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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
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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 meyer@ncsu.edu 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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