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1、Chapter 6EIGENVALUESChapter Overview1. Eigenvalues and Eigenvectors Eigenvalues and Eigenvectors 2. Systems of Linear Differential Equations Systems of Linear Differential Equations3. Diagonalization Diagonalization5. The Singular Value Decomposition5. The Singular Value Decomposition6. Quadratic Fo
2、rms Quadratic Forms7. Positive Definite Matrices Positive Definite Matrices8. Nonnegative Matrices Nonnegative Matrices 2Chapter 6 EigenvaluesChapter 6 Eigenvalues6.3. DiagonalizationIn this sectionIn this section We consider the problem of factoring an n n matrix A into a product of the form XDX-1,
3、 where D is diagonal; We will give a necessary and sufficient condition for the existence of such a factorization and look at a number of examples.3 Theorem 6.3.1Theorem 6.3.1 If 1, 2, , k are distinct eigenvalues of an n n matrix A with corresponding eigenvectors x1, x2, , xk, then x1, x2, , xk are
4、 linearly independent. 6.3.1 Linearly Independent Eigenvectors6.3 Diagonalization6.3 Diagonalization4ProofProofLet r = dim(Span(x1, x2, xk), and assume that r k.( Implies that there are at most r linearly independent vectors in x1, x2, xk ). Assume that x1, x2, xr are linearly independent without lo
5、ss of generality.Since x1, x2, xr, xr+1 are linearly dependent, there exists scalars c1, , cr, cr+1 not all zeros such thatc1x1+ + crxr + cr+1xr+1 = 0(1)6.3 Diagonalization6.3 Diagonalization5(2)Note that cr+1 cannot be 0 (if cr+1 =0, then x1, , , xr are l.d.). So cr+1xr+1 0, and hence c1, , , cr ca
6、nnot all be 0.Multiplying both sides of (1) by A, we havec1Ax1+ + crAxr + cr+1Axr+1 = 0 c11x1+ + crrxr + cr+1r+1xr+1 = 0.Multiplying (1) by r+1 and subtracting it from (2), we obtainc1(1 r+1)x1+ + cr(r r+1)xr = 0. 6.3.1 Linearly Independent Eigenvectors6.3 Diagonalization6.3 DiagonalizationProof (co
7、ntinued)Proof (continued)6Since 1, 2, , k are distinct eigenvalues, then x1, , , xr becomes linearly dependent.So it is a contradiction to the assumption thatr = dim Span(x1, , , xk) k.Hence r = k and x1, , , xk are linearly independent. 6.3.1 Linearly Independent Eigenvectors6.3 Diagonalization6.3
8、DiagonalizationProof (continued)Proof (continued)7 Definition DefinitionAn n n matrix A is said to be diagonalizable diagonalizable if there exists a nonsingular matrix X and a diagonal matrix D such thatX-1AX = Dwe say that X diagonalizesdiagonalizes A. Remarks Remarks If A is diagonalizable by X,
9、the column vectors of X areeigenvectors of A, and the diagonal elements of D arethe eigenvalues of A. The diagonalizing matrix X is not unique. 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 Diagonalization8 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 Diagonalization Theorem 6.3.2: Theo
10、rem 6.3.2:An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.9ProofProofSuppose that A has n linearly independent eigenvectors x1, x2 , , xn . Let i be the eigenvalue of A corresponding to xi for each i (some of the is may be equal). Let X = (x1, x2, , , xn),
11、then jxj is the jth column vector of AX. Therefore,6.3 Diagonalization6.3 DiagonalizationSince X has n linearly independent column vectors, it follows that X is nonsingular and hence10#Conversely, suppose that A is diagonalizable, then there exists a nonsingular matrix X such that AX = XD.If x1, x2,
12、 , , xn are the column vectors of X, thenThus, for each j, j is an eigenvalue of A and xj ( 0) is an eigenvector belonging to j.Since X is nonsingular, A has n linearly independent engienvectors. 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 DiagonalizationProof (continued)Proof (continued)11
