EIGENVALUES OF SEVERAL TRIDIAGONAL MATRICES

Applied Mathematics E-Notes, 5(005), 66-74 c ISSN 607-50 Available free at mirror sites of http://www.math.thu.edu.tw/ ame/ EIGENVALUES OF SEVERAL TRIDIAGONAL MATRICES We-Chyua Yueh Received 4 September 004 Abstract Tridiagoal matrices appear frequetly i mathematical models. I this ote, we derive the eigevalues ad the correspodig eigevectors of several tridiagoal matrices by the method of symbolic calculus i []. Itroductio There are may mathematical models which ivolves tridiagoal matrices of the form [] α + b c 0 0... 0 0 a b c 0... 0 0 A = 0 a b c... 0 0. ()..................... 0 0 0 0... a β + b I particular, whe a = c =,b = adα = β =0, the eigevalues of A has bee proved [3,4] to be kπ λ k (A )= +cos,k=,,..., ; + whe a = c =,b= adα = β =, or, whe a = c =,b=, α =adβ =0, the eigevalues have bee reported as λ k (A )= +cos kπ,k=,,..., ; or kπ λ k (A )= +cos,k=,,..., + respectively without proof. I this ote, we ited to derive the eigevalues ad the correspodig eigevectors of several tridiagoal matrices of the form A. Mathematics Subect Classificatios: 5A8 Departmet of Refrigeratio, Chi-Yi Istitute of Techology, Taichug, Taiwa, R. O. Chia 66

W. C. Yueh 67 The Eigevalue Problem Cosider the eigevalue problem A u = λu, where a, b, c ad α, β are umbers i the complex plae C. We will assume that ac = 0 sice the cotrary case is easy. Let λ be a eigevalue (which may be complex) ad (u,..., u ) a correspodig eigevector. We may view the umbers u,u,..., u respectively as the first, secod,..., ad the -thtermofaifiite (complex) sequece u = {u i } i=0. Sice A u = λu ca be writte as u 0 = 0 au 0 + bu + cu = λu + αu, au + bu + cu 3 = λu +0,... =..., au + bu + cu = λu +0, au + bu + cu + = λu + βu, u + = 0, we see that the sequece u = {u k } k=0 satisfies u 0 =0,u + =0ad au k + bu k + cu k+ = λu k + f k,k=,,..., () where f = αu ad f = βu, while f k =0fork =,. Note that u caot be 0, for otherwise from (), cu = 0 ad iductively u 3 = u 4 = = u = 0 which is cotrary to the defiitio of a eigevector. Let f = {f k } k=0 be defied above. The () ca be expressed as c {u k+ } k=0 + b {u k+} k=0 + a {u k} k=0 = λ {u k+} k=0 + {f k+} k=0. We ow recall that h = {0,, 0,...}, α = {α, 0, 0,...} ad the properties of covolutio product xy of two sequeces x = {x k } k=0 ad y = {y k} k=0 (see [] for details). The by takig covolutio of the above equatio with h = h h, ad otig that ad h {u + } = h {u,u,...} = {0,u,u,...} = u u 0 h {u + } = h {u,u 3,...} = {0, 0,u,u 3,...} = u u 0 u h, we have c (u u 0 u h)+(b λ) h (u u 0 )+a h u = h f f 0. Solvig for u, ad substitutig u 0 = f 0 =0,weobtai a h +(b λ) h + c u =(f + cu ) h. Sice c = 0,wecadividetheaboveequatio[]bya h +(b λ) h + c to obtai u = (f + cu ) h a h +(b λ) h + c. (3)

68 Eigevalues of Tridiagoal Matrices Let γ ± = (b λ) ±,ac= 0 a be the two roots of az +(b λ) z + c =0,where =(b λ) 4ac. Sice a, b, c as well as γ ±, are i the complex domai, we first itroduce the followig Lemma. LEMMA. Let z = x + iy where z C ad x, y R. The (i) si z =0ifadoly if z = x = kπ for some k Z, ad (ii) cos z = ± if ad oly if z = x = π for some Z. PROOF. If z = x = kπ, k Z, thesiz = 0, which gives the sufficiet coditio of (i). If si z =si(x + iy) =six cosh y + i (cos x sih y) =0, the si x cosh y = 0 ad cos x sih y =0. Sicecoshy = 0, hece si x =0sothat x = kπ, k Z. Cosequetly cos x = 0adsihy = 0, which yields y =0. Hece z = x = kπ, k Z. This gives the ecessary coditio of (i). To prove (ii), i a similar maer we see that if z = kπ, k Z, thacosz = ±. O the other had, if cos z =cos(x + iy) =cosx cosh y i (si x sih y) =±, the cos x cosh y = ± adsix sih y =0. If si x = 0,thesihy =0sothaty =0, cosequetly cos x = ± adx = kπ, k Z. But the six = 0 which cotradicts the assumptio si x = 0. Hece si x =0adx = kπ, k Z. The cosx = ± ad cosh y =, which demads y = 0. This completes the proof. COROLLARY. If z = kπ where k Z, thesiz = 0, cos z = ± adsi z =0, cos z =0. PROOF. If z = kπ, siz = 0 ad cos z = ± follows readily from Lemma. Sice si z =si z cos z =0,sowehavesiz = 0 ad cos z = 0. This completes the proof. Accordig to γ ± beig two differet complex umbers or two equal umbers, there aretwocasestobecosidered. Case I. Suppose γ + ad γ are two differet complex umbers. Let γ ± = p ± iq where p, q C ad q = 0. Sice γ + γ = p + q = c/a ad γ + + γ =p =(λ b) /a, we may write γ ± = p + q (cos θ ± i si θ) = ρ e±iθ, where a ρ = c, cos θ = By the method of partial fractios, u = = = γ h γ (+) p p + q = λ b, ρ, θ C. (4) ac (f + cu ) h γ + h γ (+) + (f + cu ) h a + γ + + γ + (f + cu ) h, c

