Lecture 9 eigenvalues - 5-1 & 5-2

download Lecture 9   eigenvalues -  5-1 & 5-2

of 17

  • date post

    15-Jul-2015
  • Category

    Education

  • view

    45
  • download

    4

Embed Size (px)

Transcript of Lecture 9 eigenvalues - 5-1 & 5-2

Slide 1

2012 Pearson Education, Inc.Eigenvalues and EigenvectorsEIGENVECTORS AND EIGENVALUES55.11Slide 5.1- 2 2012 Pearson Education, Inc.EIGENVECTORS AND EIGENVALUESDefinition: An eigenvector of an nxn matrix A is a nonzero vector x such that Ax=x for some scalar . A scalar is called an eigenvalue of A if there is a nontrivial solution x of Ax=x; such an x is called an eigenvector corresponding to .

Eigenvectors may NOT be 0Eigenvalue may be 0

2Eigenvectors - ExampleSlide 5.1- 3 2012 Pearson Education, Inc.Finding Eigenvectors - ExampleShow that 7 is an eigenvalue of A from previous example and find corresponding eigenvectors.Slide 5.1- 4 2012 Pearson Education, Inc.EigenspaceGeneral solution of Ax=x is same as Nul space of (A I)Subspace of RnCalled EigenspaceContains 0 and all eigenvectorsSlide 5.1- 5 2012 Pearson Education, Inc.Finding Basis for Eigenspace - ExampleSlide 5.1- 6 2012 Pearson Education, Inc.Slide 5.1- 7 2012 Pearson Education, Inc.Geometric InterpretationThe eigenspace, shown in the following figure, is a two-dimensional subspace of R3.

Slide 5.1- 8 2012 Pearson Education, Inc.EIGENVALUES of Triangular MatrixTheorem 1: The eigenvalues of a triangular matrix are the entries on its main diagonal.Proof: For simplicity, consider the 3x3 case.If A is upper triangular, the (A I) has the form

Slide 5.1- 9 2012 Pearson Education, Inc.EIGENVALUES of Triangular MatrixThe scalar is an eigenvalue of A if and only if the equation (A I) = 0 has a nontrivial solution, i.e., iff the equation has a free variable.

Because of the zero entries in (A I), it is easy to see that (A I) = 0 has a free variable if and only if at least one of the entries on the diagonal of A I is zero. This happens if and only if equals one of the entries a11, a22, a33 in A. Slide 5.1- 10 2012 Pearson Education, Inc.EIGENVECTORS Linear IndependenceTheorem 2: If v1, , vr are eigenvectors that correspond to distinct eigenvalues 1, , r of annxn matrix A, then the set {v1, , vr} is linearly independent. 0 as EigenvalueIf 0 is an eigenvalue:Ax=0x=0 has non-trivial solutionA is NOT invertible

Additions to IMT:(s) 0 is not an eigenvalue of A(t) |A| 0Slide 5.1- 11 2012 Pearson Education, Inc. 2012 Pearson Education, Inc.Eigenvalues and EigenvectorsTHE CHARACTERISTIC EQUATION55.112Finding Eigenvalues - ExampleSlide 5.1- 13 2012 Pearson Education, Inc.Characteristic Equationcharacteristic equation: |A I| = 0Scalar equationDegree n, where A is nxnSolutions are eigenvalues of ASlide 5.1- 14 2012 Pearson Education, Inc.Slide 5.2- 15 2012 Pearson Education, Inc.THE CHARACTERISTIC EQUATIONExample 2: Find the characteristic equation of

Slide 5.2- 16 2012 Pearson Education, Inc.SIMILARITYIf A and B are nxn matrices, then A is similar to B if there is an invertible matrix P such that P-1AP = B, or, equivalentlyA = PBP-1

Changing A into B is called a similarity transformation.Slide 5.2- 17 2012 Pearson Education, Inc.SIMILARITYTheorem 4: If nxn matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).