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e-vectors, e-valuesDefinition, exampleDiagonalizabilitye-vectors, e-valuesLecture 2: IntroductionAmos RonUniversity of Wisconsin - MadisonJanuary 27, 2021Amos Ron CS513, remote learning S21e-vectors, e-valuesOutline1e-vectors, e-valuesDefinition, exampleDiagonalizabilityAmos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityOutline1e-vectors, e-valuesDefinition, exampleDiagonalizabilityAmos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityEigenpairsDefinitionA m × m. (λ, v), λ ∈ C, v ∈ Cm\0, is eigenpair of A, ifAv = λv.The set of all eigenvalues is the spectrumσ(A)of A.Note: A real valued matrix might have complex eigenvalues!Reminder: The characteristic polynomial of A ispA(t) := det(A − tI).λ ∈ σ(A) ⇐⇒ pA(λ) = 0.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityEigenpairsThe eigenvectors of the matrix A is any linearly independentmaximal set of eigenvectors. The cardinality of such setdepends only on A.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityEigenpairsA =2 21 3.Then:pA(t) = (t − 2)(t − 3) − 2 = t2− 5t + 4This implies that σ(A) = {1, 4}.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityEigenpairsA =2 21 3.Then:The matrix A − 4I must be singular. Indeed,A − 4I =−2 21 −1Every non-zero vector in ker(A − 4I) is an eigenvector.Since dim ker(A − 4I) = 1, we select only one eigenvectorfrom this null space. For example, we can choose v1= ....Then, (4, v1) is an eigenpair.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityEigenpairsA =2 21 3.Then:The matrix A − I must be singular. Indeed,A − I =1 21 2Every non-zero vector in ker(A − I) is an eigenctor. Sincedim ker(A − I) = 1, we select only one eigenvector from thisnull space. For example, v2= .... Then, (1, v2) is aneigenpair.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityEigenpairsA =2 21 3.Then:Since eigenvectors associated with different e-values arealways linearly independent, (v1, v2) are independent,hence form a basis for R2.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityDiagonalizabilityDefinition: DiagonalizabilityA m × m is diagonalizable is there exists a basis for Cmmade ofe-vectors of AAmos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityDiagonalizabilityTheorem:A is square. TFCAE:1A is diagonalizable2There exist a matrix P and a diagonal matrix D such thatA = PDP−1.3Another equivalent condition deferred.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityProof of Theorem(2) =⇒ (1): We have AP = PD. We prove that each column ofP is an eigenvector of A. This proves (1), since the columns ofany m × m invertible matrix form a basis for Cm.The jth column of P is Pej. Now:A(Pej) = (AP)ej= (PD)(ej) = P(Dej) = P(D(j, j)ej) = D(j, j)(Pej).So, (D(j, j), Pej) is an eigenpair of A.Amos Ron CS513, remote learning S21e-vectors, e-valuesDefinition, exampleDiagonalizabilityProof of Theorem(1) =⇒ (2) We are given m eigenpairs (λj, vj), with (v1, . . . , vm)a basis for Cm. Let P be the matrix whose columns arev1, . . . , vj, and let D be the diagonal matrix whose diagonal isλ1, . . . , λm. We show that A = PDP−1by showing that AP = PD,i.e., by showing that, for every j,(AP)ej= (PD)ej.Now,(AP)ej= A(Pej) = Avj= λjvj= P(λjej) = P(Dej) = (PD)ej.Amos Ron CS513, remote learning

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