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Chapter 4: Problem 71

Suppose \(u\) and \(v\) belong to a vector space \(V\). Simplify each of the following expressions: (a) \(E_{1}=4(5 u-6 v)+2(3 u+v)\), (c) \(E_{3}=6(3 u+2 v)+5 u-7 v\), (b) \(E_{2}=5(2 u-3 v)+4(7 v+8)\), (d) \(E_{4}=3(5 u+2 / v)\).

Short Answer

Expert verified
The simplified expressions are: (a) \( E_{1}= 26u - 22v \) (b) \( E_{2}= 10u + 13v + 32 \) (c) \( E_{3}= 23u + 5v \) (d) \( E_{4}= 15u + \frac{6}{v} \)

Step by step solution

01

Distribute the scalar to each vector term

Multiply each scalar (4 and 2) by the corresponding vector terms: \( E_{1}= (4 \cdot 5)u - (4 \cdot 6)v + (2 \cdot 3)u + 2v \)
02

Simplify the scalar-vector multiplication

Simplify the multiplication: \( E_{1}= 20u - 24v + 6u + 2v \)
03

Combine like terms

Combine the terms that correspond to the same vector: \( E_{1}= (20u + 6u) + (-24v + 2v) = 26u - 22v \) (b) \( E_{2}=5(2 u-3 v)+4(7 v+8) \)
04

Distribute the scalar to each vector term

Multiply each scalar (5 and 4) by the corresponding vector terms: \( E_{2}= (5 \cdot 2)u - (5 \cdot 3)v + (4 \cdot 7)v + (4 \cdot 8) \)
05

Simplify the scalar-vector multiplication

Simplify the multiplication: \( E_{2}= 10u - 15v + 28v + 32 \)
06

Combine like terms

Combine the terms that correspond to the same vector: \( E_{2}= 10u + (-15v + 28v) + 32 = 10u + 13v + 32 \) (c) \( E_{3}=6(3 u+2 v)+5 u-7 v \)
07

Distribute the scalar to each vector term

Multiply the scalar (6) by the corresponding vector terms: \( E_{3}= (6 \cdot 3)u + (6 \cdot 2)v + 5u - 7v \)
08

Simplify the scalar-vector multiplication

Simplify the multiplication: \( E_{3}= 18u + 12v + 5u - 7v \)
09

Combine like terms

Combine the terms that correspond to the same vector: \( E_{3}= (18u + 5u) + (12v - 7v) = 23u + 5v \) (d) \( E_{4}=3(5 u+2 / v) \)
10

Distribute the scalar to each vector term

Multiply the scalar (3) by the corresponding vector terms: \( E_{4}= (3 \cdot 5)u + (3 \cdot \frac{2}{v}) \)
11

Simplify the scalar-vector multiplication

Simplify the multiplication: \( E_{4}= 15u + \frac{6}{v} \) In conclusion, the simplified expressions are: (a) \( E_{1}= 26u - 22v \) (b) \( E_{2}= 10u + 13v + 32 \) (c) \( E_{3}= 23u + 5v \) (d) \( E_{4}= 15u + \frac{6}{v} \)

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Key Concepts

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

Scalar Multiplication
In the context of vector spaces, scalar multiplication is a fundamental operation. It involves multiplying a vector by a scalar (a real number). The result is a new vector that points in the same or opposite direction as the original vector but has its magnitude scaled by the factor of the scalar. For instance, considering a scalar multiplication like \(4(5u - 6v)\), you multiply each element inside the parenthesis by 4. The expression becomes \(20u - 24v\). Here, both vectors \(u\) and \(v\) are scaled individually.

Some important points about scalar multiplication are:
  • Scalar multiplication is commutative and associative. This means \(a(bu) = (ab)u\).
  • If a scalar is zero, the resulting vector is a zero vector, regardless of the original vector.
  • Scalar multiplication does not change the vector's direction if multiplied by a positive scalar.
Vector Addition
Vector addition is another crucial concept in vector spaces. It involves adding two vectors to create a resultant vector. The operation is straightforward: add each corresponding component of the vectors. For example, consider combining like terms in the expression \(26u - 22v\). This step is essential in simplifying computations involving multiple vectors.

Here's how vector addition works:
  • The addition is conducted by summing up corresponding components, maintaining the dimension of the vectors.
  • Vector addition is both commutative: \(u + v = v + u\), and associative: \(u + (v + w) = (u + v) + w\).
  • Adding a zero vector leaves the vector unchanged: \(u + 0 = u\).
Linear Combinations
In vector spaces, a linear combination refers to an expression made up of vectors and scalars. This combination involves both operations鈥攕calar multiplication and vector addition. The expressions \(E_1 = 4(5u - 6v) + 2(3u + v)\) and \(E_3 = 6(3u + 2v) + 5u - 7v\) are modifications using linear combinations of the vectors \(u\) and \(v\). This type of expression helps describe any vector in terms of other vectors.

Linear combinations cover a few key characteristics:
  • A linear combination is "linear" because it involves no powers or products of the involved vectors.
  • They play a central role in defining vector spaces, allowing for linear dependencies and independencies.
  • Any vector in a vector space can be expressed as a linear combination of a set of basis vectors.

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Most popular questions from this chapter

Suppose \(\left(a_{11}, \ldots, a_{1 n}\right),\left(a_{21}, \ldots, a_{2 n}\right), \ldots,\left(a_{m 1}, \ldots, a_{m n}\right)\) are linearly independent vectors in \(K^{n},\) and suppose \(v_{1}, v_{2}, \ldots, v_{n}\) are linearly independent vectors in a vector space \(V\) over \(K .\) Show that the following vectors are also linearly independent: \\[ w_{1}=a_{11} v_{1}+\cdots+a_{1 n} v_{n}, \quad w_{2}=a_{21} v_{1}+\cdots+a_{2 n} v_{n}, \quad \ldots, \quad w_{m}=a_{m 1} v_{1}+\cdots+a_{m n} v_{n} \\]

Show that if any row is deleted from a matrix in echelon (respectively, row canonical) form, then the. resulting matrix is still in echelon (respectively, row canonical) form.

Let \(V\) be the set of ordered pairs \((a, b)\) of real numbers. Show that \(V\) is not a vector space over \(\mathbf{R}\) with addition and scalar multiplication defined by (i) \((a, b)+(c, d)=(a+d, b+c)\) and \(k(a, b)=(k a, k b)\), (ii) \((a, b)+(c, d)=(a+c, b+d)\) and \(k(a, b)=(a, b)\), (iii) \((a, b)+(c, d)=(0,0)\) and \(k(a, b)=(k a, k b)\), (iv) \((a, b)+(c, d)=(a c, b d)\) and \(k(a, b)=(k a, k b)\).

Consider the following subspaces of \(\mathbf{R}^{5}:\) \\[ U=\operatorname{span}\left(u_{1}, u_{2}, u_{3}\right)=\operatorname{span}\\{(1,3,-2,2,3), \quad(1,4,-3,4,2), \quad(2,3,-1,-2,9)\\} \\] \(W=\operatorname{span}\left(w_{1}, w_{2}, w_{3}\right)=\operatorname{span}\\{(1,3,0,2,1), \quad(1,5,-6,6,3), \quad(2,5,3,2,1)\\}\) Find a basis and the dimension of (a) \(U+W\), (b) \(U \cap W\)

Show that \((\mathrm{a})\) If \(S \subseteq T,\) then \(\operatorname{span}(S) \subseteq \operatorname{span}(T) .\) (b) \(\operatorname{span}[\operatorname{span}(S)]=\operatorname{span}(S)\).
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