Chapter Linear Equations and Vectors
Linear Algebra, Fall 6 Matrices and Systems of Linear Equations Figure.
Linear Algebra, Fall 6 Figure.
Linear Algebra, Fall 6 Figure.
Linear Algebra, Fall 6 Unique solution 5
Linear Algebra, Fall 6 No solutions 6
Linear Algebra, Fall 6 Many solutions 7
Linear Algebra, Fall 6 Definition A matri is a rectangular array of numbers. The numbers in the array are called the elements of the matri. Rows and Columns 7 5 7 5 row row 7 column 5 column column 8
Linear Algebra, Fall 6 9 Submatri A A R Q P A submatrices of matri 5 7 5 7 5 7 Size and Type column matri a row matri a square matri a matri matri matri matri 8 5 8 8 5 9 7 5 5
Linear Algebra, Fall 6 Location 7 5 The element 7 is in row, column. Wesay that it is in location (,). Identity Matrices I I
Linear Algebra, Fall 6 matri of coefficient and augmented matri coefficient augmented matri matri of 6 6
Linear Algebra, Fall 6 Elementary Transformations. Interchange two equations.. Multiply both sides of an equation by a nonzero constant.. Add a multiple of one equation to another equation. Elementary Row Operations. Interchange two rows of a matri.. Multiply the elements of a row by a nonzero constant.. Add a multiple of the elements of one row to the corresponding elements of another row.
Linear Algebra, Fall 6 Eample Solving the following system of linear equation. 6 6 6 Solution Equation Method Initial system: Analogous Matri Method Augmented matri: 8 Eq+( )Eq Eq+( )Eq 8 R+( )R R+( )R
Linear Algebra, Fall 6 5 5 Eq+( )Eq Eq+()Eq Eq+( )Eq Eq+()Eq ( 5)Eq ( 5)R Eq+( )Eq Eq+Eq R+( )R R+R The solution is.,, The solution is.,,
Linear Algebra, Fall 6 Eample 5 Solving the following system of linear equation. 8 8 5 8 8 5 6 R+( )R R+R R 6 8 R+()R R+( )R R. The solution is 8 Solution
Linear Algebra, Fall 6 Eample Solving the system Solution 8 6 8 8 6 8 R 7 7 6 8 R+( )R 7 R+R 6
Linear Algebra, Fall 6 R R R 5 6 5 R ( )R 5 R ( )R R R. The solution is,,. 7
Linear Algebra, Fall 6 8 Eample Solving the following three systems of linear equation, all of which have the same matri of coefficients. for b b b in turn,, 8 b b b 8 R+( )R R+R 5 8 5 )R ( R R R R R )R ( R.,,,,,, are solutions The
Linear Algebra, Fall 6 Gauss-Jordan Elimination Definition A matri is in reduced echelon form if. Any rows consisting entirely of zeros are grouped at the bottom of the matri.. The first nonzero element of each other row is. This element is called a leading.. The leading of each after the first is positioned to the right of the leading of the previous row.. All other elements in a column that contains a leading are zero. 9
Linear Algebra, Fall 6 In Reduced Echelon Form 9 7 8 5 9 7 8 5 6 7
Linear Algebra, Fall 6 Not in Reduced Echelon Form 8 7
Linear Algebra, Fall 6 6 5 )R ( R R R R )R ( R Eample Use the method of Gauss-Jordan elimination to find reduced echelon form of the following matri. 9 9 R R pivot R pivot 6 5 7 R R )R ( R The matri is the reduced echelon form of the given matri.
Linear Algebra, Fall 6 Eample Solve, if possible, the system of equations 7 5 7 9 7 5 7 7 5 7 9 R )R ( R )R ( R R R R R The general solution to the system is parameter). number (called a real is which, r r r r
Linear Algebra, Fall 6
Linear Algebra, Fall 6 5
Linear Algebra, Fall 6 6
Linear Algebra, Fall 6 Homogeneous System of linear Equations Theorem. A homogeneous system of linear equations in n variables always has the solution =, =., n =. This solution is called the trivial solution. Theorem. A homogeneous system of linear equations that has more variables than equations has many solutions. One of these solutions is the trivial solution. 7
Linear Algebra, Fall 6 The Vector Space R n Rectangular Coordinate System the origin:(, ) the position vector: OA the initial point:o the terminal point:a(5, ) 8
Linear Algebra, Fall 6 EXAMPLE Sketch the position vectors See Figure.6. OA (, ), OB ( 5, ) and OC (, ) Figure.6 9
Linear Algebra, Fall 6 Figure.7
Linear Algebra, Fall 6 Let ( u, u,..., un) be a sequence of n real numbers. The set of all such sequences is called n-space and is called R n. u is the first component of ( u, u,..., un), u is the second component and so on. Eample R is the collection of all sets of four ordered real numbers. For eample, (,,, ) and (,, 5., ) are element of R. R 5 is the collection of all sets of five ordered real numbers. For eample, (,,,, 9) is in this collection.
