where functions like matrices or polynomials are transformed by two operations, denoted as \( S \) and \( T \). The goal is to show that \( ST eq TS \), highlighting their non-commutative nature.

Let's break it down further:

• **Order Matters:** If you swap the order of multiplication in non-commutative operations, you get a different result. The same concept applies when applying two transformations in sequence.
• **Example in Matrices:** With matrices \( S \) and \( T \), you might find \( ST(A) \) not equal to \( TS(A) \), where \( A \) is a given matrix. This shows the transformations are non-commutative.
• **Consequences:** Understanding and identifying non-commutative operations is crucial in areas like physics and computer science, where the sequence of operations can affect outcomes.

Non-commutative operations are a fundamental concept in linear algebra, as seen in the given exercise example.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. The topic of matrix transformations, as explored in the exercise, is central to understanding linear algebra. Here's a quick overview of the key aspects:

• **Matrices as Transformations:** Matrices can represent linear transformations, which map vectors from one space to another. For example, in your exercise, each function \( S \) and \( T \) can be considered as matrix transformations in vector spaces like \( \mathbb{R}^2 \) or \( \mathbb{R}^3 \).
• **Key Properties:** Some properties essential to linear algebra include linearity (additivity and homogeneity) and the impact of non-commutative operations as described before.
• **Applications:** Linear algebra has applications in numerous fields such as engineering, physics, computer science, and economics, making it a widely applicable and crucial subject to master.

In essence, linear algebra provides a framework for understanding how things change linearly and how combinations of operations, like those in matrix transformations, work together.

Vector Spaces

Vector spaces are a primary structure studied in linear algebra, providing the setting in which vectors exist and interact. The transformations we are examining through \( S \) and \( T \) operate within such spaces. Here's what you need to know:

• **Definition:** A vector space is defined as a collection of vectors, which are objects that can be added together and multiplied by scalars (numbers), satisfying certain axioms.
• **Examples:** Common examples include \( \mathbb{R}^2 \), the plane of real numbers, and \( \mathbb{R}^3 \), which represents three-dimensional space. Notice how your exercise uses these spaces as domains and codomains for transformations.
• **Subspaces and Bases:** Key concepts include subspaces, which are subsets of vector spaces, and bases, which are sets of vectors that span the entire space.
• **Role in Transformations:** Within vector spaces, transformations like those by \( S \) and \( T \) map one vector to another, allowing for operations like rotation, scaling, and reflection.

Understanding vector spaces is essential for delving deeper into linear transformations and their applications in real-world problems.