55.3© 2012 Pearson Education, Inc.Eigenvalues and EigenvectorsDIAGONALIZATIONSlide 5.3- 2© 2012 Pearson Education, Inc.DIAGONALIZATION Example 1: Let . Find a formula for Ak, given that , whereand Solution: The standard formula for the inverse of a matrix yields 7241A1APDP1112P5003D2212111PSlide 5.3- 3© 2012 Pearson Education, Inc.DIAGONALIZATION Then, by associativity of matrix multiplication, Again, 211 11 12212()()()11 215012 1103IAPDP PDP PD P P DP PDDPPD P     3 1 2 1 21 21 31()( )IAPDPAPDPPDPPDDPPDP Slide 5.3- 4© 2012 Pearson Education, Inc.DIAGONALIZATION In general, for , A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, that is, if for some invertible matrix P and some diagonal, matrix D.1k 111 215012 110325 3 5 323 25 23 5kkkkkk kkkkkkAPDP  1APDPSlide 5.3- 5© 2012 Pearson Education, Inc.THE DIAGONALIZATION THEOREM Theorem 5: An matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.In fact, , with D a diagonal matrix, if and only if the columns of P and n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P.In other words, A is diagonalizable if and only if there are enough eigenvectors to form a basis of . We call such a basis an eigenvector basis of .nn1APDPnnSlide 5.3- 6© 2012 Pearson Education, Inc.THE DIAGONALIZATION THEOREM Proof: First, observe that if P is any matrix with columns v1, …, vn, and if D is any diagonal matrix with diagonal entries λ1, …, λn, then----(1)while----(2)nn12 1 2vv v v v vnnAP A A A A1211 2 2λ 000 λ 0λ v λ v λ v00 λnnnPD P Slide 5.3- 7© 2012 Pearson Education, Inc.THE DIAGONALIZATION THEOREM Now suppose A is diagonalizable and . Then right-multiplying this relation by P, we have. In this case, equations (1) and (2) imply that----(3) Equating columns, we find that----(4) Since P is invertible, its columns v1, …, vnmust be linearly independent.1APDPAPPD12 1122vv vλ v λ v λ vnnnAA A111 2 22v λ v, v λ v, , v λ vnnnAAA Slide 5.3- 8© 2012 Pearson Education, Inc.THE DIAGONALIZATION THEOREM Also, since these columns are nonzero, the equations in (4) show that λ1, …, λnare eigenvalues and v1, …, vnare corresponding eigenvectors. This argument proves the “only if ” parts of the first and second statements, along with the third statement, of the theorem. Finally, given any n eigenvectors v1, …, vn, use them to construct the columns of P and use corresponding eigenvalues λ1, …, λnto construct D.Slide 5.3- 9© 2012 Pearson Education, Inc.THE DIAGONALIZATION THEOREM By equation (1)–(3), . This is true without any condition on the eigenvectors. If, in fact, the eigenvectors are linearly independent, then P is invertible (by the Invertible Matrix Theorem), and implies that .APPD1APDPAPPDSlide 5.3- 10© 2012 Pearson Education, Inc.DIAGONALIZING MATRICES Example 2: Diagonalize the following matrix, if possible.That is, find an invertible matrix P and a diagonal matrix D such that . Solution: There are four steps to implement the description in Theorem 5. Step 1. Find the eigenvalues of A. Here, the characteristic equation turns out to involve a cubic polynomial that can be factored:133353331A1APDPSlide 5.3- 11© 2012 Pearson Education, Inc.DIAGONALIZING MATRICES The eigenvalues are and . Step 2. Find three linearly independent eigenvectors of A. Three vectors are needed because A is a matrix. This is a critical step. If it fails, then Theorem 5 says that A cannot be diagonalized.3220det( λ ) λ 3λ 4(λ 1)(λ 2)AI  λ 1λ 233Slide 5.3- 12© 2012 Pearson Education, Inc.DIAGONALIZING MATRICES Basis for Basis for and You can check that {v1, v2, v3} is a linearly independent set. 11λ 1: v 1121λ 2:v 10 31v01Slide 5.3- 13© 2012 Pearson Education, Inc.DIAGONALIZING MATRICES Step 3. Construct P from the vectors in step 2. The order of the vectors is unimportant. Using the order chosen in step 2, form Step 4. Construct D from the corresponding eigenvalues. In this step, it is essential that the order of the eigenvalues matches the order chosen for the columns of P.123111vvv 1 1010 1PSlide 5.3- 14© 2012 Pearson Education, Inc.DIAGONALIZING MATRICES Use the eigenvalue twice, once for each of the eigenvectors corresponding to : To avoid computing , simply verify that . Computeλ 2λ 2100020002D1PADPD133111 122353110 120331101 102AP     Slide 5.3- 15© 2012 Pearson Education, Inc.DIAGONALIZING MATRICES Theorem 6: An matrix with n distinct eigenvalues is diagonalizable. Proof: Let v1, …, vnbe eigenvectors corresponding to the n distinct eigenvalues of a matrix A. Then {v1, …, vn} is linearly independent, by Theorem 2 in Section 5.1. Hence A is diagonalizable, by Theorem 5.111100 122110020 120101002 102PD        nnSlide 5.3- 16© 2012 Pearson Education, Inc.MATRICES WHOSE EIGENVALUES ARE NOT DISTINCT It is not necessary for an