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23. Eigenvalues and Eigenvectors 23. Eigenvalues and Eigenvectors 11/17/20 Eigenvalues and eigenvectors have a variety of uses. They allow us to solve linear difference and differential equations. For many non-linear equations, they inform us about the long-run behavior of the system. They are also useful for defining functions of matrices. 23.1 Eigenvalues We start with eigenvalues. Eigenvalues and Spectrum. Let A be an m m matrix. An eigenvalue (characteristic value, proper value) of A is a number λ so that× A λI is singular. The spectrum of A, σ(A)= {eigenvalues of A}.− Sometimes it’s possible to find eigenvalues by inspection of the matrix. ◮ Example 23.1.1: Some Eigenvalues. Suppose 1 0 2 1 A = , B = . 0 2 1 2 Here it is pretty obvious that subtracting either I or 2I from A yields a singular matrix. As for matrix B, notice that subtracting I leaves us with two identical columns (and rows), so 1 is an eigenvalue. Less obvious is the fact that subtracting 3I leaves us with linearly independent columns (and rows), so 3 is also an eigenvalue. We’ll see in a moment that 2 2 matrices have at most two eigenvalues, so we have determined the spectrum of each:× σ(A)= {1, 2} and σ(B)= {1, 3}. ◭ 2 MATH METHODS 23.2 Finding Eigenvalues: 2 2 × We have several ways to determine whether a matrix is singular. One method is to check the determinant. It is zero if and only the matrix is singular. That means we can find the eigenvalues by solving the equation det(A λI)=0. (23.2.1) − When A is m m, the left hand side of equation (23.2.1) is a polynomial in λ, called the characteristic× polynomial. When A isa2 2 matrix, we have × 0 = det(A λI) − a λ a = 11 − 12 a21 a22 λ − 2 = λ (a11 + a22)λ + (a11a22 a12a21) (23.2.2) − − = p(λ) Here the characteristic equation is a polynomial of degree two, p(λ). It has at most two distinct roots. 23. EIGENVALUES AND EIGENVECTORS 3 23.3 Finding Eigenvalues: m m × Let’s consider the eigenvalues of a general m m matrix. The characteristic equation is × 0 = det(A λI) − a λ a a 11 − 12 ··· 1n a21 a22 λ a2n = . .− ···. .. a a a λ m1 m2 ··· mn − m m m−1 = ( 1) λ + a1λ + + am−1 λ + am − ··· = p(λ) for some numbers a1,...,am. The characteristic polynomial p(λ) is a polynomial of degree m. In general, the characteristic equation of an m m matrix is a polynomial of degree m, with at most m distinct roots. The spectrum× of an m m matrix has at most m elements. × In fact, the Fundamental Theorem of Algebra states that the characteristic polynomial p(λ) can be factored as p(λ) = ( 1)m(λ λ ) (λ λ ) (23.3.3) − − 1 ··· − m where the λi are the (possibly complex) roots of the polynomial, the eigenvalues. They need not be distinct, so the spectrum may include fewer than m points. A and the Characteristic Polynomial. We can also apply the characteristic polynomial p(λ) to the matrix A.1 Then m m m−1 p(A) = ( 1) A + a A + + a − A + a I − 1 ··· m 1 m = ( 1)m(A λ I) (A λ I) − − 1 ··· − m = 0. 1 This is not a proof, which is more involved. 4 MATH METHODS 23.4 Invariance of the Characteristic Polynomial One interesting thing about the characteristic polynomial is that it is the same regardless of the basis used to write the matrix. That’s another way of saying that the characteristic polynomial really belongs the transformation represented by the matrix. Recall our change of basis formula for matrices representing linear transformations in equation (31.11.1). Given a new basis B = {b1,..., bm} with basis matrix B = [b1,..., bm], the matrix in the new coordinates becomes −1 AB = B AEB. (31.11.1) Now −1 det(AB λI) = det(B AEB λI) − − −1 −1 = det(B AEB λB IB) − −1 = det B (AE λI)B − −1 = (det B ) det(AE λI) (det B) − = det(AE λI). − which shows that the characteristic polynomial of a transformation is the same in all bases. 23. EIGENVALUES AND EIGENVECTORS 5 23.5 Coefficients of the Characteristic Polynomial Since the characteristic polynomial is the same in all bases, its coefficients are the same in all bases too. Such numbers are referred to as invariants. We can rewrite equation (23.2.2) to expose two invariants. 