15. Eigenvalues, Eigenvectors

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1 5 Eigenvalues, Eigenvectors Matri of a Linear Transformation Consider a linear ( transformation ) L : a b R 2 R 2 Suppose we know that L and L Then c d because of linearit, we can determine what L does to an vector : L L( + ) L Now notice that for an vector a b c d +L, we have a + b L c + d a + c b d a + b c + d a b Then the matri acts b matri multiplication in the same wa c d that L does Call this matri the matri of L in the basis {, } Since ever linear function from R 2 R 2 can be given a matri in this wa, we see that ever such linear function has a matri in the basis {, } We will revisit this idea, and develop the notion of a basis further, and learn about how to make a matri for an arbitrar linear transformation R n R m in an arbitrar basis Invariant Directions 4 Consider the linear transformation L such that L and L 4, so that the matri of L is Recall that a vector is a direction and a magnitude; L applied to or changes both the direction 7 7 and the magnitude of the vectors given to it

2 4 + 5 Notice that L Then L fies both the magnitude and direction of the vector v Now, notice that an 5 vector with the same direction as v can be written as cv for some constant c Then L(cv ) cl(v ) cv, so L fies ever vector pointing in the same direction as v Also notice that L 2 Then L fies the direction of the vector v 2 but stretches v 2 2 b a factor of 2 Now notice that for an constant c, L(cv 2 ) cl(v 2 ) 2cv 2 Then L stretches ever vector pointing in the same direction as v 2 b a factor of 2 In short, given a linear transformation L it is sometimes possible to find a vector v and constant λ such that L(v) λv We call the direction of the vector v an invariant direction In fact, an vector pointing in the same direction also satisfies the equation: L(cv) cl(v) λcv The vector v is called an eigenvector of L, and λ is an eigenvalue Since the direction is all we reall care about here, then an other vector cv (so long as c ) is an equall good choice of eigenvector ( Returning ) to our eample of the linear transformation L with matri 4, we have seen that L enjos the propert of having two invariant 7 directions, represented b eigenvectors v and v 2 with eigenvalues and 2, respectivel It would be ver convenient if I could write an vector w as a linear combination of v and v 2 Suppose w rv + sv 2 for some constants r and s Then: L(w) L(rv + sv 2 ) rl(v ) + sl(v 2 ) rv + 2sv 2 Now L just multiplies the number r b and the number s b 2 If we could write this as a matri, it would look like: r 2 s 2

3 a b This is much slicker than the usual scenario, in which L a +b c d c + d Here, r and s give the coordinates of w in terms of the vectors v and v 2 In the previous eample, we multiplied the vector b the matri L and came up with a complicated epression In these coordinates, we can see that L is a ver simple diagonal matri, whose diagonal entries are eactl the eigenvalues of L This process is called diagonalization, and it can make complicated linear sstems much easier to analze Now that we ve seen what eigenvalues and eigenvectors are, there are a number of questions that need to be answered How do we find eigenvectors and their eigenvalues? How man eigenvalues and (independent) eigenvectors does a given linear transformation have? When can a linear transformation be diagonalized? We ll start b tring to find the eigenvectors for a linear transformation Eample Let L : R 2 R 2 such that L(, ) (2 + 2, 6 + 6) First, we can find the matri of L: L We want to find an invariant direction v L(v) λv such that or, in matri notation, λ λ λ λ λ

4 ( This is a homogeneous ) sstem, so it onl has solutions when the matri 2 λ 2 is singular In other words, 6 6 λ det ( 2 λ 2 ) 6 6 λ (2 λ)(6 λ) 2 λ 2 8λ 2 (λ )(λ + 2) For an square n n matri M, the polnomial in λ given b det(λi M) ( ) n det(m λi) is called the characteristic polnomial of M, and its roots are the eigenvalues of M In this case, we see that L has two eigenvalues, λ and λ 2 2 To find the eigenvectors, we need to deal ( with these two) cases separatel 2 λ 2 To do so, we solve the linear sstem with 6 6 λ the particular eigenvalue λ plugged in to the matri λ : We solve the linear sstem Both equations sa that 4, so an vector will do Since we 4 onl need the direction of the eigenvector, we can pick a value for Setting is convenient, and gives the eigenvector v 4 λ 2: We solve the linear sstem Here again both equations agree, because we chose λ to make the ( sstem ) singular We see that 2 works, so we can choose v 2 2 In short, our process was the following: 4

5 Find the characteristic polnomial of the matri M for L, given b det(λi M) Find the roots of the characteristic polnomial; these are the eigenvalues of L For each eigenvalue λ i, solve the linear sstem (λ i I M) to obtain an eigenvector associated to λ i References Hefferon, Chapter Three, Section III: with Matrices Representing Linear Maps Hefferon, Chapter Five, Section II: Eigenvalues and Eigenvectors Wikipedia: Eigen* Characteristic Polnomial Linear Transformations (and matrices thereof) Review Questions Consider L : R 2 R 2 with L(, ) ( cos θ + sin θ, sin θ + cos θ) i Write the matri of L on the basis, ii When θ, eplain how L acts on the plane Draw a picture iii Do ou epect L to have invariant directions? iv Tr to find eigenvalues for L b solving the equation L(v) λv v Does L have real eigenvalues? If not, are there comple eigenvalues for L, assuming that i eists? 5

6 2 2 Let M Find all eigenvalues of M Does M have two 2 independent eigenvectors? Can M be diagonalized? Let L be the linear transformation L : R R given b L(,, z) ( +, + z, + z) Let e i be the vector with a one in the ith position and zeros in all other positions i Find Le i for each i m ii Given a matri M m 2 m m 2 m 2 2 m 2, what can ou sa about m m 2 m Me i for each i? iii Find a matri M representing L Choose three non-trivial vectors pointing in different directions and show that Mv Lv for each of our choices 6

LESSON 35: EIGENVALUES AND EIGENVECTORS APRIL 21, (1) We might also write v as v. Both notations refer to a vector.

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