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6.1 Vector Spaces-Basic Properties

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Euclidean n-space Just like we have ordered pairs (n=2), and ordered triples (n=3), we also have ordered n-tuples v=(v 1,v 2,…,v n ). Just like with ordered pairs, the order matters. For two n- tuples to be equal, corresponding entries must be equal. Given and integer n ≥ 1, the set of all n-tuples with real entries is called Euclidean n-space ( n ) 1 - number line 2 - X,Y plane 3 - 3D coordinate system

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Definitions for n-space v = (v 1,v 2,…,v n ) and u = (u 1,u 2,…,u n ) 1. u + v = (u 1 +v 1,u 2 +v 2,…,u n +v n ) 2. av = (av 1,av 2,…,av n ) 3. 0 = (0,0,…,0) 4. -v = (-v 1,-v 2,…,-v n ) 5. u - v = (u 1 -v 1,u 2 -v 2,…,u n -v n )

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Properties of n 1. u + v = v + u 2. u + (v + w) = (u + v) + w 3. v + 0 = v 4. v + (-v) = 0 5. a(v+w) = av + aw 6. (a+b) v = av + bv 7. a(bv) = (ab)v 8. 1v = v

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Vector Space A vector space consists of a non-empty set V of vectors that can be added, multiplied by a scalar, and for which certain axioms hold. For u,v,w in V: Axioms for vector addition A1. If u and v are in V, then u+v is in V. A2. u + v = v + u A3. u + (v + w) = (u + v) + w A4. An element 0 exists s.t. v + 0 = v = 0 + v A5. For each v in V, an element (-v) exists in V such that -v + v = 0 = v + (-v)

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More... Axioms for scalar multiplication S1. If v is in V, then av is in V for all a in S2. a(v+w) = av + aw S3. (a + b)v = av + bv S4. a(bv) = (ab)v for all v in V S5. 1v = v We say V is closed under vector addition and scalar multiplication, which means that: 1. If we add two vectors in V, we get another vector in V 2. If we multiply a vector V by a scalar, we get a vector in V

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So... n is a vector space by the definition. Let M m n be the set of all (m x n) matrices w/ real entries, then: The set M m n is a vector space using matrix addition and scalar multiplication. (note that all axioms hold)

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Example Show that V = {(x,x,y)| x,y are real numbers} is a vector space using the operations of 3 A1-A3, S1-S5 easy to show just writing in component form. Need to show others true: A4: (0,0,0) is in V, so we show it satisfies A4 A5: (-x,-x,-y) is in V, so we show it satisfies A5

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Example 5 V is set of all ordered pairs (x,y), and define addition in V as in 2. Define scalar mult in V by a(x,y) = (ay,ax). Determine if V is a vector space with these operations. A1-A5 clearly hold (just like 2 ). Test the axioms in S and find that S4 fails.

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Polynomials Let P be the set of all polynomials and p(x) = a 0 + a 1 x + a 2 x 2 + … + a n x n q(x) = b 0 + b 1 x + b 2 x 2 + … + b n x n Note that addition is defined: p(x) + q(x) = (a 0 +b 0 ) + (a 1 +b 1 )x + … + (a n +b n )x n And scalar mult is defined: cp(x) = ca 0 + (ca 1 )x + … + (ca n )x n P is a vector space -- show that it satisfies the axioms.

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Functions F[a,b] is the set of all functions on the interval [a,b] (i.e. x, the input value, is in the interval [a,b] Two functions, f and g, are equal if f(x) = g(x) for every x in [a,b]. It is said that f and g have the same action Pointwise addition: (f+g) (x) = f(x) + g(x) Scalar multiplication: (rf)(x) = rf(x) The set F[a,b] is a vector space if pointwise addition and scalar multiplication are the operations. (0 function: 0(x) = 0) (-f)(x)=-f(x) Can show A1,S1, but others for homework

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Theorem 1-Cancellation u,v,w are vectors in V. If v + u = v + w, then u = w Proof: Given v + u = v + w -v + (v + u) = -v + (v + w) (A5) (-v + v) + u = (-v + v) + w (A3) 0 + u = 0 + w (A5) u = w(A4)

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Theorem 2 Given u and v in vector space V, x + v = u has only one solution x in V: x = u - v Proof: (Cannot prove like Thm 1 since don’t know x is in V yet.) x = u - v is a solution since: x + v = (u - v) + v = u + (-v + v) = u + 0 = u It is the only solution: assume x 1 was also a solution so: x 1 + v = u, then: x 1 + v = x + v, so x 1 = x by cancellation

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Theorem 3 Let v be a vector in vector space V, and a be a real number. 1. 0v = 0 2. a0 = 0 3. If av = 0, then either a = 0 or v = 0 4. (-1)v = -v 5. (-a)v = -(av) = a(-v) Proof: 1. 0v + 0v = (0 + 0)v =0v = 0v + 0 so Thm 1 gives 1. 3. If av = 0, we show that if a≠0, then v = 0

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Theorem 3 (proof continued) 4. -v + v = 0 by A5. Also (-1)v+v=(-1)v+1v=(-1+1)v=0v=0 so (-1)v + v = -v + v so (-1)v= -v

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Example Given vectors u,v in vector space V, find x,y in V s.t. x-4y = u 2x + 3y = v Can be done just like solving linear systems since we have shown that operations are just like those in linear systems.

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