Linear Algebra, Fall 6 Addition and Scalar multiplication Let u ( u, u,..., un) and v ( v, v,..., vn) be two elements of R n. We say that u and v are equal if u = v,, u n = v n (their corresponding components are equal.) Definition Let u ( u be elements of R n, u,..., un) and v ( v, v,..., vn) and let c be a scalar. Addition and Scalar multiplication are performed as follows. Addition: Scalar multiplication : u v ( u v,..., u cu ( cu,..., cu ) n n v n )
Linear Algebra, Fall 6 Eample Let u = (,,, 7) and v = (,,, ) be vectors in R. Find u + v and u. u v (,,, 7) (,,, ) (,,, 7) u (,,, 7) (,, 9, )
Linear Algebra, Fall 6 Eample This eample gives us a geometrical interpretation of vector addition. Consider the sum of the vectors (, ) and (, ). We get (, ) + (, ) = (6, ) In Figure.8 we interpret these vectors as position vectors. Construct the parallelogram having the vectors (, ) and (, ) as adjacent sides. The vector (6,), the sum, will be the diagonal of the parallelogram. Figure.8
Linear Algebra, Fall 6 Eample Such vectors are used in the physical sciences to describe forces. In this eample (, ) and (, ) might be forces, acting on a body at origin, O. The vectors would give the directions of the forces and their lengths would be the magnitudes of the forces,. and.6. The vector sum (6,) is the resultant force, a single force that would be equivalent to the two forces. The magnitude of the resultant force would be 6 7. 5
Linear Algebra, Fall 6 In general, if u and v are vectors in the same vector space, then u+ vis the diagonal of the parallelogram defined by u and v. See Figure.9. Figure.9 6
Linear Algebra, Fall 6 Eample This eample gives us a geometrical interpretation of scalar multiplication. Consider the scalar multiple of the vector (, ) by. We get (, ) = (6, ) Observe in Figure. that (6, ) is a vector in the same direction as (, ), and times it in the length. Figure. 7
Linear Algebra, Fall 6 In general, if u is a vector in any vector space, and c a nonzero scalar, the direction of cu will be the same as the direction of u if c>, and the opposite direction to u if c<. The length of cu is c times the length of u. See Figure. Figure. 8
Linear Algebra, Fall 6 Special Vectors The vector (,,, ), having n zero components, is called the zero vector of R n and is denoted. Negative Vector The vector ( )u is written u and is called the negative of u. Subtraction Subtraction is performed on elements of R n by subtracting corresponding components. For eample, in R (5,, -6) (,, ) = (,, -9) 9
Linear Algebra, Fall 6 Theorem. Properties of Vector Addition and Scalar Multiplication Let u, v, and w be vectors in R n and let c and d be scalars. (a) u + v = v + u Commutative property (b) u + (v + w) = (u + v) + w Associative property (c) u + = + u = u Property of the zero vector (d) u + ( u) = Property of the negative vector (e) c(u + v) = cu + cv (f) (c + d)u = cu + du Distributive properties (g) c(du) = (cd)u (h) u = u Scalar multiplication by
Linear Algebra, Fall 6 These result are proved by writing the vectors in terms of components and using the definitions of vector addition and scalar multiplication, and the properties of real numbers. We give the proofs of (a) and (e). u + v = v + u : Let u ( u, u,..., un) and v ( v, v,..., vn).then u v u,..., u v,..., v ( u ( v v u v,..., v u,..., u v u n n,..., u,..., v n n v u n n ) ) n n c( u v) c( u ( cu c( ( c( u u,..., u v,..., v v v cv,..., u ),..., c( u ),..., cu v v cv cu,..., c un cv,..., cvn cu,..., u cv,..., v cu cv n n n n n n ) n n n )) ) n )
Linear Algebra, Fall 6 Figure.
