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Eigenvalues and Eigenvectors - MIT Mathematics

Chapter 6 Eigenvalues and Introduction to EigenvaluesLinear equationsAxDbcome from steady state problems. Eigenvalues have their greatestimportance indynamic problems. The solution ofdu=dtDAuis changing with time growing or decaying or oscillating. We can t find it by elimination. This chapter enters anew part of linear algebra, based onAxD x. All matrices in this chapter are good model comes from the powersA; A2;A3;:::of a matrix. Suppose you need thehundredth powerA100. The starting matrixAbecomes unrecognizable after a few steps,andA100is very close to :6 :6I:4 :4 : :8 :3:2 :7 :70 :45:30 :55 :650 :525:350 :475 :6000 :6000:4000 :4000 AA2A3A100A100was found by using theeigenvaluesofA, not by multiplying 100 matrices. Thoseeigenvalues (here they are1and1=2) are a new way to see into the heart of a explain Eigenvalues , we first explain Eigenvectors .

P is symmetric, so its eigenvectors .1;1/ and .1; 1/ are perpendicular. The only eigenvalues of a projection matrix are 0 and 1. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvectors for D 1 (which means Px D x/ fill up the column space. The nullspace is projected to zero. The column space projects onto itself.

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Eigenvalues and Eigenvectors

math.mit.edu

Special properties of a matrix lead to special eigenvalues and eigenvectors. That is a major theme of this chapter (it is captured in a table at the very end). 286 Chapter 6.

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math.mit.edu

6.1. Introduction to Eigenvalues 287 Eigenvalues The number is an eigenvalue of Aif and only if I is singular: det.A I/ D 0: (3) This “characteristic equation” det.A I/ D 0 involves only , not x. When A is n by n,

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math.mit.edu

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math.mit.edu

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2 subject to: 5x 1 + 7x 2 8 4x 1 + 2x 2 15 2x 1 + x 2 3 x 1 0;x 2 0: Some more terminology. A solution x= (x 1;x 2) is said to be feasible with respect to the above linear program if it satis es all the above constraints. The set of feasible solutions is called the feasible space or feasible region. A feasible solution is optimal if its ...

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math.mit.edu

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[4] Computing Eigenvectors [5] Computing Eigenvalues [1] Eigenvectors and Eigenvalues Example from Di erential Equations Consider the system of rst order, linear ODEs. dy 1 dt = 5y 1 + 2y 2 dy 2 dt = 2y 1 + 5y 2 We can write this using the companion matrix form: y0 1 y0 2 = 5 2 2 5 y 1 y 2 : Note that this matrix is symmetric. Using notation ...

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1 = (1,0,3) and u 2 = (1,1,3). It is a fact that all other eigenvectors associated with λ 2 = −2 are in the span of these two; that is, all others can be written as linear combinations c 1u 1 +c 2u 2 using an appropriate choices of the constants c 1 and c 2. Example: Find the eigenvalues and associated eigenvectors of the matrix A = −1 2 0 ...

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0 3 1 0 0 3 1 A: Show that A is not diagonalizable. Solution: Use Theorem 5.2.2 and show that A does not have 3 linearly independent eigenvectors. I To nd the eigenvalues, we solve det( I A) = 1 1 1 0 + 3 1 0 0 + 3 = ( 1)( +3)2 = 0: So, = 1; 3 are the only eigenvalues of A: Satya Mandal, KU Eigenvalues and Eigenvectors x5.2 Diagonalization

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P is symmetric, so its eigenvectors .1;1/ and .1; 1/ are perpendicular. The only eigenvalues of a projection matrix are 0 and 1. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvectors for D 1 (which means Px D x/ fill up the column space. The nullspace is projected to zero. The column space projects onto itself.

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courses.physics.illinois.edu

If all 3eigenvalues are distinct →-−%≠0 Hence, /1"=0, i.e., the eigenvectors are orthogonal (linearly independent), and consequently the matrix !is diagonalizable. Note that a diagonalizable matrix !does not guarantee 3distinct eigenvalues.

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www2.math.upenn.edu

Find all of the eigenvalues and eigenvectors of A= 1 1 0 1 : The characteristic polynomial is ( 1)2, so we have a single eigenvalue = 1 with algebraic multiplicity 2. The matrix A I= 0 1 0 0 has a one-dimensional null space spanned by the vector (1;0). Thus, the geometric multiplicity of this eigenvalue is 1.

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scipp.ucsc.edu

λ2 − 1 = 0, (12) which yields the eigenvalues, λ = ±1. The interpretation of this result is immediate. The matrix R(θ) when operating on a vector ~v represents a reflection of that vector through a line of reflection that passes through the origin. In the case of λ = 1 we have R(θ)~v = ~v, which means that ~v is a

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mathcs.clarku.edu

form, so we can read o the three eigenvalues: 1 = 1, 2 = 3, and 3 = 2. (It doesn’t matter the order you name them.) Thus, the spectrum of this matrix is the set f1;2;3g. Let’s nd the 1-eigenspace. We need to solve Ax = 1x. That’s the same as solving (A 1I)x = 0. The matrix A 1Iis 2 4 0 0 0 3 2 0 3 2 1 3 5 which row reduces to 2 4 1 0 1 6 ...

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