# sec4_1 - 4.1 Vector Spaces& Subspaces Many concepts...

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Unformatted text preview: 4.1 Vector Spaces & Subspaces Many concepts concerning vectors in R n can be extended to other mathematical systems. We can think of a vector space in general, as a collection of objects that behave as vectors do in R n . The objects of such a set are called vectors. A vector space is a nonempty set V of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the ten axioms below. The axioms must hold for all u, v and w in V and for all scalars c and d. 1. u  v is in V. 2. u  v  v  u. 3. u  v  w  u  v  w 4. There is a vector (called the zero vector) 0 in V such that u  0  u. 5. For each u in V, there is vector u in V satisfying u  6. cu is in V. 7. c u  v cu cv. 8. c  d u  cu  du. 9. cd u  c du . 10. 1u  u. Vector Space Examples ab cd u  0. EXAMPLE: Let M 2 2  : a, b, c, d are real In this context, note that the 0 vector is . 1 EXAMPLE: Let n 0 be an integer and let P n  the set of all polynomials of degree at most n 0. Members of P n have the form p t  a0  a1t  a2t2    antn where a 0 , a 1 , , a n are real numbers and t is a real variable. The set P n is a vector space. We will just verify 3 out of the 10 axioms here. Let p t  a 0  a 1 t    a n t n and q t  b 0  b 1 t    b n t n . Let c be a scalar. Axiom 1: The polynomial p  q is defined as follows: p  q t  p t q t . Therefore, p  q t  p t q t  __________  __________ t    __________ t n which is also a _____________________ of degree at most ________. So p  q is in P n . Axiom 4: 0 0  0t    0t n (zero vector in P n ) p  0 t  p t 0  a 0  0  a 1  0 t    a n  0 t n  a0  a1t    antn  p t and so p  0  p Axiom 6: cp t  cp t  ________  ________ t    ________ t n which is in P n . The other 7 axioms also hold, so P n is a vector space. 2 Subspaces Vector spaces may be formed from subsets of other vectors spaces. These are called subspaces. A subspace of a vector space V is a subset H of V that has three properties: a. The zero vector of V is in H. b. For each u and v are in H, u  v is in H. (In this case we say H is closed under vector addition.) c. For each u in H and each scalar c, cu is in H. (In this case we say H is closed under scalar multiplication.) If the subset H satisfies these three properties, then H itself is a vector space. a EXAMPLE: Let H  0 b Solution: Verify properties a, b and c of the definition of a subspace. a. The zero vector of R 3 is in H (let a  _______ and b  _______). b. Adding two vectors in H always produces another vector whose second entry is ______ and therefore the sum of two vectors in H is also in H. (H is closed under addition) c. Multiplying a vector in H by a scalar produces another vector in H (H is closed under scalar multiplication). Since properties a, b, and c hold, V is a subspace of R 3 . Note: Vectors a, 0, b in H look and act like the points a, b in R 2 . : a and b are real . Show that H is a subspace of R 3 . x3 x2 H x1 3 EXAMPLE: Is H  x x1 : x is real a subspace of _______? I.e., does H satisfy properties a, b and c? x2 3 2.5 2 1.5 1 0.5 x1 − 0.5 0.5 1 1.5 2 Graphical Depiction of H Solution: All three properties must hold in order for H to be a subspace of R 2 . Property (a) is not true because ___________________________________________. Therefore H is not a subspace of R 2 . Another way to show that H is not a subspace of R 2 : Let u 1 3 x2 3 2.5 2 1.5 1 0.5 x1 0 1 and v  1 2 , then u  v  and so u  v  R2. , which is ____ in H. So property (b) fails and so H is not a subspace of x2 3 2.5 2 1.5 1 0.5 x1 −0.5 0.5 1 1.5 2 − 0.5 0.5 1 1.5 2 Property (a) fails Property (b) fails 4 A Shortcut for Determining Subspaces THEOREM 1 If v 1 , , v p are in a vector space V, then Span v 1 , , v p is a subspace of V. Proof: In order to verify this, check properties a, b and c of definition of a subspace. a. 0 is in Span v 1 , , v p since 0 _____v 1  _____v 2    _____v p b. To show that Span v 1 , Span v 1 , , v p : , v p closed under vector addition, we choose two arbitrary vectors in u a 1 v 1  a 2 v 2    a p v p and v b 1 v 1  b 2 v 2    b p v p . Then u  v  a1v1  a2v2    apvp  b1v1  b2v2    bpvp  ___v 1  ____v 1  ____v 2  ____v 2    ____v p  ____v p  a1  b1 v1  a2  b2 v2    ap  bp vp. So u  v is in Span v 1 , , vp . c. To show that Span v 1 , , v p closed under scalar multiplication, choose an arbitrary number c and an arbitrary vector in Span v 1 , , v p : v b 1 v 1  b 2 v 2    b p v p . Then cv c b 1 v 1  b 2 v 2    b p v p  ______v 1  ______v 2    ______v p So cv is in Span v 1 , , vp . , v p is a subspace of V. Since properties a, b and c hold, Span v 1 , 5 Recap 1. 2. To show that H is a subspace of a vector space, use Theorem 1. To show that a set is not a subspace of a vector space, provide a specific example showing that at least one of the axioms a, b or c (from the definition of a subspace) is violated. a subspace of R 2 ? Why or why not? EXAMPLE: Is V  a  2b, 2a 3b : a and b are real Solution: Write vectors in V in column form: a  2b 2a 3b  a 2a  2b 3b  _____ 1 2  _____ 2 3 So V Span v 1 , v 2 and therefore V is a subspace of _____ by Theorem 1. a  2b EXAMPLE: Is H  a1 a Solution: 0 is not in H since a  b  0 or any other combination of values for a and b does not produce the zero vector. So property _____ fails to hold and therefore H is not a subspace of R3. 2a b : a and b are real a subspace of R 3 ? Why or why not? EXAMPLE: Is the set H of all matrices of the form Explain. Solution: Since 2a b 3a  b 3b a  3a  b 3b a subspace of M 2 2 ? 2a 0 3a 0  0 b b 3b b . Therefore H Span 20 30 , 01 13 and so H is a subspace of M 2 2 . 6 ...
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