Linear Algebra, Fall 6 Linear Combinations of Vectors Eapmle 5 Let u = (, 5, ), v = (,, 9), and w = (,, ). Determine the vector combination u v + w. Solution u v w (, 5, ) (,, 9) (,, ) (,, 6) (,, 7) (,, ) (,, 6 7 ) (, 7, )
Linear Algebra, Fall 6 Let v, v,, v m be vectors in R n. The vector v in R n is a linear combination of v, v,, v m if there eist scalars c, c,, c m such that v can be written v = c v + + c m v m
Linear Algebra, Fall 6 Eample 6 Determine wherever the vector (8,, 5) is a linear combination of the vectors (,, ), (,, ), and (, -, ). c ( c (,, ) c, c c (,, ) c c c c c c c c, c (,,) (8,, 5) c c c c c 8 5 ) (8,, 5) c, c, c (8,, 5) (,, ) (,, ) (,,) 5
Linear Algebra, Fall 6 6 Column Vectors n n n n v u v u v v u u We defined addition and scalar multiplication of column vectors in R n in a componentwise manner: and n cu n cu u u c For eample, in R, 8 and 9 5 7
Linear Algebra, Fall 6 Subspace of R n The subset must be closed under addition and under scalar multiplication. All the other vector space properties such as the commutative property u + v = v + u, and the associative property u + (v + w) = (u + v) + w, are inherited from larger R n. A nonempty subset of the vector space R n that has all the algebraic properties of R n is called a subspace. A subset of R n is a subspace if it is closed under addition and under scalar multiplication. 7
Linear Algebra, Fall 6 EXAMPLE A subset of a vector space is often defined in terms of an arbitrary vector. Consider the subset W of R consisting of vectors of the form (a, b, a+b), where the third component is the sum of the first two. The vector (, 5, 7) is in W for eample, whereas (, 5, 9) is not. Let us show that W is a subspace of R. Let u= (a, b, a+b) and v=(c, d, c+d) be vectors in W and k be a scalar. Then 8 ) (,,,,,,,, d b c a d b c a d c b a d b c a d c d c b a b a u v
Linear Algebra, Fall 6 u+v is in W since the third component is the sum of the first two components. W is closed under addition. Further, ku k a, b, a b ka, kb, ka, kb, ku is in W. W is also closed under scalar multiplication. W is a subspace of R. It has all the algebraic properties of a vector space. Let us look at the geometrical interpretation of this subspace W. Consider the arbitrary vector u= (a, b, a+b). Separate the variable a and b as follows: k( a b) ka kb 9
Linear Algebra, Fall 6 u= (a, b, a+b) = (a,, a)+ (, b, b) = a(,, )+ b(,, ) The vectors (,, ) and (,, ) define a plane through the origin. See Figure.. Any linear combination of vectors u = a(,, )+ b(,, ) lies in this plane. W is in fact the plane determined by the vectors (,, ) and (,, ). When every vector of a vector space can be written as a linear combination of a set of vectors as here we say that vectors span the space. Thus the vectors (,, ) and (,, ) span W. The preceding eample leads to a general result. Let u,v, and w be vectors in a plane W. The sum u+v lies in the plane, and so does the scalar multiple kw. See Figure.. Thus W is a subspace. A set of vectors that form a plane is a subspace. 5
Linear Algebra, Fall 6 Figure. The plane determined by the vectors (,, ) and (,, )
Linear Algebra, Fall 6 Figure. W is closed under addition and scalar multiplication. 5
Linear Algebra, Fall 6 EXAMPLE Consider the subset V of R of vectors of the form (a, a, a), where the second component is twice of the first, and the third is three times the first. Let us show that V is a subspace of R.Let (a, a, a) and (b, b, b) be two vectors in V, and let k be a scalar. Then (a, a, a) + (b, b, b) = (a+b, a+b,a+ b) = (a+b, (a+b),(a+b) This is a vector in V since the second component is twice the first, and the third is three times the first. V is closed under addition. Further, k(a, a, a)= (ka, ka, ka) 5
Linear Algebra, Fall 6 The vector is in V. V is closed under scalar multiplication. V is a subspace of R. We now proceed to look at the geometry. Consider the arbitrary vector u=(a, a, a). We get u=(a, a, a) =a (,, ) The vectors of V are scalar multiples of the single vector (,, ). These vectors form the line through the origin defined by the vector (,, ). See Figure.5. The vector (,, ) spans V. The preceding eample leads to a general result. Let u and v be vectors on a line V. The sum u+v lies on the line, and so does the scalar multiple ku. See Figure.6. Thus V is a subspace. A set of vectors that forms a line is a subspace. 5
