Review• If Ax = λx, then x is an eigen vec tor of A with eigenvalue λ.All eigenv ectors (plus0) with eigenvalue λ form the eigenspace of λ.• λ is an eigenvalue o f Adet (A − λI)characteristic polynomial= 0.Why? Because A x = λx(A − λI )x = 0.By the way: this means that the eigenspace ofλ is just Nul(A − λI).• E.g., if A =3 2 30 6 100 0 2then det (A − λI) = (3 − λ)(6 − λ)(2 − λ).• Eigenvectors x1,, xmof A correspon ding to different eigenvalues are independent.• By the way:◦ product of eigenvalues = de terminant◦ sum of eigenvalues = “trace” (sum of diagonal entries)Example 1. Find the eigenvalues of A as well as a basis for the corresponding eigen-spaces, whereA =2 0 0−1 3 1−1 1 3.Solution.•The characteristic polynomial is:det (A − λI) =2 − λ 0 0−1 3 − λ 1−1 1 3 − λ= (2 − λ)3 − λ 11 3 − λ= (2 − λ)[(3 − λ)2− 1]=(2 − λ)(λ − 2)(λ − 4)• A has eigenvalu e s 2, 2, 4. A =2 0 0−1 3 1−1 1 3Since λ = 2 is a double root, it has (alg ebra ic) multiplicity 2.• λ1= 2:(A − λ1I)x =0 0 0−1 1 1−1 1 1x = 0Armin [email protected] independent solutions: x1=110, x2=0−11In other words: the eigenspace for λ = 2 is span(110,0−11).• λ2= 4:(A − λ2I)x =−2 0 0−1 −1 1−1 1 −1x = 0x3=011• In summary, A has eigenvalues 2 and 4:◦ eigenspace for λ = 2 has basis110,0−11,◦ eigenspace for λ = 4 has basis011.An n × n matrix A has up to n different eigenvalues.Namely, the roots of the degreen characteristic polynomial det (A − λI ).• For each eigenvalue λ, A has at least o ne eigenvec tor.That’s becauseNul(A − λI) has dimension at l east 1.• If λ has multiplicity m, then A has up to m (indep e ndent) eigen vectors for λ.Ideally, we would like to find a total of n (independent) e igenvectors of A.Why can there be no more thann eigenvectors?!Two sources of trouble: eigenvalues can be• complex numbers (that is, not enou gh real roots), or• repeated roots of the characteristic polynomial.Example 2. Find the eige nvectors and eige nvalues ofA =0 −11 0. Geometrica lly, what is the trouble?Solution. Axy=− yxi.e. multiplication with A is rotation by 90◦(counter-cl ock-wise).Which vector is parallel afte r rotation by 90◦? Trouble.Armin [email protected]: work with complex numbers!• det (A − λI) =−λ −11 −λ= λ2+ 1So, the eigenva l ues are λ1= i and λ2= −i.• λ1= i:−i −11 −ix =00x1=i1Let us check:0 −11 0i1=−1i= ii1• λ2= −i:i −11 ix =00x2=−i1Example 3. Find the eigenvectors and eigenvalu e s of A =1 10 1. What is the trouble?Solution.•det (A − λI) =1 − λ 10 1 − λ= (1 − λ)2So: λ = 1 is the on ly eigenvalue (it has multipli c ity 2).• (A − λI)x =0 10 0x = 0x1=10So: the eigenspace is spann10o. Only d imension 1!• Trouble: only 1 independent eigenvector for a 2 × 2 matrixThis kind of trouble cannot really be fixed.We have to lower our expectati ons and look for generalized eigen vectors .These are solutions to(A − λI )2x = 0, (A − λI)3x = 0,Practi ce problemsExample 4. Find the eigenvec tors and eigenvalues of A =1 2 10 −5 01 8 1.Armin
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