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Chapter 4: Problem 3

Show that (a) \(k(u-v)=k u-k v,\) (b) \(u+u=2 u\)

Short Answer

Expert verified
We proved (a) \(k(u-v)=ku-kv\) by showing that: 1. \(k(u-v)=k(u+(-v))\) 2. Applying distributive property: \(k(u+(-v))=ku+k(-v)\) 3. Associative property: \(ku+k(-v)=ku+(-kv)\) 4. Vector subtraction: \(ku+(-kv)=ku-kv\) For (b) \(u+u=2u\), we proved by: 1. Adding equal vectors: \(u+u=(u_1, u_2, u_3)+ (u_1, u_2, u_3)\) 2. Adding components: \((u_1+u_1 , u_2+u_2, u_3+u_3)\) 3. Simplifying: \((2u_1, 2u_2, 2u_3)\) 4. Scalar multiplication: \((2u_1, 2u_2, 2u_3) = 2(u_1, u_2, u_3)\)

Step by step solution

01

(a) Proving \(k(u-v)=ku-kv\)

First, we need to show that the scalar multiple of the difference of two vectors u and v is equal to the difference of the scalar multiples of these vectors. Let's start by looking at the left side of the expression \(k(u-v)\). According to the definition of vector subtraction, the difference of two vectors u and v can be written as \(u-v=u+(-v)\). Now, let's apply the scalar multiplication on this expression: \(k(u-v)=k(u+(-v))\). By the distributive property of scalar multiplication, we have: \(k(u+(-v))=ku+k(-v)\). Since scalar multiplication is associative, we can write: \(ku+k(-v)=ku+(-kv)\). Finally, using the definition of vector subtraction again, we can rewrite this expression as: \(ku+(-kv)=ku-kv\). Thus, we have proved that \(k(u-v)=ku-kv\).
02

(b) Proving \(u+u=2u\)

Now, let's prove that the sum of a vector u with itself is equal to 2 times the vector u. First, let's note that \(u+u\) is simply the addition of two equal vectors. According to the definition of vector addition, we add the components of the two vectors: \(u+u=(u_1, u_2, u_3)+ (u_1, u_2, u_3)\). Now, we can add the components: \((u_1+u_1 , u_2+u_2, u_3+u_3)\). Which can be simplified to: \((2u_1, 2u_2, 2u_3)\). Now, we can use the definition of scalar multiplication to express this vector as a scalar multiple of u: \((2u_1, 2u_2, 2u_3) = 2(u_1, u_2, u_3)\). Thus, we have proved that \(u+u=2u\).

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Key Concepts

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

Scalar Multiplication
Scalar multiplication is a fundamental concept in vector algebra. It involves multiplying a vector by a scalar, which is merely a single real number. The effect is that each component of the vector is multiplied by the same scalar.
This operation scales the vector by stretching or shrinking it without altering its direction, unless the scalar is negative. In that case, it also reverses the direction of the vector.
  • For example, if we have a vector \( \mathbf{u} = (u_1, u_2, u_3) \) and a scalar \( k \), the scalar multiplication \( k \mathbf{u} \) is calculated as \( (k \, u_1, k \, u_2, k \, u_3) \).
  • In simple terms, if \( k = 3 \) and \( \mathbf{u} = (1, 2, 3) \), the resulting vector \( 3 \mathbf{u} \) is \( (3, 6, 9) \).
This scaling is at the heart of vector transformations and forms the base of complex vector processes in physics and engineering.
Vector Addition
Vector addition is a crucial aspect of vector algebra. It allows us to combine two vectors to produce a third vector. The process involves adding corresponding components of the vectors. This operation effectively translates one vector to the end of another, acting geometrically like adding steps in a path.
  • Mathematically, if you have vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), their addition \( \mathbf{u} + \mathbf{v} \) results in \( (u_1 + v_1, u_2 + v_2, u_3 + v_3) \).
  • For instance, if \( \mathbf{u} = (1, 2, 3) \) and \( \mathbf{v} = (4, 5, 6) \), then \( \mathbf{u} + \mathbf{v} = (5, 7, 9) \).
By understanding vector addition, you can navigate through scenarios such as motion and forces in physics, vector displacement in aircraft navigation, and many more practical applications.
Distributive Property
The distributive property is a key principle in vector algebra that allows us to simplify expressions involving scalar multiplication and vector addition. It states that multiplying a scalar by a sum of vectors equals summing the scalar multiplication of each vector individually.
  • This property can be represented as \( k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v} \), where \( k \) is a scalar, and \( \mathbf{u} \) and \( \mathbf{v} \) are vectors.
  • For example, with \( k = 2 \), \( \mathbf{u} = (1,2) \), and \( \mathbf{v} = (3,4) \), applying the distributive property gives \( 2((1,2) + (3,4)) = 2(1,2) + 2(3,4) = (2,4) + (6,8) = (8,12) \).
Understanding this property is crucial because it helps in simplifying vector equations, making complex vector calculations more manageable.
Associative Property
The associative property is another essential rule in vector algebra that deals with both vector addition and scalar multiplication. It allows you to group vectors or scalars without altering the result of operations.
  • For vector addition, the associative property says that \( (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) \).
  • For instance, with vectors \( \mathbf{u} = (1, 0), \mathbf{v} = (0, 1), \mathbf{w} = (1, 1) \), you will find \( ((1, 0) + (0, 1)) + (1, 1) = (2, 2) \), just as \( (1, 0) + ((0, 1) + (1, 1)) = (2, 2) \).
  • In scalar multiplication, it indicates that any association of scalar multiplication does not affect the result, so \( a(b\mathbf{u}) = (ab)\mathbf{u} \), where \( a \) and \( b \) are scalars and \( \mathbf{u} \) is a vector.
Recognizing these properties helps in arranging and solving complex vector equations in a more seamless manner.

