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Chapter 9 Eigenvectors and Eigenvalues Chapter 9 Eigenvectors and Eigenvalues 9.1 Eigenvectors and Eigenvalues of a Linear Map Given a finite-dimensional vector space E,letf : E E ! be any linear map. If, by luck, there is a basis (e1,...,en) of E with respect to which f is represented by a diagonal matrix λ1 0 ... 0 0 λ ... D = 2 , 0 . ... ... 0 1 B 0 ... 0 λ C B nC @ A then the action of f on E is very simple; in every “direc- tion” ei,wehave f(ei)=λiei. 511 512 CHAPTER 9. EIGENVECTORS AND EIGENVALUES We can think of f as a transformation that stretches or shrinks space along the direction e1,...,en (at least if E is a real vector space). In terms of matrices, the above property translates into the fact that there is an invertible matrix P and a di- agonal matrix D such that a matrix A can be factored as 1 A = PDP− . When this happens, we say that f (or A)isdiagonaliz- able,theλisarecalledtheeigenvalues of f,andtheeis are eigenvectors of f. For example, we will see that every symmetric matrix can be diagonalized. 9.1. EIGENVECTORS AND EIGENVALUES OF A LINEAR MAP 513 Unfortunately, not every matrix can be diagonalized. For example, the matrix 11 A = 1 01 ✓ ◆ can’t be diagonalized. Sometimes, a matrix fails to be diagonalizable because its eigenvalues do not belong to the field of coefficients, such as 0 1 A = , 2 10− ✓ ◆ whose eigenvalues are i. ± This is not a serious problem because A2 can be diago- nalized over the complex numbers. However, A1 is a “fatal” case! Indeed, its eigenvalues are both 1 and the problem is that A1 does not have enough eigenvectors to span E. 514 CHAPTER 9. EIGENVECTORS AND EIGENVALUES The next best thing is that there is a basis with respect to which f is represented by an upper triangular matrix. In this case we say that f can be triangularized. As we will see in Section 9.2, if all the eigenvalues of f belong to the field of coefficients K,thenf can be triangularized. In particular, this is the case if K = C. Now, an alternative to triangularization is to consider the representation of f with respect to two bases (e1,...,en) and (f1,...,fn), rather than a single basis. In this case, if K = R or K = C,itturnsoutthatwecan even pick these bases to be orthonormal,andwegeta diagonal matrix ⌃with nonnegative entries,suchthat f(e )=σ f , 1 i n. i i i The nonzero σisarethesingular values of f,andthe corresponding representation is the singular value de- composition,orSVD. 9.1. EIGENVECTORS AND EIGENVALUES OF A LINEAR MAP 515 Definition 9.1. Given any vector space E and any lin- ear map f : E E,ascalarλ K is called an eigen- value, or proper! value, or characteristic2 value of f if there is some nonzero vector u E such that 2 f(u)=λu. Equivalently, λ is an eigenvalue of f if Ker (λI f)is nontrivial (i.e., Ker (λI f) = 0 ). − − 6 { } Avectoru E is called an eigenvector, or proper vec- tor, or characteristic2 vector of f if u =0andifthere is some λ K such that 6 2 f(u)=λu; the scalar λ is then an eigenvalue, and we say that u is an eigenvector associated with λ. Given any eigenvalue λ K,thenontrivialsubspace Ker (λI f)consistsofalltheeigenvectorsassociated2 with λ together− with the zero vector; this subspace is denoted by Eλ(f), or E(λ, f), or even by Eλ,andiscalled the eigenspace associated with λ, or proper subspace associated with λ. 516 CHAPTER 9. EIGENVECTORS AND EIGENVALUES Remark: As we emphasized in the remark following Definition 4.4, we require an eigenvector to be nonzero. This requirement seems to have more benefits than incon- venients, even though it may considered somewhat inel- egant because the set of all eigenvectors associated with an eigenvalue is not a subspace since the zero vector is excluded. Note that distinct eigenvectors may correspond to the same eigenvalue, but distinct eigenvalues correspond to disjoint sets of eigenvectors. Let us now assume that E is of finite dimension n. Proposition 9.1. Let E be any vector space of finite dimension n and let f be any linear map f : E E. The eigenvalues of f are the roots (in K) of the! polynomial det(λI f). − 9.1. EIGENVECTORS AND EIGENVALUES OF A LINEAR MAP 517 Definition 9.2. Given any vector space E of dimen- sion n,foranylinearmapf : E E,thepolynomial ! Pf (X)=χf (X)=det(XI f)iscalledthecharac- teristic polynomial of f.Foranysquarematrix− A,the polynomial PA(X)=χA(X)=det(XI A)iscalled the characteristic polynomial of A. − Note that we already encountered the