MATH 304Linear AlgebraLecture 21: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.Example. A =2 00 3.2 00 310=20= 210,2 00 30−2=0−6= 30−2.Hence (1, 0) is an eigenvector of A belonging to theeigenvalue 2, while (0, −2) is an eigenvector of Abelonging to the eigenvalue 3.Example. A =0 11 0.0 11 011=11,0 11 01−1=−11.Hence (1, 1) is an eigenvector of A belonging to theeigenvalue 1, while (1, −1) is an eigenvector of Abelonging to the eigenvalue −1.Vectors v1= (1, 1) and v2= (1, −1) form a basisfor R2. Consider a linear operator L : R2→ R2given by L(x) = Ax. The matrix of L with respectto the basis v1, v2is B =1 00 −1.Let A be an n×n matrix. Consider a linearoperator L : Rn→ Rngiven by L(x) = Ax.Let v1, v2, . . . , vnbe a nonstandard basis for Rnand B be the matrix of the operator L with respectto this basis.Theorem The matrix B is diagonal if and only ifvectors v1, v2, . . . , vnare eigenvectors of A.If this is the case, then the diagonal entries of thematrix B are the corresponding eigenvalues of A.Avi= λivi⇐⇒ B =λ1Oλ2...O λnEigenspacesLet 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. 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 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 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. T hus v1= (−1, 1, 0) is an eigenvect orassociated with the eigenvalue 0. The correspondingeigenspace is the line spann ed 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 w ith the eigenvalue 2.The corresponding eigenspace is t he plane spannedby v2and v3.Summary. A =1 1 −11 1 10 0 2.• The matrix A has two eigen values: 0 and 2.• The eigenvalue 0 is simple: the correspondingeigenspace is a line.• The eigenvalue 2 is of multiplicity 2: thecorresponding e igenspace 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 t he subseque nt 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
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