Mathematical Problems of Computer Science 58, 7–19, 2022.

doi:10.51408/1963-0088

UDC 519.6

Analytical Inversion of Tridiagonal Hermitian Matrices

Yuri R. Hakopian and Avetik H. Manukyan

Yerevan State University, Yerevan, Armenia

e-mail: yuri.hakopian@ysu.am, avetiq.manukyan1@ysumail.am

Abstract

In this paper we give an algorithm for inverting complex tridiagonal Hermitian
matrices with optimal computational efforts. For matrices of a special form and, in
particular, for Toeplitz matrices, the derived formulas lead to closed-form expressions
for the elements of inverse matrices.
Keywords: Inverse matrix, Tridiagonal matrix, Hermitian matrix, Toeplitz matrix.
Article info: Received 21 Aprile 2022; received in revised form 15 July 2022; accepted
23 August 2022.

1. Introduction

Tridiagonal matrices are encountered in many areas of applied mathematics. Such matri-
ces are of great importance in finite difference and finite element methods for differential
equations. The construction of cubic splines is reduced to solving systems with tridiag-
onal matrices. Symmetric matrices are reduced to tridiagonal matrices by the similarity
Householder transformation (see [1, 2, 3], for instance). Other examples can be cited.

There is a well-known fast numerical method for solving systems with tridiagonal matri-
ces. At the same time, the analytical matrix inversion is also of certain interest (see [4, 5, 6],
for instance). For tridiagonal matrices of special types, this leads to closed-form expressions
for the elements of inverse matrices [7, 8, 9, 10]. This is undoubtedly useful in theoreti-
cal considerations. Further, explicit formulas can be a part of more general computational
procedures. There are other reasons as well.

In this article, we focus our attention on complex Hermitian tridiagonal matrices. We
will construct a fairly simple computational procedure, consisting of a sequence of recurrence
relations, leading to the calculation of the elements of the inverse matrix. In special cases, in
particular for Toeplitz tridiagonal Hermitian matrices, the procedure can become the basis
for deriving closed-form expressions for the elements of the inverse matrix.

We note right away that throughout this article z stands for the complex conjugate of
the complex number z.

7



8 Analytical Inversion of Tridiagonal Hermitian Matrices

Let a nonsingular tridiagonal Hermitian matrix

A =




a1 b1
b1 a2 b2 0

...
...

...

0 bn−2 an−1 bn−1
bn−1 an




(1)

be given, where ai, i = 1, 2, . . . , n are real numbers and bi ̸= 0 for i = 1, 2, . . . , n − 1. In
accordance with the accepted notation, A = A∗. We assume that n > 3. The requirement
that the subdiagonal (superdiagonal) elements of the matrix be nonzero is not restrictive.
Indeed, if some of these elements are equal to zero, the problem of computing the inverse
matrix is decomposed into several similar problems for tridiagonal matrices of lower order.

2. Preliminary Calculations

Let A−1 = [xij]n×n. This matrix is also Hermitian. In our considerations we will use the
notation

X(j) ≡ [x1 j x2 j . . . xn j]T , j = 1, 2, . . . , n
for the columns of the inverse matrix.

The matrix A can be represented as a product

A = DB (2)

of the matrices
D = diag [b1, b1, b2, . . . , bn−2, bn−1] (3)

and

B =




p 1
1 f2 g2 0

1 f3 g3
...

...
...

0 1 fn−1 gn−1
1 q



, (4)

where

fi =
ai

bi−1
, gi =

bi

bi−1
, i = 2, 3, . . . , n − 1; p =

a1
b1
, q =

an

bn−1
. (5)

Having a nonsingular matrix B defined in (4), let us consider the following system of
linear algebraic equations

pµ1 + µ2 = α

µi−1 + fiµi + giµi+1 = 0, 2 ≤ i ≤ n − 1
µn−1 + qµn = 0,

(6)

where we will set the right-hand side α of the first equation a little later. It is easy to verify
that regardless of the choice of α, the recursively defined quantities

