Linear algebra. Vector spaces

Transcription

1 Linear algebra Vector spaces

2 Outline 1 Vector spaces Arithmetic vector space Properties of operations Basis of vector space V n 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 2 / 25

3 Outline 1 Vector spaces Arithmetic vector space Properties of operations Basis of vector space V n 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 2 / 25

4 Outline 1 Vector spaces Arithmetic vector space Properties of operations Basis of vector space V n 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 2 / 25

5 Outline Vector spaces 1 Vector spaces Arithmetic vector space Properties of operations Basis of vector space V n 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 3 / 25

6 Arithmetic vector space Vector spaces Arithmetic vector space Definition Let R be a set of real numbers. Let V n = R R R = R n }{{} n times be a set of all ordered n-tuples of real numbers (a 1,..., a n) (where a i R for i = 1, 2,..., n). We will call each ordered n-tupl of real numbers an arithmetic vector, n its dimension. We will write a = (a 1,..., a n). It is also possible to write a or a. Lucie Doudová (UoD Brno) Linear algebra 4 / 25

7 Arithmetic vector space Vector spaces Arithmetic vector space Definition Let a = (a 1,..., a n), b = (b 1,..., b n). We can define on set V n operations equality of vectors, sum of vectors and product of vector and real number. A1: a, b are equal (we write a = b), if holds: a 1 = b 1,..., a n = b n. A2: Sum of vectors a, b is vector a + b and it holds: a + b = (a 1 + b 1,..., a n + b n). A3: Product of vector a and real number k is vector k a and it holds: k a = (ka 1,..., ka n). Set V n with this operations we call n-tupl real arithmetic vector space of dimension n. Its elements are called arithmetic vectors. Lucie Doudová (UoD Brno) Linear algebra 5 / 25

8 Arithmetic vector space Vector spaces Arithmetic vector space Theorem Let V n V n V n be a mapping which assigns vector a + b V n to each ordered pair of vectors ( a, b) V n V n such that for every vectors a, b, c V n hods: 1 a + b = b + a, (commutative law) 2 a + ( b + c) = ( a + b) + c, (associative law for summation) 3 there exists to each vector a V n vector o V n such that a + o = a (vector o = (0,..., 0) is called a zero vector.), 4 there exists to each vector a V n vector a V n such that a + ( a) = o (vector a is called opposite vector to vector a). Lucie Doudová (UoD Brno) Linear algebra 6 / 25

9 Arithmetic vector space Vector spaces Arithmetic vector space Theorem Let R V n V n be a mapping which assigns vector r a V n to each ordered pair (r, a) R V n such that for all real numbers r, s R and every vectors a, b V n holds: 1 1 a = a, 2 r(s a) = (rs) a, (associative law for multiplication of vector by real number) 3 (r + s) a = r a + s a, (distributive law) 4 r( a + b) = r a + r b. (distributive law) Lucie Doudová (UoD Brno) Linear algebra 7 / 25

10 Arithmetic vector space Vector spaces Arithmetic vector space Notation Vectors e 1,..., e n are called unit vectors. e 1 = (1, 0, 0,..., 0) e 2 = (0, 1, 0,..., 0). e n = (0, 0,..., 0, 1) Lucie Doudová (UoD Brno) Linear algebra 8 / 25

11 Vector spaces Definition Vector u V n is called the linear combination of vectors a 1,..., a r V n (r 1) if there exist real numbers p 1,..., p r such that u = p 1 a p r a r. Examples 1 Zero vector is a linear combination of arbitrary vectors (for p 1 = = p r = 0). 2 Vector u = ( 7, 17) can be written as linear combination of vectors a 1 = (1, 2), a 2 = (3, 7) V 2. u = 2 a 1 3 a 2 = 2(1, 2) 3(3, 7) = (2, 4) (9, 21) = ( 7, 17) Lucie Doudová (UoD Brno) Linear algebra 9 / 25

12 Vector spaces Lineární kombinace a závislost vektorů Příklad In V 3 write vector u = (u 1, u 2, u 3) as a linear combination of vectors e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1). u = (u 1, 0, 0) + (0, u 2, 0) + (0, 0, u 3) = = u 1(1, 0, 0) + u 2(0, 1, 0) + u 3(0, 0, 1) = = u 1 e 1 + u 2 e 2 + u 3 e 3 Remark Arbitrary vector u V n can be written as a linear combination of vectors e 1, e 2,..., e n. Lucie Doudová (UoD Brno) Linear algebra 10 / 25