W. C. Yueh 69 where the last two equalities are due to a h = a (+) =0 ad γ +γ = c/a. Applyig De Moivre s Theorem, this may further be writte as u = i ρ + si ( +)θ (f + cu ) h. Settig f = αu, f = βu ad f =0for =,,wemayevaluateu ad obtai u = i cu ρ si θ + αu ρ si ( ) θ + H ( ) βu ρ si ( ) θ (5) for, where H (x) is the uit step fuctio defied by H(x) =ifx 0ad H(x) = 0ifx<0. I particular, i u + = cu ρ + si ( +)θ + αu ρ si θ + βu ρ si θ = cu ρ + si ( +)θ + αu ρ si θ i si θ +βρ cu ρ si θ + αu ρ si ( ) θ = cu ρ + si ( +)θ +(α + β) u ρ si θ + αβu c ρ si ( ) θ, wherewehavesubstitutedi ac si θ =.Siceρ,u =0adu + =0, we fially obtai the ecessary coditio ac si ( +)θ +(α + β) ac si θ + αβ si ( ) θ =0. (6) Sice γ + = γ, γ + γ =i c a si θ = 0. By Lemma, θ = mπ for m Z. Theby (4), we have λ = b + ac cos θθ= mπ, m Z. (7) Note that we may also obtai from (5) that u = i cu ρ si θ + αu ρ si ( ) θ = u si θ ρ si θ + α si ( ) θ ac for =,,...,. Case II. γ ± are two equal roots. I this case, q =0,or =(b λ) 4ac =0. So we have λ = b ± ac. (9) Furthermore, from (3), we have u = = (f + cu ) h c + b λ c h + a c h = (f + cu ) h c a c h + a c h ρ h ac ( ρ h) (f + cu )= (±) + ρ (f + cu ) ac (8)

70 Eigevalues of Tridiagoal Matrices The -th term ow becomes u = (±) + cu ρ +(±) αu ( )ρ ac + ac (±) + H( )βu ( ) ρ. (0) By lettig u + =0,weobtai ac (α + β) ac + αβ +(ac αβ) =0. Sice this formula must be valid for all, thus ac ± (α + β) ac + αβ =0ad ac αβ = 0. This yields the ecessary coditio α = β = ac (where α = β = ac correspods to the eigevalue λ = b + ac, adα = β = ac correspods to the eigevalue λ = b ac). The correspodig eigevectors may be obtaied from (0). Sice, wehave,ifwesetu =,u =( ρ) whe α = ac ad u = ρ whe α = ac. 3 Special Tridiagoal Matrices Now we ca apply the results of the last sectio to fid the eigevalues of several tridiagoal matrices of the form (). We will assume ac = 0adsetρ = a/c as before. Suppose α = β =0iA. Suppose λ is a eigevalue. I Case I, (6) reduces to si ( +)θ =0. Hece by Lemma, θ = kπ,k=0, ±, ±,... + Case II does ot hold sice 0 = α = β = ac is ot allowed. I other words, if λ is a eigevalue of A ad (u,u,..., u ) is a correspodig eigevector, the accordig to (7), λ = b + ac cos kπ + for some k {,..., }, ad the correspodig, accordig to (8), is give by = ρ si kπ,=,,...,. () + wherewehaveassumed =si kπ +. Coversely, we may check by reversig the argumets i Sectio that for each k {,..., }, the umber λ k = b + ac cos kπ,k=,,...,, () +