µn = 1 , µn−1 = −q ,
µi−1 = −fiµi − giµi+1 , i = n − 1, n − 2, . . . , 2

(7)



Yu. Hakopian and A. Manukyan 9

satisfy all equations of the system (6), starting with the second one. Then, we choose the
quantity α as follows:

α = pµ1 + µ2. (8)

Remark 1 Since, by assumption, the matrix B is nonsingular (it follows from (2)), then
α ̸= 0. Indeed, otherwise we would have obtained that the homogeneous system (6) has a
nontrivial solution. Further,

α =
a1
b1
µ1 + µ2 =

1

b1
(a1µ1 + b1µ2).

Therefore
a1µ1 + b1µ2 ̸= 0

as well.

Thus,
α = b−11 t

−1, (9)

where
t ≡ (a1µ1 + b1µ2)−1. (10)

Let us introduce the vector
r(1) ≡ [µ1 µ2 . . . µn]T ,

the components of which are specified in (7). As follows from (4), (6) and (9),

Br(1) = [α 0 . . . 0]T = αe(1) = b−11 t
−1e(1),

where e(1) ≡ [1 0 . . . 0]T . Further, on the basis of factorization (2) of the matrix A, we obtain
the equality

Ar(1) = DBr(1) = b−11 t
−1De(1) = t−1e(1); (11)

here we have used the obvious equality De(1) = b1e
(1) (see (3)). The equality (11) allows to

compute the first column of the inverse matrix A−1. Indeed, from this equality we find that

A−1e(1) = tr(1).

Since A−1e(1) = X(1), then X(1) = tr(1), or

xi1 = tµi, i = 1, 2, . . . , n. (12)

Thus, we have found the first column of the inverse matrix. Similarly, we can calculate
the last column of the matrix A−1. For this purpose, let us consider the linear system

pν1 + ν2 = 0

νi−1 + fiνi + giνi+1 = 0, 2 ≤ i ≤ n − 1
νn−1 + qνn = β,

(13)

where we will set the right-hand side β of the last equation later. Regardless of the choice
of β, the recursively defined quantities

ν1 = 1 , ν2 = −p ,

νi+1 = −
1

gi
(νi−1 + fiνi) , i = 2, 3, . . . , n − 1

(14)



10 Analytical Inversion of Tridiagonal Hermitian Matrices

satisfy the first n−1 equations of the system (13). Then we choose the quantity β as follows:

β = νn−1 + qνn. (15)

Since the matrix B is nonsingular, then β ̸= 0 (see Remark 1). Substituting the expression
of the quantity q given in (5) into (15) yields

β = νn−1 +
an

bn−1
νn =

1

bn−1
(bn−1νn−1 + anνn).

Thus,

β = bn−1
−1
θ−1, (16)

where
θ ≡ (bn−1νn−1 + anνn)−1.

Now let us introduce the vector

r(n) ≡ [ν1 ν2 . . . νn]T ,

the components of which are specified in (14). From (4), (13) and (16) we find that

Br(n) = [0, . . . 0 β]T = βe(n) = bn−1
−1
θ−1e(n),

where e(n) ≡ [0 . . . 0 1]T . Having the factorization (2) of the matrix A, we obtain the equality

Ar(n) = DBr(n) = bn−1
−1
θ−1De(n) = θ−1e(n).

From here,
A−1e(n) = θr(n).

Since A−1e(n) = X(n), then X(n) = θr(n), or

xin = θνi, i = 1, 2, . . . , n. (17)

Let us refine the last expression. From (12), xn1 = tµn = t. Further, according to (17),
x1n = θν1 = θ. Since A

−1 is a Hermitian matrix, then x1n = xn1. Consequently, θ = t, and
we come to the conclusion that

xin = tνi, i = 1, 2, . . . , n. (18)

So, we have found the first and the last columns of the Hermitian matrix A−1. These
are expressions (12) and (18). Taking into account that ν1 = 1 and µn = 1, we write these
elements in the form of

xi1 = tµiν1, xin = t µnνi, i = 1, 2, . . . , n. (19)

Moreover, the diagonal elements x11 = tµ1ν1 and xnn = t µnνn are real numbers. Therefore,
we can write xnn = tµnνn as well.

