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MATB24H3 (3)
Final

# Lecture3and4.pdf

6 Pages
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Department
Mathematics
Course
MATB24H3
Professor
Sophie Chrysostomou
Semester
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=⇒ = 0r1+ 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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