LINEAR ALGEBRA MATH 304
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Date Created: 10/21/15
Highlights Reminder The first examination is Friday June 9 Math 304 Linear Algebra From last time gt A set of vectors is linearly independent if none of the vectors is in the span of the others gt A set of vectors is linearly dependent if some nontrivial boastamuedu linear combination of the vectors equals the 0 vector Harold P Boas Today gt the notions of basis and dimension gt changing coordinates June 7 2006 Basis Example exercise 10 on page 151 A basis for a vector space is Extract a basis for BS from the following vectors gt a linearly independent set of vectors that spans the vector 1 2 1 2 1 Space X1 2 7 X2 5 7 X3 3 7 X4 7 7 X5 1 gt equivalently a maximal linearly independent set of vectors 2 4 2 4 o gt equivalently a minimal spanning set for the vector space Solution The vectors x1 and x2 are linearly Independent but Example The standard basis for RS consists of the three they do not form a basis for 93 because they do not span Rs 1 0 The vectors x1 x2 and x3 do not form a basis because they Vedors e1 0 1 e2 1 1 and 93 0 are linearly dependent x3 x2 7 x1 Vectors x1 x2 and x4 do 0 0 1 not form a basis because they are linearly dependent One nonstandard basis for R3 consists of the three vectors 1 2 2 1 2 1 1 4 7 det 2 5 7 0However 2 51 727 0so v1 2 v2 5 andv3 8 244 240 3 6 0 vectors x1 x2 and x5 do form a basis for R3 Dimension The dimension of a vector space is the number of vectors in a basis All bases of a vector space have the same number of vectors Examples gt The dimension of R3 equals 3 gt The dimension of the space HzX5 of 2 x 5 matrices equals gt The dimension of the space P3 of polynomials of degree less than 3 equals 3 One basis is 1 x x2 gt The function space C0 1 is infinite dimensional Any finite subset of the functions 1 x x2 x3 is linearly independent Example continued 5 v i 4 a 2 i 5 How are coordinates with respect to the u basis related to coordinates with respect to the v basis Consider another nonstandard basis v1 The matrix 2 2 whose columns are v1 and v2 transforms v coordinates to standard coordinates and the inverse matrix 7 451 transforms from standard coordinates to v coordinates Since 1 7 transforms from u coordinates to standard coordinates the product matrix 5 74 1 71 7 1 79 76 5 1 1 T 71 11 transforms from u coordinates to v coordinates 1 7 4 13 Sothevcoordlnatesofxare lt71 11 71gt 715gtV Change of basis an example Let u1 and u2 be a nonstandard basis for H2 and let e1 and eg be the standard basis vectors Let x Then x 5e1 3e2 In other words 5 and 3 are the coordinates of the vector x with respect to the standard basis Since x 4u1 7 U2 the coordinates of x with respect to the u basis are 4 and 71 Question How are the u coordinates and the standard coordinates related ltgtx4uu2lt w The transition matrix between u coordinates and standard coordinates is the matrix whose columns are the standard coordinates of the u basis vectors The inverse matrix gives the change from standard coordinates to u coordinates Math 304 Linear Algebra Harold P Boas boastamuedu June 13 2006 What is a linear transformation Matrix multiplication preserves linear combinations namely Ax y Ax Ay and Acx cAx for every scalar c Any W39 a 39 Mi p ut from one vector space to another is called linear if it satisfies the same two properties Highlights From last time gt row space and column space gt the ranknullity property Today gt examples of linear transformations Examples of linear transformations gt Geometric examples in H2 or R3 dilation rotation reflection projection gt Examples in spaces of functions differentiation integration point evaluation gt Examples in the space of matrices taking the transpose taking the trace the sum of the numbers on the main diagonal Nonexamples of linear transformations Subspaces associated to a linear transformation Suppose L V a W is a linear transformation from one vector space V to another vector space W gt The kernel of L is the set of vectors v in V such that Lv 0 Example If L P5 a H is defined by Lp p0 then the kernel of L is the set of polynomials with constant term 0 The kernel is essentially the same concept as the gt Function spaces squaring a function nullspace of a matrix gt Matrices taking the inverse The kernel is always a subspace of V L is onetoone or infective if kerL 0 The range L V is the set of all vectors w in W such that w Lv for some v in V Example If L P5 a P5 is differentiation then the range of L is P4 The range is always a subspace of W The range is analogous to the column space of a matrix L is onto or surjective if L V W gt Geometric translation V Math 304 Linear Algebra Harold P Boas boastamuedu June 12 2006 Example 1 21 21 LetA 2 4 3 71 4 8 511 3 Threepart problem Find a basis for gt the row space the subspace of H5 spanned by the rows of the matrix gt the column space the subspace of RS spanned by the columns of the matrix gt the nuspace the set of vectors x such that Ax 0 We can analyze all three parts by Gaussian elimination even though row operations change the column space Highlights From Wednesday gt basis and dimension gt transition matrix for change of basis Today gt row space and column space gt the ranknullity property Example continued 1 2 4 212111721112 4371oo 