Looking ahead, we say that in the next section we will prove that the quantities

tµiνi, i = 2, 3, . . . , n − 1 (20)

are the remaining diagonal elements of the matrix A−1. To do this, here we first establish
that the quantities (20) are real numbers (naturally, without assuming that they are somehow
related to the matrix A−1).

Let us introduce into consideration the quantities

Ri ≡ bi−1(tµiνi−1) + bi−1(tµi−1νi), i = 2, 3, . . . , n − 2. (21)



Yu. Hakopian and A. Manukyan 11

Lemma 1. The quantity R2 is a real number.

Proof. Since ν1 = 1 and ν2 = −p (see (2.13)), then

R2 = t(b1µ2ν1 + b1µ1ν2) = tb1(µ2 − pµ1).

Further, taking into account the equalities (8) and (9), we get

R2 = tb1(α − 2pµ1) = tb1α − 2pb1(tµ1) = 1 − 2a1(tµ1).

The quantities a1 and tµ1 are real numbers, so R2 is also a real number. 2

Lemma 2. The quantities Ri from (21) satisfy the relations

Ri = −Ri−1 − 2ai−1(tµi−1νi−1), i = 3, 4, . . . , n − 2. (22)

Proof. From (6) we have the equality

µi−2 + fi−1µi−1 + gi−1µi = 0.

Using formulas (5), let us write this equality in the form of

bi−2µi−2 + ai−1µi−1 + bi−1µi = 0.

Multiplying both parts of the last equality by tνi−1, we get that

bi−1(tµiνi−1) = −bi−2(tµi−2νi−1) − ai−1(tµi−1νi−1). (23)

Similarly, from (13) we have the equality

νi−2 + fi−1νi−1 + gi−1νi = 0,

which can be written as follows:

bi−2νi−2 + ai−1νi−1 + bi−1νi = 0.

Multiplying both parts of this equality by tµi−1 yields

bi−1(tµi−1νi) = −bi−2(tµi−1νi−2) − ai−1(tµi−1νi−1). (24)

The relation (22) follows directly from the equalities (23) and (24). 2

Lemma 3. The quantities tµiνi, i = 2, 3, . . . , n − 1 are real numbers.

Proof. Consider first the quantity tµ2ν2. Since pµ1 + µ2 = α and ν2 = −p (see (6) and
(14)), then

tµ2ν2 = t(pµ1 − α)p = (pp)(tµ1) − tαp.

Further, using the equality (9), we obtain that

tµ2ν2 = (pp)(tµ1) −
p

b1
= (pp)(tµ1) −

a1

b1b1
.

Thus, the quantity tµ2ν2 is a real number.



12 Analytical Inversion of Tridiagonal Hermitian Matrices

Next, consider the quantity tµ3ν3. As follows from (6) and (13),

µ3 = −
a2
b2
µ2 −

b1
b2
µ1, ν3 = −

a2

b2
ν2 −

b1

b2
ν1.

Proceeding from these equalities, we get that

tµ3ν3 =
1

b2b2

[
a22(tµ2ν2) + b1b1(tµ1ν1) + a2R2

]
.

The quantities tµ1ν1 and tµ2ν2 are real numbers. According to Lemma 1, the quantity R2
is also a real number. Therefore, tµ3ν3 is a real number as well.

Further reasoning will be carried out by the method of mathematical induction on i.
Suppose that for some value of i, where 3 ≤ i ≤ n − 2, it is already known that the
quantities tµkνk, k ≤ i and Rk, k ≤ i − 1 are real numbers. From (6) and (13) we have

µi+1 = −
ai
bi
µi −

bi−1
bi

µi−1, νi+1 = −
ai

bi
νi −

bi−1

bi
νi−1.

Then

tµi+1νi+1 =
1

bibi

[
a2i (tµiνi) + bi−1bi−1(tµi−1νi−1) + aiRi

]
.