13、If A is diagonalizable, then the column vectors of the diagonalizing matrix X areeigenvectors of A the diagonal elements of D are the correspondingeigenvalues of A The diagonalizing matrix X is not unique. Reorderingthe columns of a given diagonalizing matrix X ormultiplying them by nonzero scalars
14、will produce anew diagonalizing matrix. 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 Diagonalization12 If A is n n and A has n distinct eigenvalues, then Ahas n linearly independent eigenvectors, and henceA is diagonalizable. If the eigenvalues are not distinct, then A may or maynot be diagon
15、alizable depending on whether A has nlinearly independent eigenvectors. A has n linearly independent eigenvectors ifdimension of N(A- I) = multiplicity of for all repeated eigenvalues. 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 Diagonalization13 If A is diagonalizable, then A can be factore
16、d into a product XDX-1. Therefore, A2 = (XDX1)(XDX1) = XD2X1in general, 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 Diagonalization14 Example 1: Example 1: Factor the matrix A into XDX-1, where D is diagonal: Some Examples6.3 Diagonalization6.3 Diagonalization15SolutionSolutionThe eigenvalue
17、s of A are 1=1 and 2= - -4. Corresponding to 1 and 2, we have eigenvectors x1=(3,1)T and x2=(1,2)T. Letthenand6.3 Diagonalization6.3 Diagonalization16 Example 2: Example 2: Factor the matrix A into XDX-1, where D is diagonal: Some Examples6.3 Diagonalization6.3 DiagonalizationSolution Solution (P328
18、)(P328)17 Some Examples6.3 Diagonalization6.3 Diagonalization Example 3: Example 3: Factor the matrix A into XDX-1, where D is diagonal:Solution (P328)Solution (P328)18 If an n n matrix A has fewer than n linearly independent eigenvectors, we say that A is defectivdefective. A defective matrix is no
19、t diagonalizable.The matrixhas two distinct eigenvalues 1 = 4, 2 = 3 = 2. The eigenspace of 1 is Span(e2) and the eigenspace of 2 and 3 is Span(e3). A is a defective matrix. 6.3.2 Diagonalizable Matrices6.3 Diagonalization6.3 Diagonalization Example 4: Example 4:19 6.3.3 Geometric Multiplicity of Ei
20、genvalues6.3 Diagonalization6.3 Diagonalization Consider the eigenvectors of the matrices20 eigenvalues: 1=4, 2= 3=2eigenspaces: 1:e2, 2, 3:e3 =2 has geometric multiplicity 1 1=4, 2= 3=2 1:x1=(0,1,3)T, 2, 3:x2=(2,1,0)T, e3 geometric multiplicity 2 6.3.3 Geometric Multiplicity of Eigenvalues6.3 Diago
21、nalization6.3 Diagonalization21 Given a scalar a, the exponential ea can be expressed in terms of a power series Given an n n matrix A, we can define the matrix exponential eA in terms of the convergent power series In case of a diagonal matrix: 6.3.4 Exponential of a Matrix6.3 Diagonalization6.3 Di
22、agonalization22 If the n n matrix A is diagonalizable, then Example 5: Example 5: compute eA forSolution Solution (P338)(P338) 6.3.4 Exponential of a Matrix6.3 Diagonalization6.3 Diagonalization23 The matrix exponential can be applied to the initial value initial value problemproblem The solution to
23、 the initial value problem is simply If A is a diagonal matrix: 6.3.4 Exponential of a Matrix6.3 Diagonalization6.3 Diagonalization24 Example 6: Example 6: Use the matrix exponential to solve the initial value problem Solution Solution (P339)(P339) 6.3.4 Exponential of a Matrix6.3 Diagonalization6.3
24、 Diagonalization25 Example 7: Example 7: Use the matrix exponential to solve the initial value problem Solution Solution (P340)(P340) 6.3.4 Exponential of a Matrix6.3 Diagonalization6.3 DiagonalizationSince the matrix A is defective, we will use the definition of the matrix exponential to compute etA. Note that A3=O, sothe solution is given by26谢谢观赏！谢谢观赏！
