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## Numerical Methods

by: Kiera Considine I

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Kiera Considine I
UCD
GPA 3.9

Staff

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COURSE
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11
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KARMA
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## Popular in Engineering Applied Sci Davis

This 11 page Class Notes was uploaded by Kiera Considine I on Tuesday September 8, 2015. The Class Notes belongs to EAD 210B at University of California - Davis taught by Staff in Fall. Since its upload, it has received 44 views. For similar materials see /class/187494/ead-210b-university-of-california-davis in Engineering Applied Sci Davis at University of California - Davis.

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Date Created: 09/08/15
EGR2013 Tutorial 8 Linear Algebra Outline 0 Powers of a Matrix and Matrix Polynomial 0 Vector Algebra 0 Vector Spaces Powers of a Matrix and Matrix Polynomial If A is a square matrix then we define the nonnegative integer power of A to be A0 I Aquot 2 AA A If A is invertible then we de ne the negative integer powers to be A quot A 1quot A 1A 1uA 1 If A is a square matrix we have a matrix polynomial in A as PAa01a1AazA2anAn Example 1 HA is a nth order square matrix and Ak 0 evaluate I 471 Solution 3939I AquotI AIAA2Aquot 1I I A 11AA2Akil Definition of Vectors We often use two kinds of quantities namely scalars and vectors A scalar is a quantity that is determined by its magnitude A vector is a quantity that is determined by both its magnitude and its direction Equality of Vectors two vectors a and b are equal if they have the same length and the same direction Representations in Cartesian coordinate system the vector can be described using real numbers If the given vector a has an initial point 131x1 y1zl and a terminal point 132x2y2zz then the vector a can be described as a a 2P1 2 x2 x1y2 y1ZZ Zl And the norm lal Jalz az2 a Another representation of vectors is a a1 a2 a3 ali a2 a3k Where i j k are the standard unit vectors in the positive directions of the axes in a Cartesian coordinate system Basic Properties of Vectors gt Zero vector 0 has length 0 and no direction gt Negative vector a has the length lal and the direction is opposite to that of a gt Unit vector is a vector of norm 1 Vector Addition and Scalar Multiplication If a a1 a2 a3 and b 2 b1 b2 3 are two vectors then ailalib1 12in a3ib3 W 2 m1 kaz W3 Basic Properties gt abba abcabc gt a00aa aa0 gt kabkakb klakala gt klakla 0a0 gt 1aa 1aa 2 Inner Product or Dot Product De nition if a and b are two vectors and 9 is the angle between a and b then the dot product or Euclidean inner product ml is de ned by ab abcos6 a1b1 ale a3b3 Angle between two vectors ab b39 The angle 9 between the vectors is 0056 I a Orthogonality The inner product of two vectors is zero if and only if these vectors are perpendicular Properties of the Dot Product ab Slallbl gt Schwarz inequality gt Triangle inequality ab S lal lbl Orthogonal Projection Orhtogonal projection of a vector a in the direction of a vector b is defined by b p projba a b lblz Example 2 If vector a is perpendicular to any vector show a is a zero vector Proof a is perpendicular to any vector for any vector a 0 especially if a thus ma 2 0 lal 0 therefore a is a zero vector Zero vector is perpendicular to any vector Example 3 Show vector c and vector acbbca are orthogonal vectors m acbbcac acbcbcac acbcbcac 0 Thus c and acbbca are orthogonal vectors Vector Product or Cross Product De nitions if a and b are two vectors then the cross product or vector product is avector V agtltb 2 V1 v2 V3 The vector v can be obtained from the expansion by the first row of the symbolical thirdorder determinant i j k V23 a a 32 213a1 213211 azk 12 3b2b3b1b3 blbz b1 b2 b3 The length of the vector v is given as la gtlt bl absin9 The direction of the vector v is perpendicular to both a and b Example 4 Iftwo vectors a and b lal 10 lbl 2 and ab 12 Evaluate la gtlt bl Solution ab lallblcos c059 2 2 la bl 5 sine 2i 5 la gtlt bl lallblsin 16 4 Example 5 Given a X bC gtlt 13 gtlt Cb gtlt d show a d is parallel to bC Proof If ad is parallel to bC then a d gtlt bC should be 0 a d gtlt bCa gtlt bd gtlt Cd gtlt ba X C axbCXdbxdaxc0 therefore vector a d is parallel to vector bC Scalar Triple Product If a b and c are three vectors then the scalar triple product is de ned by 31 a2 a3 ab x c b1 b2 b3 C1 C2 C3 The absolute of the scalar triple product is the volume of the parallelepiped With a b and c as edge vectors Example 6 Given three vectors ai4j4k b5i5jk and c6ij5k show these three vector are in the plane Proof Because the absolute of the scalar triple product is the volume of the parallelepiped if the volume is zero we can say the edge vectors are in the same plane a1 a2 a3 1 4 4 1 4 4 Thus abxc b1 b2 b3 5 5 l 0 25 19 0 C1 C2 C3 6 l 5 0 25 19 Therefore vectors a b and c are in the same plane 2 If ab 0 then vector a is perpendicular to vector b 2 If a X b 0 then vector a is parallel to vector b 2 If ab gtlt c0 then vectors a b and c are in the same plane Vector Spaces In a nonempty set V we de ne two algebraic operations 1 vector addition V06 E V 3 06 E V 