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Eigenvalues and Eigenvectors October 7, 2013 EIGENVALUES AND EIGENVECTORS RODICA D. COSTIN Contents 1. Motivation 3 1.1. Diagonal matrices 3 1.2. Example: solving linear di↵erential equations 3 2. Eigenvalues and eigenvectors: definition and calculation 4 2.1. Definitions 4 2.2. The characteristic equation 5 2.3. Geometric interpretation of eigenvalues and eigenvectors 6 2.4. Digression: the fundamental theorem of algebra 6 2.5. Diagonal matrices 8 2.6. Similar matrices have the same eigenvalues 8 2.7. Projections 9 2.8. Trace, determinant and eigenvalues 9 2.9. The eigenvalue zero 10 2.10. Eigenvectors corresponding to di↵erent eigenvalues are independent 10 2.11. Diagonalization of matrices with linearly independent eigenvectors 11 2.12. Eigenspaces 13 2.13. Real matrices with complex eigenvalues; decomplexification 14 2.14. Jordan normal form 16 2.15. Block matrices 20 3. Solutions of linear di↵erential equations with constant coefficients 22 3.1. The case when M is diagonalizable. 22 3.2. Non-diagonalizable matrix 26 3.3. Fundamental facts on linear di↵erential systems 29 3.4. Eigenvalues and eigenvectors of the exponential of a matrix 30 3.5. Higher order linear di↵erential equations; companion matrix 31 3.6. Stability in di↵erential equations 34 4. Di↵erence equations (Discrete dynamical systems) 40 4.1. Linear di↵erence equations with constant coefficients 40 4.2. Solutions of linear di↵erence equations 40 4.3. Stability 41 4.4. Example: Fibonacci numbers 42 4.5. Positive matrices 43 4.6. Markov chains 43 5. More functional calculus 45 1 2RODICAD.COSTIN 5.1. Discrete equations versus di↵erential equations 45 5.2. Functional calculus for digonalizable matrices 46 5.3. Commuting matrices 49 EIGENVALUES AND EIGENVECTORS 3 1. Motivation 1.1. Diagonal matrices. Perhaps the simplest type of linear transforma- tions are those whose matrix is diagonal (in some basis). Consider for ex- ample the matrices a 0 b 0 (1) M = 1 ,N= 1 0 a2 0 b2 It can be easily checked that ↵a + βb 0 ↵M + βN = 1 1 0 ↵a2 + βb2 and 1 0 k 1 a1 k a1 0 a1b1 0 M − = 1 ,M = k ,MN= 0 0 a 0 a2b2 a2 2 Diagonal matrices behave like the bunch of numbers on their diagonal! The linear transformation consisting of multiplication by the matrix M in (1) dialates a1 times vectors in the direction of e1 and a2 times vectors in the direction of e2. In this chapter we will see that most linear transformations do have diag- onal matrices in a special basis, whose elements are called the eigenvectors of the transformation. We will learn how to find these bases. Along the special directions of the eigenvectors the transformation just dialation by a factor, called eigenvalue. 1.2. Example: solving linear di↵erential equations. Consider the sim- ple equation du = λu dt which is linear, homogeneous, with constant coefficients, and unknown func- tion u(t) R (or in C). Its general solution is, as it is well known, 2 u(t)=Ceλt. Consider now a similar equation, but where the unknown u(t) is a vector valued function: du n (2) = Mu, u(t) R ,Mis an n n constant matrix dt 2 ⇥ Inspired by the one dimensional case we look for exponential solutions. Substituting in (2) u(t)=eλtv (where λ is a number and v is a constant vector, both be determined) and dividing by eλt, we obtain that the scalar λ and the vector v must satisfy (3) λv = Mv or (4) (M λI)v = 0 − 4RODICAD.COSTIN If the null space of the matrix M λI is zero, then the only solution of (4) is v = 0 which gives the (trivial!)− solution u(t) 0. If however, we can find special values of λ for which⌘ (M λI) is not null, then we found a nontrivial solution of (2). Such valuesN of−λ are called eigenvalues of the matrix M, and vectors v (M λI), v = 0, are called eigenvectors corresponding to the eigenvalue2N − λ. 6 Of course, the necessary and sufficient condition for (M λI) = 0 is that N − 6 { } (5) det(M λI)=0 − Example. Let us calculate the exponential solutions for 1 3 (6) M = − − " 02# Looking for eigenvalues of M we solve equation (5), which for (6) is 1 3 10 1 λ 3 det − − λ = − − − =( 1 λ)(2 λ) " 02# − " 01#! 