85113R3 4R 00 R7R12121R7R12 001371 gt00 00000 00 1 1 1 0 71 2 1 3 71 0 0 0 Notice that in each step the second column is twice the first column also the second column is the sum of the third and fifth columns Row operations changethe column space but preserve linear relations among the columns The final rowechelon form shows that the column space has dimension equal to 2 and the first and third columns are linearly independent Example interpretation 12121 120712 24371LWgt001371 4 8 5113 Peraquot quots 0 0 0 0 0 gt The dimension of the row space called the rank is 2 gt The dimension of the column space also equals the rank Both equal the number of lead 1 s in the row echelon form gt One basis for the row space is the pair of vectors 120 7127 and 0013e1T gt One basis for the column space is the first and third 1 1 columns ofthe originalmatrix namely 2 and 3 4 5 Rank and nullity The example illustrates the following general principles gt rank dimension of row space dimension of column space number of lead 1 s in row echelon form gt nullity dimension of nullspace number of free variables in row echelon form gt rank nullity number of columns in matrix Example nullspace 1 2 1 2 1 1 2 0 71 2 2 4 3 7 1 0 0 1 3 e1 4 8 5 11 3 Peraquot quots 0 0 0 0 0 The lead variables are X1 and X3 The other variables X2 X4 and X5 are free variables they can take arbitrary values The nullspace consists of vectors of the form 2X2 X4 7 2X5 2 1 2 X2 1 0 0 73x4 X5 x2 0 x4 73 x5 1 X4 0 1 0 X5 0 0 1 The nullspace has dimension 3 and one basis is the vectors 721000T 107310T and 720101T Math 304 Linear Algebra Harold P Boas boastamuedu June 22 2006 Example in R3 1 Letu1 1 u2 7 andu3 1 sealer 2 W 7 These vectors form an orthonormal set each vector is orthogonal to the others and each vector has norm equal to 1 The matrix U whose columns are U1 U2 and us is called an orthogonal matrix Notice that UTU lthe identity matrix so UT Uquot This property characterizes orthogonal matrices An orthogonal matrix preserves the scalar product namely ltUx Uygt ltxygt Here is why am7UvgtUnTUvxTUTUyxTvltnv Therefore an orthogonal matrix also preserves the norm llUXll llel Highlights From last time gt inner products and norms Today gt orthonormal sets Example continued Lety2 0 73 7 2u1i3u24u3 1 m T l k 4 assr Problem Determine llyll Solution Since y U 73 and since multiplication by an b N V orthogonal matrix preserves the norm llyll 4916 29 2 73 4 In general llc1u1 02U2 Csuallz C12 Cg The analogous statement for an arbitrary orthonormal set is Parseval s formula More on the example Fourier coefficients Projection and approximation Problem Express the vector v 1 2 3T as a linear combination of u1 U2 and US In other words find coefficients Cf 0239 and CS SUCh that V Cf 1 C2 2 CS 3 or Problem Find the projection of the vector v 211T onto the plane spanned by u1 and U2 L 1 2 r 26 g Cl 1 CZ 7 39 Solution We can write i W W V ltV7U1gtU1ltV7U2gtU2 ltV7U3gtU37 I Solution No row reduction necessary Take the Inner product but Us is orthogonal to the plane so the projection equals with u1 thus v U1gt lt01u1 02u2 Q3U3 U1gt c1 by orthonormality so 01 72 Similarly 02 ltV7 U2gt 2 and Cs ltV7U3gt 8 ltvu1gtu1ltvu2gtU2 lt In general any vector v can be represented in terms of an orthonormal basis u1 U2 and us via cal minle V Mid ONl V l l ml Noni mi N l 4 x v ltvu1gtu1ltvu2gtu2 v u3gtu3 Math 304 Linear Algebra Harold P Boas boastamuedu June 28 2006 When is a matrix diagonal Suppose a linear operator L on BS is represented in a basis 2 0 uh uzu3 by the matrix 0 3 0 0 5 This means that Lu1 2u1 and Lug 3u2 and Lu3 5u3 In other words the basis vectors are eigenvectors of the operator L A square matrix A is diagonalizabe if the linear transformation x i gt Ax can be represented in some basis by a diagonal matrix in other words if there is a basis consisting of eigenvectors of A equivalently if there is an invertible matrix S such that SquotAS is a diagonal matrix Highlights From last time gt application of eigenvectors to systems of differential equations Today gt diagonalization of matrices and applications Example exercise 1b p 340 i i In other words find an invertible matrix S and a diagonal matrix D such that SquotAS D or A SDSquot Solution First find the eigenvalues and eigenvectors of A The Diagonalize the matrix A vector is an eigenvectorwith eigenvalue 1 and the vector is an eigenvectorwith eigenvalue 2 The matrix 3 2 S 72 71 Is the transttion matrix from the eigenvector basis to the standard basis and the matrix SquotAS is the 1 0 diagonal matrix 0 2 Continuation i 5 6 10 IfA7lt72 72gtfindA 1 0 1 0 1 7 7 10 7 Solution Since 8 AS 7 D 7 lt0 2 and D 7 0 210 we have 3 2 1 o 7172 10 10 71 A SDS 4 71gtlt0210gtlt2 3 732 2 763x2 27211 473x21039 More Since the exponential function is given by a power series ex 1 X x2x3xquotwealso have 9quot IA A2Se08 1 3 2 e 0 71 72 lt72 71 lt0 e2gtlt 2 3 73e 4e2 76e 6e2 lt 2e7 292 4e7 392 Application to differential equations We have two ways to solve the system of differential equations 5 6 7 From yesterday we can write the general solution as 3 2 W 016172 0292 71 Alternatively we can write the general solution as tA017t07101 yt7e CZ ise S 02 73et4e2t 76et6e2t c1 2e 72e2t 4e 73e2t Cg 39 In the second form C1 y1 0 02 Y20
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