2 scalar multiplication V06 6 VVk E R 3 ka 6 V For vector addition gt abba gt abcabc gt there is an object 0 in V such that 0aa for all a in V gt for each a in V there is an object 7a in V such that aa0 For scalar multiplication gt kabkakb gt klakala gt klakla gt for every a in V laa If V follows all the aXioms above then V is called a vector space Example 6 Let V R2 and we define the addition and scalar multiplication in V for two vectors a a1 a2 and b 11 b2 and a scalar k as follows ab a1 azb1 b2 a1 171 0 ka 2 ka1 a2 ka1 0 First Va eVa3 V and VaeVVkeRk0 eV are satisfied But there is no zero vector in V such that al a2 a1 a2 So set V is not a vector space with stated operations EGR2013 Tutorial 9 Linear Algebra Outline 0 Linear IndependenceDependence of Vectors and Subspace O Fundamental Vector Spaces of a Matrix 0 Matrix Eigenvalue Problem Linear IndependenceDependence of Vectors ALinear combination ofvectors 3132 a in a vector space V is an In expression kla1 k2212 kmam These vectors are called linear independence if the vector equation kla1 k2212 kmam 0 implies that k1 0162 0 km 2 0 Otherwise they are linear dependent Span If S 3132 air is a set of vectors in a vector space V then the subspace W ofV is called the space spanned by 31 32 ar denoted by W spanS or W Spanalazar Dimension dimV the maximum number oflinear independent vectors Basis gt The basis of V consists of a maximum possible number of linear independent vectors in V gt Every vector in V can be expressed as a linear combination of the vectors in the basis gt A basis is not unique Example 1 Given All vectors in R4 such that 11 0113 0112 V4 S 0 is this set of vectors a vector space Example 2 If vectors 3132 and 213 are linear independent determine whether the vectors 1 31 2212 33 33 al 2 a1 212212a3a3211 form a linear dependent set or a linear independent set Solution Geometric interpretation Because 3132 and 213 are linear independent so they are not in the same plane We can assume they are the edges ofa triangular pyramid O ABC C US If 0A 2 3103 2320C 33 then all212212a3 and 213211 are the edges of the triangle ABC respectively so they are in the same plane in other words they are linear dependent But 31 azaza3 and 213a1 are in the plane OAB CBC and OCA respectively obviously they are not in the a plane so they are linear independent 1 ell212212213 and a3 a1 are linear dependent Because we have 1211 az 1212 a3 1213 a1 0 2 211a2 aza3 and 213 31 are linear independent The vector equation k1aiaz 93233 13331 0 can be written as k1k3 a1 k1kZ a2 k2 k3a3 0 2 because 3132 and 213 are linear independent then we have k1 k3 0 k1 k2 0 k2 k3 0 l 0 1 because the determinant l l 2 0 0 l 1 thus the system has trivial solutions k1 2 k2 2 k3 0 so 211 212212a3 and 213a1 are linear independent Fundamental Vector Spaces of a Matrix A is a m X n matriX then we define Row space subspace of Rm spanned by row vectors Column space subspace of Rquot spanned by column vectors Nullspace the solution space of AX 0 Basic Concepts of Fundamental Matrix Space Row operation doesn t change the fundamental spaces ofa matriX Basis can be obtained by reducing the matriX A to its echelon form dim row space dim column space rank A The dimension ofthe nullspace ofA is called the nullity ofA and rankAnullityAn n is the number of column of A Example 3 VVVV A is a nth order matrix ifA is invertibleCgt IAI 7t 0 Cgt rank An Example 4 Given rankA3 evaluate a and b A mix0 Solution Row operations 1 1 1 1 1 1 1 1 0 1 1 b 0 1 1 b A gt gt 0 1 a Z 2 0 0 a l 2 12 0 2 2 4 0 0 0 4 212 because rank A3 therefore a i 11 2 or a 11 7t 2 Example 5 Given A is a m X n matrix B is a n X s matrix if AB0 show rankA rankB S n Proof Matrix B can be expressed as B 2 B1 Bz liv then we have 332 935 DABZDDABJ 030930 213 0 jl2s Therefore each column of B is a solution of AX0 The dimension of nullspace ofA is equal to n rankA B is some vectors in this nullspace so rankB rankB1 32 Bs S n rankA Then we have rankA rankB S n Matrix Eigenvalue Problem Eigenvalue and Eigenvetor For an n X n matrix A and a vector X ifa value of 3 exists such that Ax 2 ix has nonzero solution vector x 7quot 0 then A is de ned as an eigenvalue of matrix A the corresponding solution vector x is eigenvector Method for Finding Eigenvalues and Eigenvectors gt Characteristic equation p0 detA XI 1quot Aquot Cquot 0 gt The eigenvector corresponding to l is the solutions of A Jx 0 Example 6 0 l A 2 1 0 nd the eigenvalues and eigenvectors ofA Solution 1 The Characteristic equation ofA is l 1 detA M02 1 l zlz10 2tit i02t1 2412 2139 2 Eigenvectors are the solutions of A AUX 0 For 21 i the linear system becomes 139 1 x1 139 1 x1 x1 139 i 0 2 0 2 s so that 1 1 x2 0 0 x2 x2 1 1 is a eigenvector corresponding 21 i For 21 i the linear system becomes i 1 x1 i 1 x1 x1 i i 0 2 0 2 s so that 1 i x2 0 0 x2 x2 1 1 is a eigenvector corresponding 21 i Example 7 If A is the eigenvalue of matriX A evaluate the eigenvalue of A2 2A 1 Solution If A is the eigenvalue of matriX A and X is the eigenvector corresponding then we have Ax 2 ix Multiply both sides by matriX A then we get AAx Alx 2 Azx le 12x thus 32 is eigenvalue of matriX A2 with the corresponding eigenvector x In the same way we can get the eigenvalue of A2 2A 1 is 32 2 l

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