13 Vector spaces Example Verify if vector u = ( 2, 1, 6) is a linear combination of vectors a 1 = (4, 0, 1) and a 2 = (2, 0, 5). u = p 1 a 1 + p 2 a 2 2 = 4p 1 + 2p 2 1 = 0p 1 + 0p 2 6 = 1p 1 + 5p 2 This system of equations is insolvable (see second equation). This means that vector u is not a linear combination of vectors a 1, a 2. Lucie Doudová (UoD Brno) Linear algebra 11 / 25

14 Vector spaces Definition Let a 1,..., a r be vectors from vector space V n, r 1. We say that vectors a 1,..., a r are linear dependent (LD) if there exists real numbers p 1,..., p r such that at least one of them is different from zero and p 1 a p r a r = o. If this equality is fulfilled only for p 1 = = p r = 0, vectors a 1,..., a r are called linearly independent (LI). Lucie Doudová (UoD Brno) Linear algebra 12 / 25

15 Vector spaces Example Find out if vectors e 1 = (1, 0), e 2 = (0, 1) are linearly dependent or independent. p 1 e 1 + p 2 e 2 = o p 1(1, 0) + p 2(0, 1) = (0, 0) (p 1, 0) + (0, p 2) = (0, 0) (p 1, p 2) = (0, 0) p 1 = p 2 = 0 It means that given vectors are linearly independent. Remark Vectors e 1,..., e n V n are linearly independent. Lucie Doudová (UoD Brno) Linear algebra 13 / 25

16 Vector spaces Theorem Group of vectors a 1,..., a r V n, r 2, is linearly dependent if and only if one of vectors a 1,..., a r is a linear combination of other vectors. Proof I. II. p 1 0 : p 1 a p r a r = o p 1 a 1 = p 2 a 2... p r a r a 1 = p2 p 1 a 2... pr p 1 a r a 1 = p 2 a p r a r o = a 1 + p 2 a p r a r from it p 1 = 1 0. Lucie Doudová (UoD Brno) Linear algebra 14 / 25

17 Vector spaces Example Is it possible to find a group of n + 1 linearly independent vectors in vector space V n? p 1 a p n a n + p n+1 a n+1 = o (1) Vector equation (1) represents homogenous system of n equations. It has n + 1 unknowns (p 1,..., p n+1). This system is solvable and it has infinitely many solutions so it has nonzero solution too vectors are linearly dependent. Remark If there is the zero vector among vectors a 1..., a r then they are linearly dependent. Theorem Maximum number of linearly independent vectors in V n is n. (Group of n + 1 vectors is always linearly dependent; there does not exist more than n linearly independent vectors in V n.) Lucie Doudová (UoD Brno) Linear algebra 15 / 25

18 Basis of vector space V n Vector spaces Basis of vector space Vn Definition Let a 1,..., a r be vectors from vector space V. If it is possible uniquely express arbitrary vector u V as a linear combination of vectors a 1,..., a r, we say that vector space is generated by vectors a 1,..., a r. This set is called set of generators of vector space V. Definition Set of n linearly independent generators of vector space V n is called a basis of the vector space V n. Definition Number of vectors in arbitrary basis of vector space V is called dimension of vector space V (we write dimv ). Lucie Doudová (UoD Brno) Linear algebra 16 / 25

19 Basis of vector space V n Vector spaces Basis of vector space Vn Theorem: It holds: 1 There exists infinitely many bases in V n. 2 Each basis in V n has exactly n vectors. 3 Each vector u V n can be uniquely expressed as linear combination of basis vectors. 4 If vectors a 1,..., a n form the basis, then u = p 1 a p n a n. Numbers p 1,..., p n are called coordinates of vector u in basis B. We write u = (p 1,..., p n) B. Remark Orthonormal basis: B 0 = e 1,..., e n. If u = (u 1, u 2,..., u n), then u = u 1 e u n e n, i. e. u = (u 1, u 2,..., u n) B0 If the index of basis is not written, it means that this coordinates are in orthonormal basis B 0. Lucie Doudová (UoD Brno) Linear algebra 17 / 25

20 Vector spaces Basis of vector space Vn Basis of vector space V n Example Find the coordinates of vector u = (7, 4) in basis B = a 1, a 2, where a 1 = (1, 3), a 2 = ( 1, 2). u = p 1 a 1 + p 2 a 2 (7, 4) = p 1(1, 3) + p 2( 1, 2) 7 = p 1 p 2 4 = 3p 1 + 2p 2 therefore p 1 = 2, p 2 = 5. Coordinates of vector u in basis B are: u = (2, 5) B Lucie Doudová (UoD Brno) Linear algebra 18 / 25

21 Outline Examples 1 Vector spaces Arithmetic vector space Properties of operations Basis of vector space V n 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 19 / 25

