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Linear Vector Space
Kristen Zybura
February 4, 2004
Contents
1 Linear Vector Space Definition 1
2 Example 1
3 Linear Combination 2
4 Example of a linear combination 2
5 Linear Dependence 2
1 Linear Vector Space Definition
Let V be a set on which two operations, addition and scalar multiplication, have been defined. If u and v are in V , the sum of u and v is denoted by u + v, and if c is a scalar, the scalar multiple of u by c is denoted by cu. If the following axioms hold for all u,v, and w in V and for all scalars c and d, then V is called a vector space and its elements are called vectors.
10 Axioms of a Vector Space
- u + v is in V (closure under addition)
- u + v = v + u (commutativity)
- (u + v) + w = u + (v + w) (associativity)
- There exists an element 0 in V , called a zero vector, such that u+ 0 =u.
- For each u in V , there is an element −u in V such that u + (−u)= 0.
- cu is in V (closure under scalar multiplication)
- c(u + v) = cu + cv (distributivity)
- (c + d)u = cu +du (distributivity)
- c(du) = (cd)u
- 1u = u
2 Example
Determine whether the given set is a vector space: Let V be the set of all positive real numbers x with x + x′^ = xx′, and kx = xk
- If x and y are positive reals, so is x + y = xy ( closure under addition)
- x + y = xy = yx = y + x (commutativity)
- x + (y + z) = x(yz) = (xy)z = (x + y) + z (associativity)
- 1 + x = 1x = x = x1 = x + 1 for all positive reals x (zero vector)
- x + ( (^) x^1 ) = x( (^1) x ) = 1 = 0 = 1 = ( (^1) x )x = ( (^1) x ) + x cu is in V (closure under scalar multiplication)
- If k is real and x is a positive real, then kx = xk^ is again a positive real.
- k(x + y) = (xy)k^ = xkyk^ = kx + ky (distributivity)
- (c + d)x = xc+d^ = xcxd^ = cx + dx (distributivity)
- c(dx) = (dx)c^ = (xd)c^ = xdc^ = xcd^ = (cd)x
- 1x = x^1 = x
3 Linear Combination
A vector v is a linear combination of vectors v 1 , v 2 , ..., vk if there are scalar coefficients c 1 , c 2 , ..., ck such that c 1 v 1 + c 2 v 2 + ckvk = v
4 Example of a linear combination
u= [1, 2 , −1] v= [6, 4 , 2] in R^3. Show that w= [9, 2 , 7] is a linear combination of u and v. c 1 [1, 2 , −1] + c 2 [6, 4 , 2] = [9, 2 , 7] [c 1 , 2 c 1 , − 1 c 1 ] + [6c 2 , 4 c 2 , 2 c 2 ] = [9, 2 , 7] c 1 + 6c 2 = 9 2 c 1 + 4c 2 = 2 −c 1 + 2c 2 = 7 c 1 = − 3 c 2 = 2
5 Linear Dependence
A set of vectors v 1 , v 2 , ...vk is linearly dependent if there are scalars c 1 , c 2 , ..., ck at least one of which is not zero, such that c 1 v 1 + c 2 v 2 + ...ckvk = 0
ie We say that a set of vectors is linearly dependent if one of them can be written as a linear combination of the others. A set of vectors which is not linearly dependent is linearly independent. Example: Let u= [1, 0 , 3] v= [− 1 , 1 , −3] w= [1, 2 , 3]