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Most popular questions from this chapter

Express \(M\) as a linear combination of the matrices \(A, B, C\), where $$ M=\left[\begin{array}{ll} 4 & 7 \\ 7 & 9 \end{array}\right], \quad \text { and } \quad A=\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right], \quad B=\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right], \quad C=\left[\begin{array}{ll} 1 & 1 \\ 4 & 5 \end{array}\right] $$ Set \(M\) as a linear combination of \(A, B, C\) using unknown scalars \(x, y, z\); that is, set \(M=x A+y B+z C\). This yields $$ \left[\begin{array}{ll} 4 & 7 \\ 7 & 9 \end{array}\right]=x\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]+y\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]+z\left[\begin{array}{ll} 1 & 1 \\ 4 & 5 \end{array}\right]=\left[\begin{array}{cc} x+y+z & x+2 y+z \\ x+3 y+4 z & x+4 y+5 z \end{array}\right] $$ Form the equivalent system of equations by setting corresponding entries equal to each other: $$ x+y+z=4, \quad x+2 y+z=7, \quad x+3 y+4 z=7, \quad x+4 y+5 z=9 $$ Reducing the system to echelon form yields $$ x+y+z=4, \quad y=3, \quad 3 z=-3, \quad 4 z=-4 $$ The last equation drops out. Solving the system by back-substitution yields \(z=-1, y=3, x=2\). Thus, \(M=2 A+3 B-C\)

Prove Theorem 4.3: The intersection of any number of subspaces of \(V\) is a subspace of \(V\).

Find a basis and dimension of the subspace \(W\) of \(\mathbf{R}^{3}\) where (a) \(W=\\{(a, b, c): a+b+c=0\\}\), (b) \(W=\\{(a, b, c):(a=b=c)\\}\) (a) Note that \(W \neq \mathbf{R}^{3}\), because, for example, \((1,2,3) \notin W\). Thus, \(\operatorname{dim} W<3\). Note that \(u_{1}=(1,0,-1)\) and \(u_{2}=(0,1,-1)\) are two independent vectors in \(W\). Thus, \(\operatorname{dim} W=2\), and so \(u_{1}\) and \(u_{2}\) form a basis of \(W\) (b) The vector \(u=(1,1,1) \in W\). Any vector \(w \in W\) has the form \(w=(k, k, k)\). Hence, \(w=k u\). Thus, \(u\) spans \(W\) and \(\operatorname{dim} W=1\).

Show that the vectors \(u=(1+i, 2 i)\) and \(w=(1,1+i)\) in \(C^{2}\) are linearly dependent over the complex field \(\mathbf{C}\) but linearly independent over the real field \(\mathbf{R}\).

Prove Theorem 4.12: Every basis of a vector space \(V\) has the same number of elements. Suppose \(\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) is a basis of \(V\), and suppose \(\left\\{v_{1}, v_{2}, \ldots\right\\}\) is another basis of \(V\). Because \(\left\\{u_{i}\right\\}\) spans \(V\), the basis \(\left\\{v_{1}, v_{2}, \ldots\right\\}\) must contain \(n\) or less vectors, or else it is linearly dependent by Problem 4.35-Lemma 4.13. On the other hand, if the basis \(\left\\{v_{1}, v_{2}, \ldots\right\\}\) contains less than \(n\) elements, then \(\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) is linearly dependent by Problem 4.35. Thus, the basis \(\left\\{v_{1}, v_{2}, \ldots\right\\}\) contains exactly \(n\) vectors, and so the theorem is true.
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