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by: Evert Christiansen


Evert Christiansen
Texas A&M
GPA 3.92

Yaroslav Vorobets

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Yaroslav Vorobets
Study Guide
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This 17 page Study Guide was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Study Guide belongs to MATH 304 at Texas A&M University taught by Yaroslav Vorobets in Fall. Since its upload, it has received 51 views. For similar materials see /class/226002/math-304-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15
MATH 304 Linear algebra Lecture 39 Review for the final exam Topics for the final exam Part I 0 Systems of linear equations elementary operations Gaussian elimination back substitution 0 Matrix of coefficients and augmented matrix Elementary row operations row echelon form and reduced row echelon form 0 Matrix algebra Inverse matrix 0 Determinants explicit formulas for 2x2 and 3x3 matrices row and column expansions elementary row and column operations Topics for the final exam Part II 0 Vector spaces vectors matrices polynomials functional spaces 0 Subspaces Nullspace column space and row space of a matrix 0 Span spanning set Linear independence 0 Bases and dimension Rank and nullity of a matrix 0 Change of coordinates transition matrix 0 Linear mappingstransformationsoperators Range and kernel of a linear mapping 0 Matrix of a linear mapping Similar matrices Topics for the final exam Part Ill 0 Norms Inner products 0 Orthogonality Orthogonal complement 0 Least squares problems 0 Orthogonal and orthonormal bases The Gram Schmidt orthogonalization process 0 Eigenvalues eigenvectors and eigenspaces Characteristic polynomial o Bases of eigenvectors Diagonalization Topics for the final exam Part IV 0 Matrix exponentials 0 Complex eigenvalues and eigenvectors Symmetric matrices o Orthogonal matrices Rotations in space 0 Orthogonal polynomials Bases of eigenvectors Let A be an ngtltn matrix with real entries 0 A has n distinct real eigenvalues gt a basis for R formed by eigenvectors of A o A has complex eigenvalues gt no basis for R formed by eigenvectors of A o A has n distinct complex eigenvalues gt a basis for CC formed by eigenvectors of A o A has multiple eigenvalues gt further information is needed 0 an orthonormal basis for R formed by eigenvectors of A ltgt A is symmetric AT A Problem For each of the following matrices determine whether it allows a a basis of eigenvectors for R b a basis of eigenvectors for C c an orthonormal basis of eigenvectors for R A 1 2 agtbcgt yes 8 8 3 agtbgtcgt no Problem For each of the following matrices determine whether it allows a a basis of eigenvectors for R b a basis of eigenvectors for C c an orthonormal basis of eigenvectors for R 66 abgtyes c no 03 3 mes axe no Problem Let V be the vector space spanned by functions f1X xsinX f2X xcosx f3X sin X and f4X cosx Consider the linear operator D V gt V D ddx a Find the matrix A of the operator D relative to the basis 137137137151 b Find the eigenvalues of A C Is the matrix A diagonalizable in R4 in C4 A is a 4x4 matrix whose columns are coordinates of functions Dfi 17 relative to the basis 137137137151 f1X xsinX xcosx sinX f2X f3X f2X xcosx Xsinxcosx f1X f4X f3X sin X cosx f4X x cosx sinX f3X 0 1 1 0 Thus A i 1 0 0 1 I OOO OI OO Eigenvalues of A are roots of its Characteristic polynomial x 1 0 0 1 x 0 0 detA 1 0 1 0 1 1 x Expand the determinant by the lst row A 0 0 1 0 0 detA 0 x 1 1 1 x 1 1 1 x 0 1 x A20 1 A2 1 A2 12 The eigenvalues are i and i both of multiplicity 2 Complex eigenvalues gt A is not diagonalizable in R4 lfA is diagonalizable in C4 then A UXU l where U is an invertible matrix with complex entries and i0 00 0 0 0 x0070 00 04 This would imply that A2 UXZU l But X2 7 so that Nw4w44 0710 0 4 0 0 0 N71 00 0 i 0710 0 1 0071 0724 0 0 11 0 2 0 071 Since A2 31 7 the matrix A is not diagonalizable in C4 Problem Consider a linear operator L R3 gt R3 defined by Lv v0 x v where V0 35 0 45 a Find the matrix B of the operator L b Find the range and kernel of L C Find the eigenvalues of L d Find the matrix of the operator L2008 L applied 2008 times Lv v0 x v v0 3570 45 Let v X7y72 xe1ye2 ze3 Then e1 82 83 Lv v0 x v 35 0 45 X y z yel x 062 Eyes In particular Le1 ge2 Le2 gel ge3 Le3 e2 0 45 0 Therefore B 45 0 35 0350 0 45 0 B 45 0 35 0 35 0 The range of the operator L coincides with the column space of the matrix B It follows that RangeL is the plane spanned by vectors v1010 and v2 4 0 3 The kernel of L is the solution set for the equation Bx0 0 45 0 1 0 34 45 0 35 gt 0 1 0 0 35 0 0 0 0 gt xzy0 gt Xt 347071 Alternatively the kernel of L is the set of vectors v E R3 such that Lv v0 x v 0 It follows that kerL is the line spanned by v0 35 0 45 Characteristic polynomial of the matrix B A 45 0 detB 45 A 35 0 35 A 3 352 452 3 21 The eigenvalues are 0 i and i The matrix of the operator L2008 is B2008 Since the matrix B has eigenvalues 0 i and i it is diagonalizable in C3 Namely B UDU l where U is an invertible matrix with complex entries and 000 D0i0 00 i Then B2008 UD2008U 1 We have that D2008 d1ag0i2008 i2008 diag011 D2 Hence 064 0 048 B2008 U D2U 1 B2 0 1 0 048 0 036


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