University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2023
Views:

187

Pages:

2
Author:

safisme6gf

Summary: Eigenvalues and Eigenvectors Given a system of two expressions in two unknowns, which can be written equivalently as either: " #" # ax +by a b x or or Ax cx +dy c d y We are interested in finding a scalar λ and a non-zero vector v so that: " Av = λv or a b c d #" # v1 v2 " =λ v1 v2 # av1 +bv2 = λv1 cv1 +dv2 = λv2 or Any scalar λ and vector v that satisfies this relationship are called an eigenvalue and an eigenvector (respectively) of the matrix. Rewriting this system, we find that we would like to solve the following system of equations for the scalar λ and vector v so that: " (a − λ) v1 + b v2 =0 c v1 + (d − λ)v2 = 0 or a−λ b c d−λ #" v1 v2 # " = 0 0 # Using Cramer’s Rule, we would find that if (a−λ)(d−λ)−bc 6= 0, then the (unique) solution to this system is v1 = 0 and v2 = 0. However, by definition, an eigenvector should not be zero. Therefore, we conclude that, to get a non-zero eigenvector, the determinant should be zero: (a − λ)(d − λ) − bc = 0 Expanding this gives the characteristic equation for the matrix A (which is also the characteristic equation for the second order DE we would get from 7.1): λ2 − (a + d)λ + (ad − bc) = 0 or λ2 − T r(A)λ + det(A) = 0 Summary To find the eigenvalues of the matrix " a b c d # we first find the eigenvalues by solving the characteristic equation: λ2 − (a + d)λ + (ad − bc) = 0 For each eigenvalue, solve the system of equations for v: (a − λ) v1 + b v2 =0 c v1 + (d − λ)v2 = 0 " or a−λ b c d−λ #" v1 v2 # " = 0 0 # Worked Examples (Also see the HW Solutions) 1. " A= # 1 −2 3 −4 The trace is 1−4 = −3 and the determinant is −4+6 = 2. The characteristic equation is: λ2 + 3λ + 2 = 0 which we solve for λ. In this case, the quadratic factors and the eigenvalues are: λ = −2, −1 Now we get the eigenvectors: • For λ = −2, we have the system: " 3 −2 3 −2 #" v1 v2 # " = 0 0 # or 3v1 − 2v2 = 0 Choose any non-zero representative of this set to be the eigenvector. For example, v1 = 2, v2 = 3. In this case, " λ = −2 v= 2 3 # • For λ = −1, we have: " 2 −2 3 −3 #" v1 v2 # " = 0 0 # or 2v1 − 2v2 = 0 Therefore, one representative eigenvector would be v1 = 1, v2 = 1, and " λ = −1, v= 1 1 # 2. Eigenvalues and eigenvectors may also be complex. For example, " A= 1 2 −2 1 # The trace is 2 and the determinant is 5. The characteristic equation is: λ2 − 2λ + 5 = 0 (λ − 1)2 + 4 = 0 which we solve for λ. In this case, completing the square seems faster than the quadratic formula, and: λ = 1 ± 2i. For λ = 1 + 2i, solve the system on the left to get: " −2i 2 −2 −2i #" v1 v2 # " = 0 0 # v2 = iv1 " You should verify that for λ = 1 − 2i, v = " # 1 . −i v= 1 i #

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