0= λ2 λ tr A + det A − where tr A is the trace, the sum of the diagonal elements, tr A = i aii. Both the trace and determinant are independent of the basis used. The coefficient on the λm−1 term is always ( 1)m−1 tr A and the constant term is always det A. P We can use equation− (23.5.4) to write the trace and determinant in terms of eigen- values. p(λ) = ( 1)m(λ λ ) (λ λ ) (23.5.4) − − 1 ··· − m m m = ( 1)m λm λ λ + + ( 1)m λ . − − i ··· − i i i ! X=1 Y=1 It follows that tr A = λ + + λ and det A = λ λ . 1 ··· m 1 ··· m 6 MATH METHODS 23.6 Finding Eigenvectors For each eigenvalue λ there is an eigenspace. It is ker(A λI)= {v Rm : (A λI)v = 0} = {v Rm : Av = λv}. − ∈ − ∈ The non-zero elements of the eigenspace are called eigenvectors. We can find elements of the eigenspace for eigenvalue λ by solving the equation (A λI)v = 0 or Av = λv − ◮ Example 23.6.1: Some Eigenvectors. Consider the matrices from Example 23.1.1. 1 0 2 1 A = , B = . 0 2 1 2 We set 0 0 v 0 0 = (A I)v = 1 = . − 0 1 v2 v2 Since v2 = 0, the eigenspace consists of vectors of the form (v1, 0) for λ = 1. Now for λ = 2, we set 1 0 v v 0 = (A 2I)v = − 1 = − 1 . − 0 0 v2 0 Now v1 = 0, so vectors of the form (0, v2) comprise the eigenspace for λ = 2. As for B, σ(B)= {1, 3} so we examine 1 1 v v + v 0 = (B I)v = 1 = 1 2 − 1 1 v2 v1 + v2 Here v = v so eigenvectors for λ = 1 have the form v (1, 1). As for λ = 3, 2 − 1 1 − 1 1 v v + v 0 = (B 3I)v = − 1 = − 1 2 − 1 1 v2 v1 v2 − − and the eigenvectors have the form v1(1, 1). ◭ 23. EIGENVALUES AND EIGENVECTORS 7 23.7 Complex Eigenvectors Even when we are dealing with a real matrix, the eigenvalues and eigenvectors can be complex. Consider 0 1 A = , which obeys A2 = I. 1 0 − − The characteristic polynomial is λ 1 p(λ) = det(A λI)= A = − = λ2 +1. − 1 λ − − It follows that the eigenvalues are purely imaginary number s, σ(A)= {+i, i}. To find the corresponding eigenvectors, we first set − i 1 iu + u 0 (A iI)u = − u = − 1 2 = − 1 i u1 iu2 0 − − − − with solution u = (i, 1). Then set i 1 iv + v 0 (A + iI)u = u = 1 2 = 1 i v1 + iv2 0 − − with solution v = ( i, 1). Notice that we found− eigenvectors that are complex conjugates of one another. It is always possible to do this, although it is not necessary because we can always multiply one of them by any non-zero complex number to get another eigenvector. Theorem 23.7.1. Suppose a matrix A has real entries. If λ1 is a complex eigenvalue, then so is λ1. If u is a eigenvector corresponding to λ1, then u is an eigenvector corresponding to λ1. Proof. Since A has real entries, the characteristic polynomial p(λ) has real coefficients. Then p(λ1)=0= p(λ1) showing that λ1 is an eigenvalue if λ1 is an eigenvalue. Now suppose u is an eigenvector for λ1. Then Au = λ1u. It follows that Au = λ1u by conjugation Au = λ1u since A is real showing that u is a eigenvector for λ1. 8 MATH METHODS 23.8 The Leslie Population Model Simon and Blume employ a form of Leslie population model to show how eigenvalues and eigenvectors can be used. We too will examine that, but do things a bit differently to highlight the role of eigenvalues and eigenvectors. This simple version of the Leslie model involves an organism that lies for exactly two th years. Let bi be the birth rate to individuals in their i year, i =1, 2, and di the death rate in each year. Survival in each year is (1 d ) and the fact that the organism lives − i at most two years means that d2 = 1. Let xn be the number of individuals in their first year in year n and yn be the number in their second (and final) year in year n. The following difference equations describe the evolution of the system. xn+1 = b1xn + b2yn y = (1 d )x + (1 d )y = (1 d )x . n+1 − 1 n − 2 n − 1 n since d2 = 1. Write this in matrix form as x b b x n+1 = 1 2 n . yn+1 1 d1 0 yn − Consider the case b1 = 1, b2 = 4, d1 = 1/2, and of course d2 = 1. The system becomes x 1 4 x x n+1 = n = A n . (23.8.5) yn+1 1/2 0 yn yn Notice that the solution to this dynamical system is already clear.
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