M427K Handout: Linear Systems: Preliminaries & Systems ofLinear First Order ODEsSalman ButtApril 6, 2006Notes1. Check to make sure you have your homework and not someone else’s.2. Office hours this Friday will be adjusted to 3-3:30 and 4:30-5.ExerciseIs the set of vectors {(1, 2, −1), (2, 1, 3), (−4, 1, −11)} linearly independent or dependent?Eigenvalues and EigenvectorsWe say ~x (not the zero vector) is an eigenvector of a matrix A if there exists a number λ suchthat A~x = λ~x. That is, A takes the vector ~x to a multiple of itself. In this case, we say λ is aneigenvalue of A. Eigenvectors and eigenvalues are important in many applications and findingthem is an important technique to know in this class and beyond. We rewrite our defining equationA~x = λ~x asThis has a nonzero solution if and only if λ is chosen so thatNote that for a given eigenvalue, there is an infinite number of associated eigenvectors – namelyscalar multiples of the vector. Let’s do some examples. Consider the m atrices 3 −14 −2! 1 −23 −4!1Systems of First Order Linear Differential EquationsSuppose we have a system of first order linear differential equations. For illustrative purposes,we will consider systems of two equations:x01= p1(t)x1+ q1(t)x2+ g1(t)x02= p2(t)x1+ q2(t)x2+ g2(t)We write this system asWe begin by studying the constant coefficient homogeneous case, i.e. where ~g(t) = 0. So wehave the matrix equation ~x0= A~x where A is a matrix of constants. We begin by guessing thatour solution is of the form ~x = ~vertwhere ~v is a constant vector. Differentiating and plugging intoour linear system, we findDividing by ertand rewriting our equation, we findSo in order solve the differential equation, we need to solve this algebraic equation. Observethat this is precisely the eigenvalue equation, so our r’s are m erely the eigenvalues of A and ~v is theeigenvector of A associated to r. Thus our proposed solution x = ~vertis a solution precisely whenr and ~v are eigenvalues and associated eigenvectors of A. In the 2 × 2 case, we find two differentvalues of r. As in the past, the nature of our eigenvalues r (i.e. if they are real and distinct,repeated, or complex) will be critical in determining our final solution. We begin with the simplecase: real, distinct roots. In this case, the two eigenvectors we find will be linearly independent,hence our general solution simply becomes ~x = c1~v1er1t+ c2~v2er2t.2Let’s see some examples: Find the solutions to the systems~x0= 1 14 1!~x ~x0= 3 −22