02λ − − − − with solutions λ1 = 1 and λ2 = 2. We next determine− an eigenvector corresponding to the eigenvalue λ = λ = 1: looking for a nozero vector v such that (M λ I)v = 0 we solve 1 − 1 − 1 1 0 3 x1 0 − = " 03#"x2 # " 0 # giving x2 = 0 and x1 arbitrary; therefore the first eigenvector is any scalar T multiple of v1 =(0, 1) . Similarly, for the eigenvalue λ = λ = 2 we solve (M λ I)v = 0: 2 − 2 2 3 3 y1 0 − − = " 00#"y2 # " 0 # which gives y2 = y1 and y1 arbitrary, and the second eigenvector is any − T scalar multiple of v2 =(1, 1) . − t T We found two particular solutions of (2), (6), namely u1(t)=e− (0, 1) and u (t)=e2t(1, 1)T . These are functions belonging to the null space 2 − of the linear operator Lu = du Mu, therefore any linear combination of dx − these two solutions also belongs to the null space: any C1u1(t)+C2u2(t)is also a solution, for and constants C1,C2. A bit later we will show that these are all the solutions. 2. Eigenvalues and eigenvectors: definition and calculation 2.1. Definitions. Denote the set of n n (square) matrices with entries in ⇥ F (= R or C) (F )= M M =[M ] ,M F Mn { | ij i,j=1,...n ij 2 } EIGENVALUES AND EIGENVECTORS 5 n A matrix M n(F ) defines an endomorphism the vector space F (over the scalars F ) by2M usual multiplication x Mx. 7! Note that a matrix with real entries can also act on Cn, since for any x Cn also Mx Cn. But a matrix with complex non real entries cannot 2 2 act on Rn, since for x Rn the image Mx may not belong to Rn (while 2 certainly Mx Cn). 2 Definition 1. Let M be an n n matrix acting on the vector space V = F n. Ascalarλ F is an eigenvalue⇥ of M if for some nonzero vector v V , v = 0 we have2 2 6 (7) Mv = λv The vector v is called eigenvector corresponding to the eigenvalue λ. Of course, if v is an eigenvector corresponding to λ, then so is any scalar multiple cv (for c = 0). 6 2.2. The characteristic equation. Equation (7) can be rewritten as Mv λv = 0, or (M λI)v = 0, which means that the nonzero vector v belongs− to the null space− of the matrix M λI, and in particular this matrix is not invertible. Using the theory of matrices,− we know that this is equivalent to det(M λI)=0 − The determinant has the form M11 λM12 ... M1n M − M λ... M 21 22 − 2n det(M λI)= . − . Mn1 Mn2 ... Mnn λ − This is a polynomial in λ, having degree n. To understand why this is the case, consider first n = 2 and n = 3. For n = 2 the characteristic polynomial is M11 λM12 − =(M11 λ)(M22 λ) M12 M21 M21 M22 λ − − − − 2 = λ (M11 + M 22)λ +(M11M22 M12 M21) − − which is a quadratic polynomial in λ; the dominant coefficient is 1. For n = 3 the characteristic polynomial is M11 λM12 M13 M − M λM 21 22 − 23 M13 M23 M33 λ − and expanding along say, row 1, 1+1 M22 λM23 1+2 M21 M23 =( 1) (M11 λ) − +( 1) M12 − − M23 M33 λ − M13 M33 λ − − 1+3 M21 M22 λ +( 1) M13 − − M13 M23 6RODICAD.COSTIN = λ3 +(M + M + M )λ2 + ... − 11 22 33 which is a cubic polynomial in λ; the dominant coefficient is 1. It is easy to show by induction that det(M λI) is polynomial− in λ, having degree n, and that the coefficient of λn is− ( 1)n. − Definition 2. The polynomial det(M λI) is called the characteristic polynomial of the matrix M, and the− equation det(M λI)=0is called the characteristic equation of M. − Remark. Some authors refer to the characteristic polynomial as det(λI M); the two polynomial are either equal or a 1 multiple of each other,− since det(λI M)=( 1)n det(M λI). − − − − 2.3. Geometric interpretation of eigenvalues and eigenvectors. Let M be an n n matrix, and T : Rn Rn defined by T (x)=Mx be the corresponding⇥ linear transformation. ! If v is an eigenvector corresponding to an eigenvalue λ of M: Mv = λv, then T expands or contracts v (and any vector in its direction) λ times (and it does not change its direction!). If the eigenvalue/vector are not real, a similar fact is true, only that multiplication by a complex (not real) scalar cannot be easily called an expansion or a contraction (there is no ordering in complex numbers), see the example of rotations, 2.13.1. The special directions of§ the eigenvectors are called principal axes of the linear transformation (or of the matrix). 2.4. Digression: the fundamental theorem of algebra. 2.4.1. Polynomials of degree two: roots and factorization. Consider polyno- mials of degree two, with real coefficients: p(x)=ax2 + bx + c.Itiswell b pb2 4ac known that p(x) has real solutions x = − ± − if b2 4ac 0 1,2 2a − ≥ 2 (where x1 = x2 when b 4ac = 0), and p(x) has no real solutions if b2 4ac < 0.
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