Linear Algebra, Fall 6 Figure.5 55
Linear Algebra, Fall 6 Figure.6
Linear Algebra, Fall 6 57
Linear Algebra, Fall 6 Eample Consider the following homogeneous system of linear equations: 5 There are many solutions, r 5, r r. 58
Linear Algebra, Fall 6 Eample Consider the following homogeneous system of linear equations: 5 There are many solutions r s, r, s. 59
Linear Algebra, Fall 6 Basis and Dimension in R n Consider the vectors (,, ), (,, ), (,, ) in R. These vectors have two very important properties: (i) They are said to span R. That is, we can write an arbitrary vector (, y, z) as a linear combination of the vectors as follows: (, y, z) = (,, ) + y(,, ) + z(,, ) (ii) They are said to be linearly independent. That is, the identity p(,, ) +q(,, ) + r(,, ) = for scalars p, q, and r has the unique solution p =, q =, r =. 6
Linear Algebra, Fall 6 A finite set of vectors {v,, v m } is called a basis for a vector space V if the set spans V and is linearly independent. The set of n vectors {(,,, ), (,,, ),, (,, )} is a basis for R n. This basis is called the standard basis for R n. The set {(,,, ), (,,, ),, (,,, )} of n vector is the standard basis of R n. The dimension of R n is n. 6
Linear Algebra, Fall 6 Span, Linear Independence, and Basis Span: The vectors v, v, v n span a space if every vector I in the space can be epressed as a linear combination of them, v = c v + c v + + c n v n. Linear Independence: The vectors v, v, v n are linearly independent if the identity c v + c v + + c n v n = is only true for c =, c =,, c n =.If there are nonzero values of c, c,, c n for which this identity holds, the set is said to be linearly dependent. Basis and Dimension: A set of vectors is a basis for a space if the vectors (i) span the space and (ii) are linearly independent. The number of vectors in a basis is unique. This number is called the dimension of the space. 6
Linear Algebra, Fall 6 Eample Consider the subspace W of R of vectors of the form (a, b, a+b) that was discussed in the previous section. Let u = (a, b, a+b) be an arbitrary vector in W. We know that we can write u in the form u = a(,, ) + b(,, ) The vector (,, ) and (,, ) thus span W. Further, the identity p(,, ) + q(,, ) = (,, ) leads to p =, q =. These vectors are linear independent. The set {(,, ), (,, )} is a basis for W. The dimension of W is. We know that W is a plane through the origin, fitting in with our intuitive thinking of a plane as being a two-dimensional space. 6
Linear Algebra, Fall 6 Figure.8
Linear Algebra, Fall 6 65
Linear Algebra, Fall 6 66
Linear Algebra, Fall 6 Eample Consider the following homogeneous system of linear equation: 7 7 It can be shown that the general solution to this system in vector form is (r s, r t, r, s, t) and that this set is closed under addition and under scalar multiplication. It is a subspace W of R 5. Let us find a basis and the dimension of this subspace. Write (r s, r t, r, s, t) r(,,,, ) s(,,,, ) t(,,,,) 5 5 5 67
Linear Algebra, Fall 6 Eample The vectors (,,,, ), (-,,,, ), (, -,,, ) span W. Let us check for linear independence. Consider the following identity for arbitrary variables p, q, h. p(,,,, ) + q(-,,,, ) + h(, -,,, ) = (,,,, ) We get ( p q, p h, p, q, h) (,,, ) This implies that p =, q =, h =. thus, the vectors are linearly independent. The set {(,,,, ), (-,,,, ), (, -,,, )} is a basis for W. the set of solution is a three-dimensional vector space embedded in a five-dimensional space. 68
Linear Algebra, Fall 6 Eample (i) Determine whether the following sets of vectors are linear dependent or independent in R. (a) {(,, ), (,, -), (,, )}. (b) {(,, ), (-,, ), (, -, -9)}. c (,, ) c(,, ) c (,, ) c c c c c c c (,, ) The system has the unique solution c =, c =, c =. Thus, the vectors are linearly independent. 69
Linear Algebra, Fall 6 Eample (ii) We eamine c (,, ) + c (-,, ) + c (, -, -9) = (,, ). This identity leads to the following system of linear equations: c c c c c c c 9c This system has many solutions, c = r, c = r, c = r, where r is any real number. The set is thus linearly dependent/ For eample, when r = we get (,, ) + (-,, ) + (, -, -9) = (,, ). 7
Linear Algebra, Fall 6 Dot Product, Norm, Angle, and Distance Let u ( u, u,..., un) and v ( v, v,..., vn) be two vectors in R n. The dot product of u and v is denoted u v and is defined by u v uv u n v n The dot product assigns a real number to each pair of vectors. For eample, if u = (,, ) and v = (,, ) their dot product is u v ( ) ( ) ( ) 8 7