Hence, by virtue of the assumptions made and taking into account the assertion of Lemma
2, we arrive at a conclusion that the quantity tµi+1νi+1 is a real number. 2

Remark 2 We have established that the quantities tµiνi, i = 1, 2, . . . , n are real numbers.
Therefore, tµiνi = t µiνi.

3. The Elements of the Inverse Matrix

Above we obtained the expressions (19) for the elements of the first and the last columns
of the inverse matrix, as well as some auxiliary statements. Based on these results, here we
derive formulas for the remaining elements of the inverse matrix.

Let 2 ≤ j ≤ n − 1. We introduce into consideration the vector

r(j) ≡ [ t µjν1 , . . . , t µjνj−1, tµjνj, tµj+1νj , . . . , tµnνj]T , (25)

where the quantities µi and νi are specified in (7) and (14), respectively. Multiplying the
matrix B defined in (4) and the vector r(j) yields

Br(j) = z(j), (26)

where the components of the vector

z(j) = [z
(j)
1 z

(j)
2 . . . z

(j)
j−1 δj z

(j)
j+1 . . . z

(j)
n−1 z

(j)
n ]

T

are calculated as follows:

z
(j)
1 = t µj(pν1 + ν2),

z
(j)
i = t µj(νi−1 + fiνi + giνi+1), 2 ≤ i ≤ j − 1,
δj = t µjνj−1 + fj(tµjνj) + gj(tµj+1νj),

z
(j)
i = t(µi−1 + fiµi + giµi+1)νj, j + 1 ≤ i ≤ n − 1,

z(j)n = t(µn−1 + qµn)νj.



Yu. Hakopian and A. Manukyan 13

Having equations (6) and (13), we conclude that z
(j)
i = 0 for 1 ≤ i ≤ j −1 and j +1 ≤ i ≤ n.

Thus,
z(j) = [0 . . . 0 δj 0 . . . 0]

T = δje
(j), (27)

where e(j) = [0 . . . 0 1 0 . . . 0]T (the unit is located on jth place).
It remains to clarify the quantity δj. Taking into account Remark 2, we have

δj = t µjνj−1 + fj(t µjνj) + gj(tµj+1νj)

= t µj(νj−1 + fjνj) + gj(tµj+1νj).
(28)

Since νj−1 + fjνj = −gjνj+1 (see (13)), then

δj = gj(tµj+1νj − t µjνj+1), 2 ≤ j ≤ n − 1. (29)

Let us get one more representation of the quantity δj. Since gjµj+1 = −µj−1 −fjµj (see (6)),
then from(28) it follows that

δj = t µjνj−1 − tµj−1νj + fj(t µjνj − tµjνj).

From here, according to Remark 2, we obtain

δj = t µjνj−1 − tµj−1νj, 2 ≤ j ≤ n − 1. (30)

Assuming that 3 ≤ j ≤ n − 1, we can write the expression (30) in the form of

δj =
1

gj−1
gj−1(tµjνj−1 − t µj−1νj).

Comparing with the record (29), we arrive at the relation

δj =
1

gj−1
δj−1, 3 ≤ j ≤ n − 1. (31)

Based on the relation (31), one can easily show that

δj =


 bj−1

−1
b1δ2 , if j is odd,

bj−1
−1
b1δ2 , if j is even.

(32)

Finally, let us calculate the quantity δ2. According to the representation (30), we have

δ2 = t µ2ν1 − tµ1ν2 = t µ2 + tµ1p

= t µ2 + t µ1 p = t (µ2 + p µ1) = t α = b1
−1
,

(33)

(see (6) and (9)). Thus, from (32) and (33) we conclude that

δj = bj−1
−1
, j = 2, 3, . . . , n − 1. (34)

Summing up the results, from (27) and (34) we come to the equality

z(j) = bj−1
−1
e(j). (35)



14 Analytical Inversion of Tridiagonal Hermitian Matrices

Proceeding from the factorization (2) of the matrix A and using the equalities (26) and
(35), we have

Ar(j) = DBr(j) = Dz(j) = bj−1
−1
De(j) = e(j)

(note that De(j) = bj−1e
(j), which follows from (3)). Further,

A−1e(j) = r(j).

