Special Symmetrical Mtrix 1

Transcription

1 Special Symmetrical Mtrix Eigenvalues and eigenvectors for Special Symmetrical Matrix We shall consider the following special symmetrical matrix with n rows and n columns denoted y S: S = () Let us consider the characteristic polynomial of S. If we denote the eigenvalue of S as λ, It is given y λ 0 0 λ 0 λ 0 λ =0, () where this is a determinant of n n matrix and a monic polynomial of degree n. we shall denote this type of n n matrix as D n+,which also depends upon λ. If we expand this determinant along the first column, we otain the recursive relations 0 0 λ D n+ =λd n + = λd n D n for n =,,3, (3) λ where we set D =, D 0 =0. To solve these recursive relations for D n, we deform them as D n+ αd n = β(d n αd n ) for n =,,3, (4) Comparing the Eq.(3) and Eq.(4), we have α + β=λ and αβ =. (5) These two constants α, β are solutions of the quadratic equation for t t λt + = 0. (6) From Eq.(4),we have successive equations D n+ αd n = β(d n αd n ) = β (D n αd n ) = = β n (D αd 0 ). Then we otain

2 Kiyoto Hira D n+ αd n = β n, (7) where we have used D =, D 0 =0. Swapping α for β in Eq.(4), similarly we also otain D n+ βd n = α n. (8) From Eq.(7) and Eq.(8), if α β we otain the solutions of the recursive relations for D n (n = 0,,,3 ) D n = βn α n β α (n = 0,,,3 ). (9) And also if α = β, then Eq.(5) yields λ =,and we otain D n+ = (n+)α n (n = 0,,,3 ). () Now, let us consider the D n+ when λ >. If we solve the quadratic equation (6) for t, t = λ± λ 4, () where, since λ 4 >0, then t are real numers. Here we take α and β as β = λ+ λ 4 and α = λ λ 4. (3) We shall show in the ellow that the D n+ is not equal to zero when λ. In other words the asolute values of the roots for the characteristic polynomial of S must e less than. Note that well known formula (a + ) N = N K K a N K K, where coefficients N are inomial ones. K We have β n+ = ( λ + λ 4 ) n+ = n+ 0 (λ )n+ + n+ (λ )n ( λ 4 ) + n+ (λ )n ( λ 4 ) + n+ 3 (λ )n ( λ 4 ) 3 + n+ 4 (λ )n 3 ( λ 4 ) 4 + n+ 5 (λ )n 4 ( λ 4 ) 5 +,

3 Special Symmetrical Mtrix 3 α n+ = λ n+ λ 4 = n+ 0 (λ )n+ n+ λ n λ 4 + n+ λ n λ 4 n+ 3 (λ )n ( λ 4 ) 3 + n+ 4 (λ )n 3 ( λ 4 ) 4 n+ 5 (λ )n 4 ( λ 4 ) 5 +, β n+ α n+ = λ n+ + λ 4 λ n+ λ 4 = n+ (λ )n ( λ 4 ) + n+ 3 (λ )n ( λ 4 ) 3 + n+ 5 (λ )n 4 ( λ 4 ) 5 +, β α = λ 4. Then we otain,using the aove two equations, D n+ = βn+ α n+ β α for β α = n+ (λ )n + n+ 3 (λ )n ( λ 4 ) + n+ 5 (λ )n 4 ( λ 4 ) 4 +. (4) From this equation we can conclude as follows: If n is even numer, ( λ )n, ( λ )n, ( λ )n 4 are positive numers and D n+ is positive definite for λ >. And if n is odd numer and λ <, ( λ )n, ( λ )n, ( λ )n 4 are negative numers and D n+ is negative definite for odd n and λ <. if n is odd numer and λ >, D n+ is positive definite for odd n and λ >. From these, the D n+ is not equal to zero for λ >. We shall also show that the D n+ is not equal to zero when λ =. From Eq.(6), λ = yields α = β = + and λ = yields α = β =. From Eq.() we have D n+ = (n+)α n = (n+)(±) n 0 (n = 0,,,3 ), where α = + with λ = + and α = with λ =. From the these fact, the eigenvalue λ of the matrix S must e λ <. From Eq.(6), for the eigenvalue λ t = λ± λ 4 = λ±i 4 λ =e ±iθ, (5)