Linear Algebra, Fall 6 7 Properties of the Dot Product Let u, v, and w be vectors in R n and let c be a scalar. Then. u v = v u. (u + v) w = u w + v w. cu v = c(u v) = u cv. u u, and u u = if and only if u = Proof. u v v u v u numbers real by the commutative property of We get )....,,, ( and )...,,, ( Let n n n n n n u v v u v u u v v v v u u u. ) ( ) (.. u u u u u u u u i i i n n n u u u u u u u u u
Linear Algebra, Fall 6 Norm of a Vector in R n Figure. 7
Linear Algebra, Fall 6 The norm (length or magnitude) of a vector u = (u,, u n ) in R n is denoted u and defined by u u un Note: The norm of a vector can also be written in terms of the dot product u u u For eample, if u = (,, 5) of R and v = (,,, ) of R, then u () () (5) 9 5 5 v () () () () 9 6 6 7
Linear Algebra, Fall 6 A unit vector is a vector whose norm is. If v is a nonzero vector, then the vector u v v is a unit vector in the direction of v. This procedure of constructing a unit vector in the same direction as a given vector is called normalizing the vector. 75
Linear Algebra, Fall 6 Eample Find the norm of the vector (,, ). Normalize this vector. (,, ) ( ). The norm of (,, ) is The normalized vector is (,, ) The vector may also be written,,. This vector is a unit vector in the direction of (,, ).. 76
Linear Algebra, Fall 6 Algebra Between Vectors Figure. 77
Linear Algebra, Fall 6 The Cauchy-Schwartz Inequality. If u and v are vectors in R n, then u v u v Here u v denotes the absolute value of the number u v. Let u and v be two nonzero vectors in R n. The cosine of the angle between these vectors is u v cos u v 78
Linear Algebra, Fall 6 Eample Determine the angle between the vectors u = (,, ) and v = (,, ) in R. u v (,, ) (,,) u v Thus, cos u v u v, The angle between u and v is π/(or 5). 79
Linear Algebra, Fall 6 THEOREM. Two nonzero vectors u and v are orthogonal if and only if u v =. For eample, the vector (,, ) and (,, ) are orthogonal since (,, ) (,, ) = ( ) + ( ) + ( ) = 6 + =. 8
Linear Algebra, Fall 6 Properties of Standard Basis of R n The standard basis for R n, {(,,, ), (,,, ),, (,,, )}, is an orthonormal set. 8
Linear Algebra, Fall 6 8
Linear Algebra, Fall 6 Figure. Figure. 8
Linear Algebra, Fall 6 Distance between Points Let (,,..., n ) and y ( y, y,..., yn ) be two points in R n. The distance between and y is denoted d(, y) and is defined by d (, y) ( - y) ( n - yn) Note: We can also write this distance formula as d(, y) y 8
Linear Algebra, Fall 6 Figure. 85
Linear Algebra, Fall 6 Eample Determine the distance between the points = (,,, ) and y = (,,, 5) in R. d( y) ( ) ( ) ( ) ( 5) 9 6 5 7 86
Linear Algebra, Fall 6 87 Eample 5 Prove that the distance in R n has the following symmetric property: d(, y) = d(y, ). ),...,, ( and ),...,, ( Let n n y y y y We get ), ( ) ( ) ( ) ( ) ( ), ( y y d y y y y d n n n n
Linear Algebra, Fall 6 88
Linear Algebra, Fall 6 89
Linear Algebra, Fall 6 9
Linear Algebra, Fall 6 Theorem.6 Let u and v be vectors in R n. (a) Triangle Inequality: u v The length of one side of a triangle cannot eceed the sum of the lengths of the other two sides. (a) Pythagorean theorem: If u v =, then u + v = u + v. The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. u v 9
Linear Algebra, Fall 6 9
Linear Algebra, Fall 6 Figure. 9
Linear Algebra, Fall 6 Curve Fitting, Electrical Networks, and Traffic Flow Curve Fitting A set of data points ( is given and it, y), (, y),, ( n, yn) is necessary to find a polynomial whose graph passes through the points. The points are often measurements in an eperiment. The -coordinates are called base points. It can be shown that if the base points are all distinct, then a unique polynomial of degree n (or less) y a a can be fitted to the points. a n n a n n 9
Linear Algebra, Fall 6 Figure.5 95
Linear Algebra, Fall 6 96 Eample Determine the equation of the polynomial of degree two whose graph passes through the points (, 6), (, ), (, ). Let the polynomial be, and substituting =, y = 6; =, y = ; =, y =, we have a a a y 9 6 a a a a a a a a a Solving this system for a, a, and a using Gauss-Jordan elimination. 8 6 9 6 )R ( R )R ( R 9 )R ( R )R ( R
Linear Algebra, Fall 6 97 9 R 6 )R ( R R R We get a =, a = 6, a =. The parabola that passes through these points is y = 6 +. Such as Figure,.6. Figure.6