Linear Algebra and Its Applications, Exercise 2.1.5

Exercise 2.1.5. There are eight rules for vector addition and scalar multiplication operations in a vector space:

  1. x+y = y+x
  2. x+(y+z) = (x+y)+z
  3. There is a zero vector (0) such that x+0 = x for all x
  4. For each x there exists one and only one vector -x such that x+(-x)=0
  5. 1x = x for all x
  6. (c_1c_2)x = c_1(c_2x)
  7. c(x+y) = cx+cy
  8. (c_1+c_2)x = c_1x+c_2x

(a) Define addition in \mathbf{R}^2 such that it adds an extra one to each component, e.g., (1, 3) + (4, 5) = (6, 9) instead of (5,8). Assume the rule for scalar multiplication is left unchanged. Which of the above rules is broken by this redefinition of addition?

(b) Define vector addition on the set of positive real numbers such that x+y = xy and define scalar multiplication on the same set so that cx = x^c. Show that the set of positive real numbers is a vector space under the operations thus defined, and describe the zero vector.

Answer: (a) For any two vectors u = (u_1, u_2) and v = (v_1, v_2) rule 1 is satisfied:

u + v = (u_1+v_1+1, u_2+v_2+1)

= (v_1+u_1+1, v_2+u_2+1) = v+u

Rule 2 is satisfied as well:

u+(v+w) = (u_1, u_2) + (v_1+w_1+1,v_2+w_2+1)

= (u_1+(v_1+w_1+1)+1, u_2+(v_2+w_2+1)+1)

= ((u_1+v_1+1)+w_1+1, (u_2+v_2+1)+w_2+1)

= (u_1+v_1+1, u_2+v_2+1) + (w_1,w_2) = (u+v)+w

Rule 3 is satisfied with (-1,-1) as the zero vector:

(u_1,u_2) + (-1,-1) = (u_1-1+1,u_2-1+1) = (u_1,u_2)

Rule 4 is satisfied with -x = (-x_1-2,-x_2-2):

(u_1,u_2) + (-u_1-2,-u_2-2) = (u_1-u_1-2+1,u_2-u_2-2+1) = (-1,-1)

Rule 5 and rule 6 are satisfied since scalar multiplication was not redefined:

1u = 1(u_1,u_2) = (1u_1,1u_2) = (u_1,u_2) = u

(c_1c_2)u = ((c_1c_2)u_1,(c_1c_2)u_2) = c_1(c_2u_1,c_2u_2) = c_1(c_2u)

However rule 7 is not satisfied since for u = v = (1, 1) we have

2(u+v) = 2(1+1+1,1+1+1) = 2(3,3) = (6,6)

while

2u + 2v = 2(1,1) + 2(1,1) = (2,2) + (2,2)

= (2+2+1,2+2+1) = (5,5)

Rule 8 is also not satisfied since for u = (1, 1) we have

(2+2)u = 4(1,1) = (4,4)

while

2u + 2u = 2(1,1) + 2(1,1) = (2,2) + (2,2)

= (2+2+1,2+2+1) = (5,5)

Thus under the redefinition of addition rules 1-6 are satisfied while rules 7 and 8 are not.

(b) We use the symbols \oplus and \otimes to denote the following definitions of vector addition and scalar multiplication on the set of positive real numbers:

x \oplus y = xy and c \otimes x = x^c

Rules 1 and 2 are satisfied:

x \oplus y = xy = yx = y \oplus x

x \oplus (y \oplus z) = x(yz) = (xy)z = (x \oplus y) \oplus z

Rule 3 is satisfied with 1 as the zero vector:

x \oplus 1 = x \cdot 1 = x

Rule 4 is satisfied with 1/x as -x:

x \oplus 1/x = x (1/x) = 1

Rules 5 through 8 are also satisfied:

1 \otimes x = x^1 = x

(c_1c_2) \otimes x = x^{c_1c_2} = x^{c_2c_1} = (x^{c_2})^{c_1} = c_1 \otimes (c_2 \otimes x)

c \otimes (x \oplus y) = (xy)^c = x^cy^c = (c \otimes x) \oplus (c \otimes y)

(c_1+c_2) \otimes x = x^{c_1+c_2} = x^{c_1}x^{c_2} = (c_1 \otimes x) \oplus (c_2 \otimes x)

Since all eight rules are satisfied the set of positive real numbers is a vector space under the vector addition and scalar multiplication operations defined above, with 1 as the zero vector.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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2 Responses to Linear Algebra and Its Applications, Exercise 2.1.5

  1. leytpapas says:

    I think Rule 4 in (a) should be -x=(-x-1,-y-1) in order to get (0,0) and not (-1,-1)

    • hecker says:

      Rule 4 refers to the zero vector *as defined for the supposed vector space*, not the specific vector (0,0) as we know it from R^2. In question (a) the definition of vector addition for this would-be vector space is that u + v = (u_1 + v_1 + 1, u_2 + v_2 + 1). So in this would-be vector space the zero vector must be (-1, -1) in order to satisfy rule 3, that a vector v plus the zero vector must equal v itself. Since the zero vector in this would-be vector space is (-1, -1), the negative of a vector x must satisfy the equality -x + x = (-1, -1). If x = (x_1, x_2) then that requires -x = (-x_1 – 2, -x_2 – 2).

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