4 4 Kiyoto Hira where we have taken cosθ = λ, 4 λ sinθ =. (6) Hence, if we take α and β as β = e +iθ and α = e iθ, we have D n+ = βn+ α n+ β α = e+i(n+)θ e i(n+)θ sin (n+)θ = e +iθ e iθ sinθ. (7) Now we can determine the eigenvalues for the matrix S. Since we must have D n+ = 0 for eigenvalue λ,we otain sin(n + ) θ = 0 and sinθ 0. sin(n + ) θ = 0 yields,corresponding the numer of the solutions for D n+, (n + )θ k = kπ (k =,, n), (8) where k = 0 is ruled out ecause sinθ = 0.Now we can otain the eigenvalues λ k (k =,,, n) for the matrix S. With Eq.(6) λ k = cosθ k, with the help of Eq.(8), we otain the eigenvalues λ k = cosθ k =cos ( kπ ) (k =,,, n), (9) n+ Next we shall determine the eigenvectors for the matrix S. If λ k (k =,,, n) are the eigenvalues for the S and v k v k = i v k are eigenvectors for λk (k =,,3,, n). v n k The matrix equation determining the eigenvectors for λ k (k =,,, n) are given v k v k 0 i 0 0 v k =λk i v k. (0) 0 0 v n k v n k From this equation, we have n simultaneous equtions v k =λ k v k v k + v k 3 =λ k v k v k + v k 4 =λ k v k 3

5 Special Symmetrical Mtrix 5 v k i + v k i+ =λ k v k i v k n + v k n =λ k v k n v k n =λ k v k n. These equations are put into the recursive relations v k l + v k l+ =λ k v k l (l =,, n), () where we set v k 0 =0, v k =, and v k n+ =0. These are the same types as Eq.(3), hence we can solve these the same as efore, y Eq.(6) t λ k t + = 0 yields t = λ k±i 4 λ k = cosθ k±i 4 4cos θ k = cosθ k±isinθ k = e ±iθ k, () where λ k = cosθ k as in Eq. (9). if we take β = e +iθ k and α = e iθ k and we can solve the recursive relation Eq.() the same as efore, we otain v k l = βl α l β α = e+ilθ k e ilθ k e ±iθ k e ±iθ k =sin (lθ k) sinθ k (l=,,,n), where,with Eq.(8), v n+ k = sin ((n+)θ k) = sinθ k kπ sin ((n+) n+ ) sinθ k =0,required condition. The square of the norm of eivenvector v k for λ k is given as v k = (v k ) + (v k ) + (v k 3 ) + + (v k n ) = sin (θ k ) sin θ k + sin (θ k ) sin θ k + sin (3θ k ) sin θ k + + sin (nθ k ) sin θ k = sin θ k {sin (θ k ) + sin (θ k ) + sin (3θ k ) + + sin (nθ k ) } = { cos (θ k) + cos ((θ k )) + cos (3(θ k )) sin θ k + + cos (n(θ k )) = sin θ k {n+ [ + cos(θ k ) + cos (θ k )+cos 3(θ k )+ + cos n(θ k )]}

6 6 Kiyoto Hira = sin θ k {n+ cos[(n + )θ k ] sin ((n+)θ k) sin θ k } = n+ sin θ k, (3) where we have used sin ((n + )θ k )=0 from Eq.(8) and the fact that + cos x + cos x + + cos nx=re{ + e ix + + e nix } = Re ei(n+)x e ix n+ sin ( x) =Re{ei(n+)x/ sin ( x ) }. Now if we denote the orthonormal eigenvector of v k as e k, we otain sin (lθ k ) e l k = v k l v k = sinθ k n + sin θ k = n+ sin (lθ k)= n+ klπ sin ( ) (k,l=.,, n), (4) n+ where we have used Eq.(3) and Eq.(8). And We otain with the help of the orthogonality properties of the S matrix e k l =e l k (k,l=.,, n), (5) n n e k e m = e l l l= k e m = l= e l k e m l = δ mk. (k, m =,,, n) (6) Finaly we shall otain the eigenvalues and corresponding orthonormal eigenvectors for the n n symmetrical matrix S a given y a 0 0 a a S a = 0 a. a 0 a S a = a 0 0 = I + as. (7) This yields, y using the facts stated aove, Se k = λ k e k, so that

