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Mathematics

MATB24H3

Sophie Chrysostomou

Winter

Description

University of Toronto at Scarborough
Department of Computer & Mathematical Sciences
MAT B24S Fall 2011
Lectures 3 and 4
Notation: From now on P will denote the vector space of all polynomials
n
with degree less than or equal to n along with the zero polynomial.
definition 0.1 A Subspace W of a vector space V A subset W of a vector
space V is a subspace of V if W with addition and scalar multiplication as
deﬁned for V is itself a vector space.
example 0.2 Both P and2P are ve3tor spaces (with pointwise addition
and scalar multiplication) over the reals and P2⊂ P . 3herefore P is 2
subspace of P3.
theorem 0.3 Test for a Subspace: Let V,⊕,⊗ is a vector space. A subset
W of a vector space V (over the ﬁeld R) is a subspace of V ⇐⇒ W satisﬁes
the following conditions: Under the same operations ⊕ and ⊗ as in V:
1. W is nonempty.
2. If v,w ∈ W =⇒ v ⊕ w ∈ W.
3. If r ∈ R and v ∈ W =⇒ r ⊗ v ∈ W.
proof. (=⇒) Assume that W is a vector space. Then
1. it contains the zero vector from property A3 of the properties of vector
spaces. Thus it is nonempty.
2. it is closed under addition and
1 3. it is closed under scalar multiplication
(⇐=) Assume that W satisﬁes the three properties above. Then W
satisﬁes the ﬁrst two conditions of vector spaces. Since W is a subset of
V then any vector addition or scalar multiplication using vectors from W
will give the same vector (in W) as it does in V. In addition, if v ∈ W
then 0 ⊗ v = z ∈ W and therefore A1 - A4 and S1 - S4 are satisﬁed. Thus
W,⊕,⊗ is a vector space also.
definition 0.4 Let X be a set of vectors in a vector space V over the ﬁeld
F. X is linearly independent if and only if for any subset {v ,v ,1..,2 } of k
X:
r1v1+ r 2 +2··· + r v k 0kimplies that r = r = 1·· =2r = 0 k
Conversely if there is a nontrivial solution to r v 1 r1v + 2··2+ r v = 0 k k
then we say that X is linearly dependent.
example 0.5 Let p (x) 1 x +2x+1, p (x) = x +2, p = x+1, p = 3 +1 4 3
and X = {p ,p1,p2,p 3. 4 ∈ P . Is 3 linearly independent or dependent?
solution. Since p (x) − p (x) − p (x) = 0 then X is linearly dependent.
1 2 3
▯
Note: The set C = {f ▯ f is a continuous function on R} with pointwise
addition and scalar multiplication is a vector space over the ﬁeld R.
example 0.6 Determine if the subset X = {sinx,e ,cosx} of C is linearly
independent.
solution. We want to determine if
r sinx + r e + r cosx = 0 (∗)
1 2 3
has a nontrivial solution.
Method 1: Evaluate (*) for diﬀerent x to get 3 equations in 3 unknowns at
x = 0 r2+ r 3 0 (1)
π π/2
x = r1+ r 2 = 0 (2)
2
x = π r e − r = 0 (3)
2 3
2 ▯ π/2 ▯
0 1 1 ▯ 0 R1←→R 2 1 e 0 ▯0
1 e π/2 0 ▯ 0 =⇒ 0 1 1 ▯0
0 eπ −1 ▯ 0 0 eπ −1 ▯0
1 eπ/2 0 ▯0 1 1 e π/2 0 ▯ 0
R3−e R2 ▯ R3(−1−e ) ▯
−−−−−→ 0 1 1 ▯0 −−−−−− → 0 1 1 ▯ 0
0 0 −1 − eπ ▯0 0 0 1 ▯ 0
▯ ▯
1 e π/2 0 ▯0 1 0 0 ▯0
−−−R→ 0 1 0 ▯0 −−−−−−→ 2 0 1 0 ▯0
▯ ▯
0 0 1 0 0 0 1 0
∴ r = r = r = 0. Therefore X is linearly independent.
1 2 3
Method 2: Diﬀerentiate (*) and evaluate at 0.
x
(∗) r1sinx + 2 e + 3 cosx = 0 r2+ r 3 = 0
1st derivative of (∗) r1cosx + 2 e − r3sinx = 0 at=⇒ = 0r1+ r2 = 0
x
2nd derivative of (∗) −r1sinx + r2e − r3cosx = 0 r2− r 3 = 0
0 1 1 ▯ 0 1 1 0 ▯ 0 1 1 0 ▯0
▯ R1←→R 2 ▯ R3−R2 ▯
−→ 1 1 0 ▯ 0

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