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# The eigenvalues of b constitute the remaining

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Unformatted text preview: matrix that bears his name, Toeplitz is known for his general theory of inﬁnite-dimensional spaces developed in the 1930s. Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 7.2 Diagonalization by Similarity Transformations http://www.amazon.com/exec/obidos/ASIN/0898714540 515 for j = 1, 2, . . . , n, and conclude that A is diagonalizable. Solution: For an eigenpair (λ, x), the components in (A − λI) x = 0 are cxk−1 +(b−λ)xk +axk+1 = 0, k = 1, . . . , n with x0 = xn+1 = 0 or, equivalently, It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] xk+2 + b−λ a xk+1 + c xk = 0 a for k = 0, . . . , n − 1 with x0 = xn+1 = 0. These are second-order homogeneous diﬀerence equations, and solving them is similar to solving analogous diﬀerential equations. The technique is to seek solutions of the form xk = ξrk for constants ξ and r. This produces the quadratic equation r2 + (b − λ)r/a + c/a = 0 with roots r1 and r2 , and it can be argued that the general solution of xk+2 + ((b − λ)/a)xk+1 + (c/a)xk = 0 is k k αr1 + βr2 αρk + βkρk xk = if r1 = r2 , if r1 = r2 = ρ, D E T H where α and β are arbitrary constants. For the eigenvalue problem at hand, r1 and r2 must be distinct—otherwise xk = αρk + βkρk , and x0 = xn+1 = 0 implies each xk = 0, which is impossible k k because x is an eigenvector. Hence xk = αr1 + βr2 , and x0 = xn+1 = 0 yields 0=α+β n n 0 = αr1 +1 + βr2 +1 IG R =⇒ r1 r2 n+1 = −β = 1 =⇒ α r1 = ei2πj/(n+1) , r2 so r1 = r2 ei2πj/(n+1) for some 1 ≤ j ≤ n. Couple this with r2 + Y P c (b − λ)r + = (r − r1 )(r − r2 ) =⇒ a a to conclude that r1 = O C r1 r2 = c/a r1 + r2 = −(b − λ)/a c/a e−iπj/(n+1) , and c/a eiπj/(n+1) , r2 = λ = b + a c/a eiπj/(n+1) + e−iπj/(n+1) = b + 2a c/a cos jπ n+1 . Therefore, the eigenvalues of A must be given by λj = b + 2a c/a cos jπ n+1 , j = 1, 2, . . . ,...
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