Vector Space Examples
Math 203
Spring 2014
Myers
Example:
S = set of all doubly infinite sequences of real numbers = {{yk } : k ∈ Z, yk ∈ R}
Define {yk } + {zk } = {yk + zk } and c{yk } = {cyk }.
Example: yk = 2k , zk = cos(k(π/2)), c = −2
{yk } = (. . . , 14 , 12 , 1, 2, 4, . . .)
{zk } = (. . . , cos(−π), cos(−π/2), cos(0), cos(π/2), cos(π), . . .) = (. . . , −1, 0, 1, 0, −1, . . .)
{yk } + {zk } = (. . . , − 34 , 12 , 2, 2, 3, . . .)
1
3 {yk }
1 1 1 2 4
= (. . . , 12
, 6 , 3 , 3 , 3 , . . .)
Claim: The set S of all doubly infinite sequences of real numbers is a vector space.
Proof: Let {uk }, {vk }, and {wk } be any three doubly infinite sequences in S, and let c and d be any two
scalars.
1. {uk } + {vk } is in S since {uk + vk } is a doubly infinite sequence.
2. {uk } + {vk } = {vk } + {uk } because:
{uk } + {vk }
=
=
=
{uk + vk }
{vk + uk }
{vk } + {uk }
definition of addition in S
addition is commutative in R
definition of addition in S
3. ({uk } + {vk }) + {wk } = {uk } + ({vk } + {wk }) because:
({uk } + {vk }) + {wk }
=
=
=
=
=
{uk + vk } + {wk }
{(uk + vk ) + wk }
{uk + (vk + wk )}
{uk } + {vk + wk }
{uk } + ({vk } + {wk })
definition of addition in S
definition of addition in S
addition is associative in R
definition of addition in S
definition of addition in S
4. Let {0} = {. . . , 0, 0, 0, 0, 0, . . .} denote the doubly infinite sequence of zeros. Then {uk } + {0} = {uk }
because:
{uk } + {0}
=
=
{uk + 0}
{uk }
definition of addition in S
property of real numbers
5. Let −{uk } = (−1){uk }. Then {uk } + (−{uk }) = {0} because:
{uk } + (−{uk })
{uk } + (−1){uk }
{uk } + {(−1)uk }
{uk } + {−uk }
{uk + (−uk )}
{0}
=
=
=
=
=
definition of −{uk }
definition of scalar multiplication in S
property of real numbers
definition of addition in S
property of real numbers
6. c{uk } is in S since {cuk } is a doubly infinite sequence.
7. c({uk } + {vk }) = c{uk } + c{vk } because:
c({uk } + {vk })
c{uk + vk }
{c(uk + vk )}
{cuk + cvk }
{cuk } + {cvk }
c{uk } + c{vk }
=
=
=
=
=
definition of addition in S
definition of scalar multiplication in S
distributive property of R
definition of addition in S
definition of scalar multiplication in S
8. (c + d){uk } = c{uk } + d{uk } because:
(c + d){uk }
=
=
=
=
{(c + d)uk }
{cuk + duk }
{cuk } + {duk }
c{uk } + d{uk }
definition of scalar multiplication in S
distributive property of R
definition of addition in S
definition of scalar multiplication in S
9. c(d{uk }) = (cd){uk } because:
c(d{uk })
=
=
=
=
c{duk }
{c(duk )}
{(cd)uk }
(cd){uk }
definition of scalar multiplication in S
definition of scalar multiplication in S
multiplication is associative in R
definition of scalar multiplication in S
10. 1{uk } = {uk } because:
1{uk }
=
=
{1uk }
{uk }
definition of scalar multiplication in S
property of real numbers
Since all ten vector space axioms hold, we can conclude that S is a vector space.