W. C. Yueh 7 is a eigevalue ad the vector =(,u(k),..., u(k) ) a correspodig eigevector of A. Before proceedig further, we itroduce the followig Lemma. LEMMA. Let β + b c 0 0... 0 0 a b c 0... 0 0 B = 0 a b c... 0 0,..................... 0 0 0 0... a α + b which is obtaied from A by iterchagig the umbers α ad β. The the eigevalues, of B are the same as A, ad the correspodig eigevectors v (k) = v (k),..., v(k) k =,...,, are give by v (k) = ρ +,k=,,..., (3) where =,..., u(k),k=,...,, are the eigevectors of A. PROOF. Let λ be a eigevalue ad u =(u,..., u ) a correspodig eigevector of A.Let 0 0... 0 ρ 0 0... ρ 4 0 R =................ 0 ρ... 0 0 ρ 0... 0 0 The sice A u = λu, wehave R A R R u = λr u or A u = λu,whereu = R u = ρ u, ρ 4 u,..., ρ u ad A = R A R. By otig that ρ c = a ad ρ a = c, itisotdifficult to see that A = B.Letv = u,thewehaveb v = λv. Thus B has the same eigevalues λ as A, ad the correspodig eigevectors v = (v,...,v ) are give by v = ρ u +. This completes the proof. Now suppose α =0adβ = ac = 0. This yields αβ =0adα + β = ac. I Case I, (6) becomes si ( +)θ +siθ =0. or si ( +)θ cos θ =0. Sice θ = mπ, m Z, by Corollary of Lemma, cos θ (+)θ =0,wehavesi =0. Thus θ = kπ,k=0, ±, ±,... + Case II does ot hold sice α = 0 = ac. By reasos similar to the case where α = β = 0 above, we may ow see the followig.

7 Eigevalues of Tridiagoal Matrices THEOREM. Suppose α =0adβ = ac = 0. The the eigevalues λ,..., λ of A are give by λ k = b + ac cos The correspodig eigevectors = kπ,k=,,...,. (4) +, k =,..., are give by,..., u(k) = ρ si kπ,=,,...,. + We remark that i case β =0adα = ac = 0, Lemma says that the eigevalues are the same as give by (4). The correspodig eigevector v (k) = v (k),..., v(k), k =,...,, i view of (3), are v (k) = ρ si k ( ) π,=,,...,. (5) + The eigevalues ad the correspodig eigevectors of the other case αβ = 0adα+ β = ac ca be obtaied i a similar way. I Case I, ow (6) becomes si ( +)θ si θ =0or ( +)θ cos si θ =0. Sice θ = mπ, m Z, by Corollary of Lemma, si θ (+)θ =0,wehavecos =0. Thus (k ) π θ =,k= ±, ±, ±3,... + THEOREM. Suppose α =0adβ = ac = 0. The the eigevalues λ,..., λ of A are give by λ k = b + (k ) π ac cos,k=,, 3,...,. (6) +, The correspodig eigevectors =,..., u(k) k =,...,, are give by = ρ si (k ) π, =,,...,. + I case β =0adα = ac = 0, the eigevalues are give by (6) ad the correspodig eigevectors by v (k) = ρ cos (k ) ( ) π,=,,...,. ( +) Next, suppose α = β = ac = 0, the (6) reduces to si ( +)θ si ( ) θ =0

W. C. Yueh 73 or Sice si θ = 0,thuscosθ =0,sothat cosθ si θ =0. (k ) π θ =,k= ±, ±, ±3,... Cases II does ot hold as before. THEOREM 3. Suppose α = β = ac = 0. The the eigevalues λ,...,λ of A are give by λ k = b + (k ) π ac cos,k=,, 3,...,. (7) The correspodig eigevectors = are give by = ρ si,,..., u(k) k =,...,, accordigto(8), (k ) ( ) π,=,,...,. 4 I case α = β = ac = 0, the eigevalues are give by (7) ad the correspodig eigevectors by v (k) = ρ cos (k ) ( ) π,=,,...,. 4 Next, suppose α = β = ac = 0,orα = β = ac = 0. Ifλ is a eigevalue of A, the i Case I, (6) reduces to or siθ (cos θ +)=0. siθ (cos θ ) = 0 respectively. Sice θ = mπ, m Z, by Corollary of Lemma, cos θ ± = 0,wehave si θ =0. Thus θ = kπ,k=0, ±, ±, ±3,.... Sice θ = mπ for m Z ad sice cos θ is eve ad periodic, we obtai λ = b + ac cos kπ,k=,, 3,...,. I Case II, by (9), we have λ = b + ac = b + ac cos 0 if α = β = ac, ad λ = b ac = b + ac cos π if α = β = ac. THEOREM 4. Suppose α = β = ac = 0. The the eigevalues λ,..., λ of A are give by λ k = b + ac cos kπ,k=,, 3,...,,

74 Eigevalues of Tridiagoal Matrices ad the correspodig eigevectors = for k =,,..., ad = ρ si,..., u(k) are give by k ( ) π,=,,...,, =( ρ),=,,...,, for k =. THEOREM 5. Suppose α = β = ac = 0. The the eigevalues λ,...,λ of A are give by λ k = b + (k ) π ac cos,k=,, 3,...,, ad the correspodig eigevectors =,..., u(k) are give by = ρ,=,,..., for k =ad for k =, 3,...,. = ρ cos (k ) ( ) π,=,,...,, Refereces [] S. S. Cheg, Partial Differece Equatios, Taylor ad Fracis, Lodo ad New York, 003. [] S.S.Cheg,M.Gil ad C.J.Tia,Sychroizatio i a discrete circular etwork, Proceedigs of the 6-th Iteratioal Coferece o Differece Equatios, i press. [3] R. T. Gregory ad D. Karey, A collectio of matrices for testig computatioal algorithm, Wiley-Itersciece, 969. [4] J. F. Elliott, The characteristic roots of certai real symmetric matrices, Mater thesis, Uiv. of Teessee, 953.