22 Řešené příklady Examples Example Express vector u = (0, 1) as linear combination of vectors a 1 = ( 5, 2), a 2 = (3, 3). Solution: u = p 1 a 1+p 2 a 2 From there we get 5p 1 + 3p 2 = 0 2p 1 + 3p 2 = 1 7p 1 = 1 p 1 = 1 7 and p2 = 0 p2 = 5 21 Therefore u = 1 7 a a2 Lucie Doudová (UoD Brno) Linear algebra 20 / 25

23 Řešené příklady Examples Example Are vectors u = (1, 1, 1), v = (3, 6, 9) and w = (0, 2, 7) linearly dependent or independent? Solution: p 1 u + p 2 v + p 3 w = o p 1 + 3p 2 = 0 p 1 + 6p 2 + 2p 3 = 0 p 1 + 9p 2 + 7p 3 = 0 From the first equation we have p 1 = 3p 2. And 3p 2 + 2p 3 = 0 6p 2 + 7p 3 = 0 Thus p 3 = 0. Now we can get easily p 2 and p 1. This system has unique solution p 1 = p 2 = p 3 = 0. It means that vectors are linear independent. Lucie Doudová (UoD Brno) Linear algebra 21 / 25

24 Řešené příklady Examples Example Find the coordinates of vector a = (3, 2, 4) in basis B u, v, w, where u = (1, 1, 0), v = (0, 1, 1), w = (1, 0, 1). p 1 + p 3 = 3 p 1 + p 2 = 2 p 2 + p 3 = 4 From there we get: next step: p 1 + p 3 = 3 p 2 p 3 = 1 p 2 + p 3 = 4 p 1 + p 3 = 3 p 2 p 3 = 1 2p 3 = 5 From the last equation we have p 3 = 5 2. Then p2 = 3 2 and p1 = 1 2. Thus the coordinates of vector a in basis B are a = (1/2, 3/2, 5/2) B. Lucie Doudová (UoD Brno) Linear algebra 22 / 25

25 Příklady k procvičení Examples 1 Find vector v = a + 2 b 3 c, which is a linear combination of vectors a = (1, 1, 3), b = (4, 0, 1), c = (2, 3, 1). 2 Find vector x for which holds: 2( x + u) = 3 v + w, where u = (1, 3, 0), v = (0, 2, 1), w = (0, 0, 2) 3 Write vector a as a linear combination other vectors. a) a = (5, 1, 11), b = (3, 2, 2), c = (2, 3, 1), d = (1, 1, 3) b) a = (1, 2, 4), u = (1, 1, 1), v = (3, 6, 9), w = (0, 2, 7) c) a = (1, 4, 6, 4), b = (1, 3, 2, 1), c = (1, 1, 3, 2), d = (2, 1 1, 3), e = (3, 2, 1, 1) 4 Write the coordinates of vector a in given basis B. a) a = (3, 2, 4), B = u, v, w, u = (1, 1, 0), v = (0, 1, 1), w = (1, 0, 1) b) a = (4, 6, 12, 6), B = u 1, u 2, u 3, u 4, where u 1 = (2, 4, 8, 3), u 2 = (2, 3, 5, 3), u 3 = ( 1, 1, 3, 2), u 4 = (1, 2, 4, 2) Lucie Doudová (UoD Brno) Linear algebra 23 / 25

26 Outline List of tasks for students 1 Vector spaces Arithmetic vector space Properties of operations Basis of vector space V n 2 Examples 3 List of tasks for students Lucie Doudová (UoD Brno) Linear algebra 24 / 25

27 List of tasks for students List of tasks for students Example 1: Find vector u for which holds: a) u = 3( 1, 2, 0) + (1, 3, 2) + 2(0, 3, 1), b) a + 2 u = 3( b u) 2 a, where a = (0, 1, 2, 3) and b = (2, 3, 2, 0), Example 2: Are given vectors linearly dependent or independent? a) u = (0, 2, 1), v = (3, 2, 2), w = (1, 4, 2), b) u = (1, 1, 1), v = (3, 6, 9), w = (0, 2, 7), Example 3: Write vector a as a linear combination of vectors u, v, w a) a = (0, 2, 7), u = (1, 1, 1), v = (3, 6, 9), w = (1, 2, 4), b) a = ( 1, 4, 2), u = (1, 2, 4), v = (1, 1, 1), w = (2, 2, 4), Example 4: Write the coordinates of vector a in basis B a) a = (4, 11, 11), B = u, v, w, u = (2, 3, 3), v = ( 1, 4, 2), w = ( 1, 2, 4), b) a = (3, 2, 3, 1), B = u 1, u 2, u 3, u 4, u 1 = (1, 3, 2, 3), u 2 = (1, 1, 3, 2), u 3 = (0, 1, 1, 1), u 4 = ( 2, 2, 1, 4), Lucie Doudová (UoD Brno) Linear algebra 25 / 25