Homework Exercise 1:
Pn = set
of all polynomials of degree at most n with real
coefficients
= a0 + a1 t + a2 t2 + . . . + an tn : a1 , a2 , . . . , an ∈ R
Given any two polynomials
p(t) = a0 + a1 t + a2 t2 + . . . + an tn and q(t) = b0 + b1 t + b2 t2 + . . . + bn tn
and any scalar c, define
p(t) + q(t) = (a0 + b0 ) + (a1 + b1 )t + (a2 + b2 )t2 + . . . + (an + bn )tn
and
c(p(t)) = (ca0 ) + (ca1 )t + (ca2 )t2 + . . . + (can )tn .
Example: p(t) = 1 + 2t + 3t2 , q(t) = 2 + 4t, c = 2
p(t) + q(t) = (1 + 2) + (2 + 4)t + (3 + 0)t2 = 3 + 6t + 3t2
2(p(t)) = (2 · 1) + (2 · 2)t + (2 · 3)t2 = 2 + 4t + 6t2
Claim: The set Pn of all polynomials of degree at most n with real coefficients is a vector space.
Proof: Let u(t) = u0 + u1 t + u2 t2 + . . . + un tn , v(t) = v0 + v1 t + v2 t2 + . . . + vn tn , and w(t) = w0 + w1 t +
w2 t2 + . . . + wn tn be any three polynomials in Pn , and let c and d be any two scalars.
Homework: Fill in the details of the proof on a separate piece of paper. (For help see Example 4 on page
192.)
1. u(t) + v(t) is in Pn since ???.
2. u(t) + v(t) = v(t) + u(t) because: ???
3. (u(t) + v(t)) + w(t) = u(t) + (v(t) + w(t)) because: ???
4. Let 0 denote the polynomial in Pn with all coefficients equal to zero. Then u(t) + 0 = u(t) because: ???
5. Let −u(t) = (−1)u(t). Then u(t) + (−u(t)) = 0 because: ???
6. cu(t) is in Pn since ???.
7. c(u(t) + v(t)) = cu(t) + cv(t) because: ???
8. (c + d)u(t) = cu(t) + du(t) because: ???
9. c(du(t)) = (cd)u(t) because: ???
10. 1u(t) = u(t) because: ???
Since all ten vector space axioms hold, we can conclude that Pn is a vector space.
Homework Exercise 2:
F = set of all real-valued functions defined on a set D
= {f : D → R}
Given any two functions f and g, and any scalar c, define f + g to be the function whose value at the point
t in D is f (t) + g(t), and define cf to be the function whose value at the point t in D is c(f (t)).
Example: D = (0, ∞), f (t) = t2 , g(t) = ln(t), c =
1
2
(f + g) : (0, ∞) → R is defined by (f + g)(t) = t2 + ln(t)
1
2f
: (0, ∞) → R is defined by
1
2f
(t) = 21 t2
Note: to say f = g means f (t) = g(t) for all t in D.
Claim: The set F of all real-valued functions defined on a set D is a vector space.
Proof: Let f , g, and h be any three functions in F, and let c and d be any two scalars.
Homework: Fill in the details of the proof on a separate piece of paper. (For help see Example 5 on page
192.)
1. f + g is in F since ???.
2. f + g = g + f because: ???
3. (f + g) + h = f + (g + h) because: ???
4. Let 0 denote the function whose value is zero for all t in D. Then f + 0 = f because: ???
5. Let −f = (−1)f . Then f + (−f ) = 0 because: ???
6. cf is in F since ???.
7. c(f + g) = cf + cg because: ???
8. (c + d)f = cf + df because: ???
9. c(df ) = (cd)f because: ???
10. 1f = f because: ???
Since all ten vector space axioms hold, we can conclude that F is a vector space.
Homework Exercise 3:
M2×2 = set of all 2 × 2 matrices with real entries =
a11
a21
a12
a22
: a11 , a12 , a21 , a22 ∈ R
Homework: Show that the set M2×2 of all 2 × 2 matrices with real entries is a vector space.