7 Special Symmetrical Mtrix 7 S a e k =(I + as)e k =( + aλ k )e k (k =,,, n). Then the ( + aλ k ) and e k (k =,,3,, n) are the eigenvalues and the corresponding normalized eigenvectors for S a. More generally we shall otain the eigenvalues and corresponding eigenvectors for the n n matrix T a given y a 0 0 c a T a = 0 c a (c 0). (8) 0 c a Note that there exist a diagonal matrix P such that P T a P = ai + c S, c 0 where P = c 0 0 c, (9) 0 n 0 0 c with an aritorary constant c. If v j (j=,,,n) are the eigenvectors of the Matrix S,then we have P T a P v j = ai + c S v j, T a P v j = P a + c λ k v j = a + c λ k P v j. Hence, P v j (j =,,, n) are the eigenvectors of the Matrix T a. Let the l component of P v j P v j l.then we have P v j l = n P lk (v k= j ) = N c j ( c k )l δ lk sin ( kjπ k= ) = c n+ j ( c )l sin ( ljπ ). (30) n+

8 8 Kiyoto Hira We shall solve the previous prolem in the way different from the stated aove. We denote the following matrix ast c : c 0 T c = 0 c 0. 0 c 0 The matrix equation determining the eigenvector λ for T c is v v c 0 0 c 0 v i =λ v i (c 0). () 0 c 0 v n v n From this equation, we have n simultaneous equtions v =λv cv + v 3 =λv cv + v 4 =λv 3 cv i + v i+ =λv i cv n + v n =λv n cv n =λv n. These equations are put into the recursive relations cv i + v i+ =λv i (i =,, n), () where we set v 0 =0, v =, and v n+ =0. To solve this recursive relations for v i, we deform this equations as v i+ αv i = β(v i αv i ) (i =,, n), (3) Comparing the Eq.() and Eq.(3), we have α + β= λ and αβ = c. (4) These two constants α, β are solutions of the quadratic equation for t t λt + c = 0. (5)

9 Special Symmetrical Mtrix 9 From Eq.(3),we have the successive relations v i+ αv i = β(v i αv i ) = β (v i αv i ) = = β i (v αv 0 ), Then,with the intial conditions, v 0 =0 and v =, we have v i+ αv i = β i. (6) Similarly swapping β for α,we have v i+ βv i = α i. (7) With these two Equations we otain the solutions for Eq.() β i α i v i = β α (α β) for for i =,,3. (8) iα i (α = β) When α = β, Eq.(5) yields λ 4c=0, λ = ± c. With Eq.(4) α + β= λ yields α = β = ± c for α = β, and Eq.(8) leads to v n+ = (n+)α n =(n+)(± c )n 0. We see that α = β does not compatile with the conditions v n+ = 0. Hence the required solutions for () are v i = βi α i β α (α β) (i = 0,,,, n + ). From this equation, we must have v n+ = 0, yielding β α n+ =. Then we have β = e jπ n+ (j =,, n), α where j = 0 is ruled out ecause this means α = β. With αβ = c within Eq.(4), we otain

10 0 Kiyoto Hira β = α c β = e jπ n+ (j =,, n). And we otain β j = c e jπ n+ (j =,, n). We also otain another solutions α j corresponding to β j α j = c β j = c e jπ n+ (j =,, n). We may otain the eigenvalue λ j corresponding j,from α j + β j = λ j λ j = α j + β j = c e jπ n+ + e jπ n+ = c cos ( jπ n+ ). (9) We otain the i component v i for the eigenvector for eigenvalue λ j, v i = βi α i i ijπ c β α = (e n+ e n+) ijπ c jπ (e n+ e n+) jπ i = c sin ijπ n + sin jπ n + i = c j c sin ( ijπ ), (0) n+ where c j (j =,, n) are constants and we suppressed the suffix j for v i for the simplicity.