MATH 304Linear AlgebraLecture 22:Eigenvalues and eigenvectors (continued).Characteristic polynomial.Eigenvalues and eigenvectors of a matrixDefinition. Let A be an n×n matrix. A numberλ ∈ R is called an eigenvalue of the matrix A ifAv = λv for a nonzero column vector v ∈ Rn.The vector v is called an eigenvector of Abelonging to (or associated with) the eigenvalue λ.Remarks. • Alternative notation:eigenvalue = characteristic value,eigenvector = characteristic vector.• The zero vector is never considered aneigenvector.Diagonal matricesLet A be an n×n matrix. Then A is diagonal if andonly if vectors e1, e2, . . . , enof the standard basisfor Rnare eigenvectors of A.If this is the case, then the diagonal entries of thematrix A are the corresponding eigenvalues:A =λ1Oλ2...O λn⇐⇒ Aei= λieiEigenspacesLet A be an n×n matrix. Let v be an eigenvectorof A belonging to an eigenvalue λ.Then Av = λv =⇒ Av = (λI )v =⇒ (A − λI )v = 0.Hence v ∈ N(A − λI ), the nullspace of the matrixA − λI .Conversely, if x ∈ N(A − λI ) then Ax = λx.Thus the eigenvectors of A belonging to theeigenvalue λ are nonzero vectors from N(A − λI ).Definition. If N(A − λI ) 6= {0} then it is calledthe eigenspace of the matrix A corresponding tothe eigenvalue λ.How to find eigenvalues and eigenvectors?Theorem Given a square matrix A and a scalar λ,the following statements are equivalent:• λ is an eigenvalue of A,• N(A − λI ) 6= {0},• the matrix A − λI is singular,• det(A − λI ) = 0.Definition. det(A − λI ) = 0 is called thecharacteristic equation of the matrix A.Eigenvalues λ of A are roots of the characteristicequation. Associated eigenvectors of A are nonzerosolutions of the equation (A − λI )x = 0.Example. A =a bc d.det(A − λI ) =a − λ bc d − λ= (a − λ)(d − λ) − bc= λ2− (a + d)λ + (ad − bc).Example. A =a11a12a13a21a22a23a31a32a33.det(A − λI ) =a11− λ a12a13a21a22− λ a23a31a32a33− λ= −λ3+ c1λ2− c2λ + c3,where c1= a11+ a22+ a33(the trace of A),c2=a11a12a21a22+a11a13a31a33+a22a23a32a33,c3= det A.Theorem. Let A = (aij) be an n×n matrix.Then det(A − λI ) is a polynomial of λ of degree n:det(A − λI ) = (−1)nλn+ c1λn−1+ · · · + cn−1λ + cn.Furthermore, (−1)n−1c1= a11+ a22+ · · · + annand cn= det A.Definition. The polynomial p(λ) = det(A − λI ) iscalled the characteristic polynomial of the matrix A.Corollary Any n×n matrix has at most neigenvalues.Example. A =2 11 2.Characteristic equation:2 − λ 11 2 − λ= 0.(2 − λ)2− 1 = 0 =⇒ λ1= 1, λ2= 3.(A − I )x = 0 ⇐⇒1 11 1xy=00⇐⇒1 10 0xy=00⇐⇒ x + y = 0.The general solution is (−t, t) = t(−1, 1), t ∈ R.Thus v1= (−1, 1) is an eigenvec tor associatedwith the eigenvalue 1. Th e correspondingeigenspace is the line spanned by v1.(A − 3I )x = 0 ⇐⇒−1 11 −1xy=00⇐⇒1 −10 0xy=00⇐⇒ x − y = 0.The general solution is (t, t) = t(1, 1), t ∈ R.Thus v2= (1, 1) is an eigenvec tor associated withthe eigenvalue 3. The corresponding eigenspace isthe line spanned by v2.Summary. A =2 11 2.• The matrix A has two eigenvalues: 1 and 3.• The eigenspace of A associated with theeigenvalue 1 is the line t(−1, 1).• The eigenspace of A associated with theeigenvalue 3 is the line t(1, 1).• Eigenvectors v1= (−1, 1) and v2= (1, 1) ofthe matrix A form an orthogonal basis for R2.• Geometrically, the mapping x 7→ Ax is a stretchby a factor of 3 away from the line x + y = 0 inthe orthogonal direction.Example. A =1 1 −11 1 10 0 2.Characteristic equation:1 − λ 1 −11 1 − λ 10 0 2 − λ= 0.Expand th e determinant by the 3rd row:(2 − λ)1 − λ 11 1 − λ= 0.(1 − λ)2− 1(2 − λ) = 0 ⇐⇒ −λ(2 − λ)2= 0=⇒ λ1= 0, λ2= 2.Ax = 0 ⇐⇒1 1 −11 1 10 0 2xyz=000Convert th e matrix to reduced row echelon form:1 1 −11 1 10 0 2→1 1 −10 0 20 0 2→1 1 00 0 10 0 0Ax = 0 ⇐⇒x + y = 0,z = 0.The gene ral solution is ( −t, t, 0) = t(−1, 1, 0),t ∈ R. T hus v1= (−1, 1, 0) is an eigenvectorassociated with t he eigenvalue 0. Thecorresponding eigenspace is th e line spanned by v1.(A − 2I )x = 0 ⇐⇒−1 1 −11 −1 10 0 0xyz=000⇐⇒1 −1 10 0 00 0 0xyz=000⇐⇒ x − y + z = 0.The gene ral solution is x = t − s, y = t, z = s,where t, s ∈ R. Equivalently,x = (t − s, t, s) = t(1, 1, 0) + s(−1, 0, 1).Thus v2= (1, 1, 0) and v3= (−1, 0, 1) areeigenvectors associated with th e eigenvalue 2.The c orresponding eigenspace is the plane spannedby v2and v3.Summary. A =1 1 −11 1 10 0 2.• The matrix A has two eigenvalues: 0 and 2.• The eigenvalue 0 is simple: the correspondingeigenspace is a line.• The eigenvalue 2 is of multiplicity 2: thecorresponding eigenspace is a plane.• Eigenvectors v1= (−1, 1, 0), v2= (1, 1, 0), andv3= (−1, 0, 1) of the matrix A form a basis for R3.• Geometrically, the map x 7→ Ax is the projectionon the plane Span(v2, v3) along the lines parallel tov1with the sub sequent scaling by a factor of 2.Eigenvalues and eigenvectors of an operatorDefinition. Let V be a vector space and L : V → Vbe a line ar operator. A number λ is called aneigenvalue of the operator L ifL(v) = λv for anonzero vect or v ∈ V . The vector v is called aneigenvector of L associated with the eigenvalue λ.(If V is a functional space then eigenvectors are alsocalled eigenfunctions.)If V = Rnthen t he linear operator L is given byL(x) = Ax, where A is an n×n matrix.In th is case, eigenvalues and eigenvectors of theoperator L are precisely eigenvalues andeigenvectors of the matrix A.EigenspacesLet L : V → V be a linear operator.For any λ ∈ R, let Vλdenotes th e set of allsolutions of the equation L(x) = λx.Then Vλis a subspace of V since Vλis the kernelof a linear operator given by x 7→ L(x) − λx.Vλminus the zero vector is the set of alleigenvectors of L associated with the eigenvalue λ.In particular, λ ∈ R is an eigenvalue of L if
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