Problem 56 - Precalculus : Building Concepts and Connections

Question

This set of exercises will draw on the ideas presented in this section and your general math background. Prove the following for any vector \(\mathbf{u}\) and any real number \(k\) : \((k \mathbf{u}) \cdot(\mathbf{v})=k(\mathbf{u} \cdot \mathbf{v})=\mathbf{u} \cdot(k \mathbf{v})\)

Step-by-Step Solution

Answer

The results for \(k \mathbf{u} \cdot \mathbf{v}\), \(k (\mathbf{u} \cdot \mathbf{v})\), and \(\mathbf{u} \cdot k \mathbf{v}\) are all obtained as \(k \sum_i u_i b_i\). Therefore, the initial equality holds true.

Step 1: Break Down the Equation

We have three expressions to prove equal. Let's denote \(k \mathbf{u}\) as \(\mathbf{a}\) and \(\mathbf{v}\) as \(\mathbf{b}\). So our equations are: \(\mathbf{a} \cdot \mathbf{b} = k (\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot (k\mathbf{b})\).

Step 2: Dot Product of a and b

Firstly, let's find \(\mathbf{a} \cdot \mathbf{b}\). For two vectors, the dot product is the sum of the products of their respective entries. Therefore, \(\mathbf{a} \cdot \mathbf{b} = \sum_i a_i b_i\). Since \( \mathbf{a} = k \mathbf{u} \), we can write this as \(\sum_i k u_i b_i = k\sum_i u_i b_i\).

Step 3: Scalar Multiplication of u and v

Next, let's compute \(k (\mathbf{u} \cdot \mathbf{v})\). The dot product \(\mathbf{u} \cdot \mathbf{v} = \sum_i u_i v_i\). Therefore, \(k (\mathbf{u} \cdot \mathbf{v}) = k\sum_i u_i v_i\).

Step 4: Dot Product of u and kb

Lastly, we need to find \(\mathbf{u} \cdot (k\mathbf{b}) = \sum_i u_i (k b_i) = k\sum_i u_i b_i\).

Step 5: Compare the Equations

If we compare the results obtained in Steps 2, 3, and 4, they all equal \(k \sum_i u_i b_i\), thus proving the original equation.

Key Concepts

Scalar Multiplication of Vectors

In the realm of vector algebra, the operation known as scalar multiplication involves the combination of a vector with a scalar (a real number) to produce a new vector. It's important to understand that when you multiply a vector \textbf{u} by a scalar k, each component of the vector is multiplied by k.

For instance, if we have a vector \textbf{u} = \begin{bmatrix}u_1\u_2\ \.\ \.\ \.\u_n \end{bmatrix} and a scalar k, the scalar multiplication of \textbf{u} by k is written as k\textbf{u} and is computed as:\[k\textbf{u} = k \begin{bmatrix}u_1\u_2\ \.\ \.\ \.\u_n \end{bmatrix} = \begin{bmatrix}ku_1\ku_2\ \.\ \.\ \.\ku_n \end{bmatrix}\]This means each entry of the vector \textbf{u} has been scaled by the factor of k. This concept is pivotal as it interplays with other operations in vector algebra, such as the dot product, which we see in the given problem.

Associative Property in Vector Multiplication

When discussing multiplication in the context of vector algebra, it's crucial to clarify the kind of multiplication we're referring to. Unlike scalar multiplication, the dot product (also known as scalar product) is not truly associative. However, when we mix scalar multiplication with the dot product, we observe an associative-like pattern.

In other words, when a scalar multiplies a dot product, the scalar may associate with either of the vectors in the dot product without changing the result. As demonstrated in the exercise, multiplying the vector \textbf{u} by the scalar k before taking the dot product with \textbf{v} results in the same value as if we first computed \textbf{u} dot \textbf{v} and then multiplied by k:\[ k(\textbf{u} \boldsymbol{\bullet} \textbf{v}) = (k\textbf{u}) \boldsymbol{\bullet} \textbf{v} = \textbf{u} \boldsymbol{\bullet} (k\textbf{v}) \]This property is especially useful in simplifying complex vector equations and understanding the distributive nature of scalar multiplication over the dot product.

Proofs in Vector Algebra

Proofs play a significant role in mathematics, confirming the truth of conjectures and theorems. In vector algebra, proofs like the one in the exercise help us understand the fundamental properties of vector operations. To prove a vector identity, we often need to go back to the definitions and manipulate the equations step by step, maintaining logical coherence.

The proof provided in the step-by-step solution is an excellent example of utilizing the definitions of vector operations to establish the equality of three expressions. It shows that regardless of the order in which you perform the scalar multiplication and the dot product, the outcome remains unchanged. This concept is part of the fundamental framework that underpins the operations in vector spaces and is critical for students to master in order to progress in linear algebra or any field that uses vectors.