MATH 304Linear AlgebraLecture 31:Eigenvalues and eigenvectors (continued).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.If λ is an eigenvalue of A then the nullspaceN(A − λI ), which is nontrivial, is called theeigenspace of A corresponding to λ. Theeigenspace consists of all eigenvectors belonging tothe eigenvalue λ plus the zero vector.Characteristic equationDefinition. Given a square matrix A, the equationdet(A − λI ) = 0 is called the characteristicequation of A.Eigenvalues λ of A are roots of the characteristicequation.If A is an n×n matrix then p(λ) = det(A −λI ) is apolynomial of degree n. It is called thecharacteristic polynomial of A.Theorem Any n×n matrix has at most neigenvalues.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.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 eigenvector associatedwith the eigenvalue 1. The 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 eigenvector 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 the 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 the 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 general solution is (−t, t, 0) = t(−1, 1, 0),t ∈ R. Thus v1= (−1, 1, 0) is an eigenvectorassociated with the eigenvalue 0. The correspondingeigenspace is the 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 general 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 the eigenvalue 2.The corresponding 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 subsequent scaling by a factor of 2.Eigenvalues and eigenvectors of an operatorDefinition. Let V be a vector space and L : V → Vbe a linear operator. A number λ is called aneigenvalue of the operator L ifL(v) = λv for anonzero vector 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 the linear operator L is given byL(x) = Ax, where A is an n×n matrix.In this 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 the 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 andonly if Vλ6= {0}.If Vλ6= {0} then it is called the eigenspace of Lcorresponding to the eigenvalue λ.Example. V = C∞(R), D : V → V , Df = f′.A function f ∈ C∞(R) is an eigenfunction of theoperator D belonging to an eigenvalue λ iff′(x) = λf (x) for all x ∈ R.It follows that f (x) = ceλx, where c is a nonzeroconstant.Thus each λ ∈ R is an eigenvalue of D.The corresponding eigenspace is spanned by eλx.Example. V = C∞(R), L : V → V , Lf = f′′.Lf = λf ⇐⇒ f′′(x) − λf (x) = 0 for all x ∈ R.It follows that each λ ∈ R is an eigenvalue of L andthe corresponding eigenspace Vλis two-dimensional.If λ > 0 then Vλ= Span(exp(√λ x), exp(−√λ x)).If λ < 0 then Vλ= Span(sin(√−λ x), cos(√−λ x)).If λ = 0 then Vλ= Span(1, x).Let V be a vector space and L : V → V be a linearoperator.Proposition 1 If v ∈ V is an eigenvector of theoperator L then the associated eigenvalue is unique.Proof: Suppose that L(v) = λ1v and L(v) = λ2v. Thenλ1v = λ2v =⇒ (λ1− λ2)v = 0 =⇒ λ1− λ2= 0 =⇒ λ1= λ2.Proposition 2 Suppose v1and v2are eigenvectorsof L associated with different eigenvalues λ1and λ2.Then v1and v2are linearly independent.Proof: For any scalar t 6= 0 the vector
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