Review• If Ax = λx, then x is an eigen vec tor of A with eigenvalue λ.• EG: x =1−2is an eigenvector of A =0 −2−4 2with eigenvalue 4because Ax =0 −2−4 21−2=4−8= 4x• Multiplicat ion with A =0 11 0is reflection through the line y = x.◦ A11= 1 ·11So: x =11is an eigenvector with eigenvalue λ = 1.◦ A−11=1−1= −1 ·−11So: x =−11is an eig envector with eigenvalue λ = −1.Example 1. Use your geometric understanding to findthe e igenvectors and eigenvalues ofA =1 00 0.Solution. Axy=x0i.e. multiplicati on with A is projection onto the x-axis.• A10= 1 ·10So: x =10is an eigenvector with eige nvalue λ = 1.• A01=00= 0 ·01So: x =01is an eigenvector with eige nvalue λ = 0.Armin [email protected] 2. Let P be the projection matrix corresponding to orthogonal projectiononto t he subspaceV . What are the eigenvalues and eigenvectors of P ?Solution.•For every vector x in V , Px = x.These are the eigenvectors with eigenvalue1.• For every vector x orthogonal to V , Px = 0.These are the eigenvectors with eigenvalue0.Definition 3. Given λ, the se t of all e igenvectors with eigenva lue λ is c a lled theeigenspace ofA corresponding to λ.Example 4. (continued) We saw that the projection matrix P has the two eigenvaluesλ = 0, 1.• The eigenspace of λ = 1 is V .• The eigenspace of λ = 0 is V⊥.How to solve Ax = λxKey observation:Ax = λxAx − λx = 0(A − λI)x = 0This has a nonzero solutiondet (A − λI) = 0Recipe. To fin d eigenvect ors and eigenvalues of A.• First, find the eigenval ues λ using:λ is an eigenvalue o f Adet (A − λI) = 0• Then , for each eigenvalue λ, find corresponding eigenvectors bysolving(A − λI)x = 0.Example 5. Find the eigenvectors and eigenvalues ofA =3 11 3.Armin [email protected].• A − λI =3 11 3− λ1 00 1=3 − λ 11 3 − λ• det (A − λI) =3 − λ 11 3 − λ= (3 − λ)2− 1= λ2− 6λ + 8 = 0λ1= 2, λ2= 4This is the characteristic polynomial of A. Its roots are the eigenvalues of A.• Find eigenvector s with eigenvalue λ1= 2:A − λ1I =1 11 1(A =3 11 3)Solutions to1 11 1x = 0have basis−11.So: x1=−11is an eigenvector with eigenvalue λ1= 2.All other eige nvectors withλ = 2 are multiples of x1.spann−11ois the eigenspace f or eigenvalue λ = 2.• Find eigenvector s with eigenvalue λ2= 4:A − λ2I =−1 11 −1(A =3 11 3)Solutions to−1 11 −1x = 0have basis11.So:x2=11is an eigenvector w ith eigenvalue λ2= 4.The eigenspace for ei genvalue λ = 4 is spann11o.Example 6. Find the eigenvectors and the eigenvalues ofA =3 2 30 6 100 0 2.Solution.•The characteristic polynomial is:det (A − λI) =3 − λ 2 30 6 − λ 100 0 2 − λ= (3 − λ)(6 − λ)(2 − λ)• A has eigenvalu e s 2, 3, 6. A =3 2 30 6 100 0 2The eigenvalues of a triangular matrix are its dia go n al entries.• λ1= 2:(A − λ1I)x =1 2 30 4 100 0 0x = 0x1=2−5/21Armin [email protected]• λ2= 3:(A − λ2I)x =0 2 30 3 100 0 −1x = 0x2=100• λ3= 6:(A − λ3I)x =−3 2 30 0 100 0 −4x = 0x3=230• In summary, A has eigenval ues 2, 3, 6 with corresponding eigenvectors2−5/21,100,2/310.These three vectors are independent. By the next result, this is always so.Theorem 7. If x1,, xmar e eigenvectors of A corresponding to different eigenvalues,then they are indepe ndent.Why?Suppose, for contradiction, thatx1,, xmare dependent.By kick ing out some of the vectors, we may assume tha t there is (up to multiple s) only one linearrelation:c1x1++ cmxm= 0.Multiply this relation withA:A(c1x1++ cmxm) = c1λ1x1++ cmλmxm= 0This is a second independent relation! Contradiction.Practi ce pr ob lemsExample 8. Find the eigenvectors and eigenvalues of A =0 −2−4 2.Example 9. What are the eigenvalues of A =2 0 0 0−1 3 0 0−1 1 3 00 1 2 4?No calc ulations!Armin
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