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Chapter 1: Problem 3

Prove that \(\frac{1}{2} \mathbf{v}+\frac{1}{2} \mathbf{v}=\mathbf{v}\)

Short Answer

Expert verified
The statement \(\frac{1}{2} \mathbf{v} + \frac{1}{2} \mathbf{v} = \mathbf{v}\) is proven to be true. This is a result of the distributive property of scalar multiplication over vector addition.

Step by step solution

01

Distributive Property of Scalar Multiplication

The distributive property of scalar multiplication over vector addition states that for any scalar \(a\) and \(b\), \(a\mathbf{v}+b\mathbf{v} = (a+b)\mathbf{v}\). Let's apply this property to the given vectors where \(a = b = 1/2\). Hence, \(\frac{1}{2} \mathbf{v} + \frac{1}{2} \mathbf{v} = (1/2 + 1/2)\mathbf{v}\)
02

Addition of Scalars

In the first step, we have \((1/2 + 1/2)\mathbf{v}\). Now performing basic addition, this equals to \(1\mathbf{v}\).
03

Simplification

A scalar \(1\) times any vector \(\mathbf{v}\) equals the vector itself. Hence, \(1\mathbf{v} = \mathbf{v}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distributive Property of Scalar Multiplication
The distributive property of scalar multiplication is a fundamental concept that makes calculations in linear algebra more manageable. It allows us to multiply a scalar with the sum of two vectors, distributing the scalar to each vector and adding the results. To fully grasp this property, let's consider a scalar, let's say 'c', and two vectors \textbf{a} and \textbf{b}. The distributive property is mathematically expressed as:
\[ c(\textbf{a} + \textbf{b}) = c\textbf{a} + c\textbf{b} \]
Thus, multiplying each vector by the scalar 'c' separately and then adding them gives the same result as first adding the vectors and then multiplying the sum by the scalar. This property is crucial for simplifying complex linear algebra expressions and solving equations involving vectors and scalars efficiently.
Scalar Multiplication
Scalar multiplication refers to the operation of multiplying a vector by a scalar (a real number), resulting in a new vector that is in the same (or opposite, if the scalar is negative) direction as the original, but scaled in length according to the scalar value. For example, if we multiply a vector \textbf{v} by the scalar 3, we get a new vector that points in the same direction as \textbf{v} but its magnitude is three times as large:
\[ 3 \textbf{v} \]
Scalar multiplication is not just about changing the magnitude. It's the means by which we can shrink, stretch, flip, or maintain vector magnitude during transformations. When we multiply a vector by 1, the vector remains unchanged in both direction and magnitude. Conversely, multiplying by 0 results in the zero vector, regardless of the original vector's magnitude or direction.
Simplification in Linear Algebra
Simplification in linear algebra entails reducing expressions to their most basic form without changing their value or meaning. This process helps in understanding complex problems by breaking them down into simpler components. For instance, consider the vector addition exercise involving vector \textbf{v} and scalar multiples:
\[ \frac{1}{2} \textbf{v}+\frac{1}{2} \textbf{v} \]
To simplify, we observe that both terms involve the same vector multiplied by a scalar. Using scalar multiplication and distributive property, we combine the scalars:
\[ (\frac{1}{2} + \frac{1}{2}) \textbf{v} \]
Given that the sum of the scalars equals 1, we're left with a straightforward result:
\[ 1\textbf{v} = \textbf{v} \]
Hence, simplification transformed the expression into a single vector, \textbf{v}, demonstrating the vector's property under scalar multiplication. This form of simplification is a critical tool in linear algebra, making complex operations more digestible and solvable.

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Most popular questions from this chapter

Let \(X\) be any nonempty set and let \(n\) be a positive integer. Let \(V\) denote the set of all functions \(f: X \rightarrow \mathbb{R}^{n}\). Any element of \(V\) corresponds to an ordered list of functions \(f_{i}: X \rightarrow \mathbb{R}\) for \(i=1,2, \ldots, n\) such that \(f(x)=\) \(\left(f_{1}(x), f_{2}(x), \ldots, f_{n}(x)\right)\) for all \(x \in X\). Define addition and scalar multiplication on \(V\) in terms of the corresponding operations on the coordinate functions. Show that with these operations, \(V\) is a vector space.

Is it possible to add two sets? For sets of real numbers, we might define addition of sets in terms of addition of the elements in the sets. Let us introduce the following meaning to the symbol \(\oplus\) for adding a set \(A\) of real numbers to another set \(B\) of real numbers: \(A \oplus B=\\{a+b \mid a \in A\) and \(b \in B\\}\). a. List the elements in the set \(\\{1,2,3\\} \oplus\\{5,10\\}\). b. List the elements in the set \(\\{1,2,3\\} \oplus\\{5,6\\}\). c. List the elements in the set \(\\{1,2,3\\} \oplus \varnothing\). d. If a set \(A\) contams \(m\) real numbers and a set \(B\) contains \(n\) real numbers, can you predict the number of elements in \(A \oplus B\) ? If you run into difficulties, can you determine the minimum and naximum numbers of elements possible in \(A \oplus B\) ? e. Does the commutative law hold for this new addition? That is, does \(A \oplus\) \(B=B \oplus A\) ? f. Reformulate other laws of real-number addition in terms of this new addition of sets. Which of your formulas are true? Can you prove them or provide counterexamples? g. What about laws that combine set addition with union and intersection? For example, does \((A \cup B) \oplus C=(A \oplus C) \cup(B \oplus C)\) ? h. Is there any hope of extending the other operations of arithmetic to sets of real numbers? What about algebra? Limits? Power series?

Determine whether we obtain a vector space from \(\mathbb{R}^{2}\) with operations defined by $$ \begin{aligned} \left(v_{1}, v_{2}\right)+\left(w_{1}, w_{2}\right) &=\left(v_{1}+w_{1}, v_{2}+w_{2}+1\right) \\ r\left(v_{1}, v_{2}\right) &=\left(r v_{1}, r_{2}\right) \end{aligned} $$ Notice that the additive identity vector is \((0,-1)\) since $$ \begin{aligned} \left(v_{1}, v_{2}\right)+(0,-1) &=\left(v_{1}+0, v_{2}+(-1)+1\right) \\ &=\left(v_{1}, v_{2}\right) \end{aligned} $$ as required by Axiom \(3 .\)

Write the \(y\)-axis as a line in \(\mathbb{R}^{2}\).

Write the line in \(\mathbb{R}^{2}\) that is tangent to the unit circle at the point \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) in the form \(\\{r+x \mid r \in \mathbb{R}\\}\). (Suggestion: The tangent line will intersect the axes at points \((a, 0)\) and \((0, a)\) equidistant from the origin. You do not need to know the value of \(a\) to find a direction vector.)
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