characteristic poly- nomial in Section 3.7; see Definition 3.9. Given any basis (e1,...,en), if A = M(f)isthematrix of f w.r.t. (e1,...,en), we can compute the characteristic polynomial χf (X)=det(XI f)off by expanding the following determinant: − X a a ... a − 11 − 12 − 1 n a21 X a22 ... a2 n det(XI A)= − . −. − . . − . ... a a ... X a − n 1 − n 2 − nn If we expand this determinant, we find that n n 1 χ (X)=det(XI A)=X (a + + a )X − A − − 11 ··· nn + +( 1)n det(A). ··· − 518 CHAPTER 9. EIGENVECTORS AND EIGENVALUES The sum tr(A)=a11+ +ann of the diagonal elements of A is called the trace··· of A. Since the characteristic polynomial depends only on f, tr(A)hasthesamevalueforallmatricesA representing f.Welettr(f)=tr(A)bethetrace of f. Remark: The characteristic polynomial of a linear map is sometimes defined as det(f XI). Since − det(f XI)=( 1)n det(XI f), − − − this makes essentially no di↵erence but the version det(XI f)hasthesmalladvantagethatthecoefficient of Xn is− +1. If we write n n 1 χA(X)=det(XI A)=X ⌧1(A)X − − k n−k n + +( 1) ⌧ (A)X − + +( 1) ⌧ (A), ··· − k ··· − n then we just proved that ⌧1(A)=tr(A)and⌧n(A)=det(A). 9.1. EIGENVECTORS AND EIGENVALUES OF A LINEAR MAP 519 If all the roots, λ1,...,λn,ofthepolynomialdet(XI A) belong to the field K,thenwecanwrite − det(XI A)=(X λ ) (X λ ), − − 1 ··· − n where some of the λismayappearmorethanonce. Consequently, n n 1 χA(X)=det(XI A)=X σ1(λ)X − − k n−k n + +( 1) σ (λ)X − + +( 1) σ (λ), ··· − k ··· − n where σk(λ)= λi, I 1,...,n i I ✓{XI =k } Y2 | | the kth symmetric function of the λi’s. From this, it clear that σk(λ)=⌧k(A) and, in particular, the product of the eigenvalues of f is equal to det(A)=det(f)andthesumoftheeigenvalues of f is equal to the trace, tr(A)=tr(f), of f. 520 CHAPTER 9. EIGENVECTORS AND EIGENVALUES For the record, tr(f)=λ + + λ 1 ··· n det(f)=λ λ , 1 ··· n where λ1,...,λn are the eigenvalues of f (and A), where some of the λismayappearmorethanonce. In particular, f is not invertible i↵it admits 0 has an eigenvalue. Remark: Depending on the field K,thecharacteristic polynomial χA(X)=det(XI A)mayormaynothave roots in K. − This motivates considering algebraically closed fields. For example, over K = R,noteverypolynomialhasreal roots. For example, for the matrix cos ✓ sin ✓ A = , sin ✓ −cos ✓ ✓ ◆ the characteristic polynomial det(XI A)hasnoreal roots unless ✓ = k⇡. − 9.1. EIGENVECTORS AND EIGENVALUES OF A LINEAR MAP 521 However, over the field C of complex numbers, every poly- nomial has roots. For example, the matrix above has the i✓ roots cos ✓ i sin ✓ = e± . ± Definition 9.3. Let A be an n n matrix over a field, K.Assumethatalltherootsofthecharacteristicpoly-⇥ nomial χA(X)=det(XI A)ofA belong to K,which means that we can write − det(XI A)=(X λ )k1 (X λ )km, − − 1 ··· − m where λ1,...,λm K are the distinct roots of det(XI A)and2k + + k = n. − 1 ··· m The integer, ki,iscalledthealgebraic multiplicity of the eigenvalue λi and the dimension of the eigenspace, E =Ker(λ I A), is called the geometric multiplicity λi i − of λi.Wedenotethealgebraicmultiplicityofλi by alg(λi) and its geometric multiplicity by geo(λi). By definition, the sum of the algebraic multiplicities is equal to n but the sum of the geometric multiplicities can be strictly smaller. 522 CHAPTER 9. EIGENVECTORS AND EIGENVALUES Proposition 9.2. Let A be an n n matrix over a field K and assume that all the roots⇥ of the charac- teristic polynomial χ (X)=det(XI A) of A belong A − to K. For every eigenvalue λi of A, the geometric multiplicity of λi is always less than or equal to its algebraic multiplicity, that is, geo(λ ) alg(λ ). i i Proposition 9.3. Let E be any vector space of fi- nite dimension n and let f be any linear map. If u1,...,um are eigenvectors associated with pairwise distinct eigenvalues λ1,...,λm, then the family (u1,...,um) is linearly independent. Thus, from Proposition 9.3, if λ1,...,λm are all the pair- wise distinct eigenvalues of f (where m n), we have a direct sum E E λ1 ⊕···⊕ λm of the eigenspaces Eλi. Unfortunately, it is not always the case that E = E E . λ1 ⊕···⊕ λm 9.1. EIGENVECTORS AND EIGENVALUES OF A LINEAR MAP 523 When E = E E , λ1 ⊕···⊕ λm we say that f is diagonalizable (and similarly for any matrix associated with f). Indeed, picking a basis in each Eλi,weobtainamatrix which is a diagonal matrix consisting of the eigenvalues, each λi occurring a number of times equal to the dimen- sion of Eλi.
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