Since A−1e(j) = X(j), then X(j) = r(j). The components of the vector r(j) are given in (25).
Thus,

xij = t µjνi, i = 1, 2, . . . , j − 1 and xij = tµiνj, i = j, j + 1, . . . , n. (36)

Combining formulas (36) with those of (12) and (18) yields

xij =




t µjνi, i = 1, 2, . . . , j − 1,

tµiνj, i = j, j + 1, . . . , n
for j = 1, 2, . . . , n. (37)

Note the following. Since the matrix A−1 is also Hermitian, then in reality we only need
to calculate the lower triangular part of this matrix.

Summarizing the considerations of Sections 2 and 3, let us write the following procedure
to calculate the elements of the inverse matrix A−1 = [xij]n×n for nonsingular matrix A given
in (1).

Procedure Inv 3d Hermitian

1. Input elements a1, a2, . . . , an and b1, b2, . . . , bn−1 of the matrix A (see (1)).

2. Calculate the quantities fi, gi, p and q (see (5)):

fi =
ai

bi−1
, gi =

bi

bi−1
, i = 2, 3, . . . , n − 1; p =

a1
b1
, q =

an

bn−1
.

3. Calculate recursively the quantities µi (see (7)):

µn = 1 , µn−1 = −q ,

µi = −fi+1µi+1 − gi+1µi+2 , i = n − 2, n − 3, . . . , 1.

4. Calculate recursively the quantities νi (see (14)):

ν1 = 1 , ν2 = −p ,

νi = −
1

gi−1
(νi−2 + fi−1νi−1) , i = 3, 4, . . . , n.

5. Calculate the quantity t (see (10) and Remark 1):

t = (a1µ1 + b1µ2)
−1.

6. Calculate the lower triangular part of the matrix A−1 (see (37)):

xij = tµiνj, i = j, j + 1, . . . , n ; j = 1, 2, . . . , n .



Yu. Hakopian and A. Manukyan 15

7. Set the upper triangular part of the matrix A−1 (see (37)):

xij = xji, i = 1, 2, . . . , j − 1 ; j = 2, 3, . . . , n .

8. Output the matrix A−1 = [xij]n×n.

End procedure

The procedure Inv 3d Hermitian can be useful for the following purposes. Firstly, it can
be used as a basis of numerical algorithms for inverting nonsingular tridiagonal Hermitian
matrices. In this case, it is easy to make sure that computing the lower triangular part of the
matrix A−1 requires 0.5n2 + O(n) arithmetical operations with complex numbers. Secondly,
for matrices of special types, the procedure can be used for deriving closed form expressions
for the elements of inverse matrices. The next section is devoted to this issue.

4. Toeplitz Tridiagonal Hermitian Matrices

Let us consider a matrix

A =




a b
b a b 0

...
...

...

0 b a b
b a




(38)

of order n, where a is a real number and b ̸= 0. Additionally, we assume that

|a| ≥ 2|b|. (39)

Condition (39) ensures the nonsingularity of the matrix (38) (see [11], for instance).
For the matrix we are considering, the quantities calculated in Item 2 of the procedure

Inv 3d Hermitian are as follows:

fi =
a

b
, gi =

b

b
, i = 2, 3, . . . , n − 1; p =

a

b
, q =

a

b
.

Further, in Item 3 of the procedure, the quantities µi are calculated. In our case, we have
second-order recurrent relations

bµi + aµi+1 + bµi+2 = 0 , i = n − 2, n − 3, . . . , 1,

where µn = 1, µn−1 = −a/b. The solution of this problem is well known (see [2, 6], for
instance). As a result of calculations, we get that

µi = (−1)n−i
b

r

[(
a + r

2b

)n+1−i
−

(
a − r
2b

)n+1−i]
, i = 1, 2, . . . , n if |a| > 2|b| (40)

and

µi = (−1)n−i (n + 1 − i)
(
a

2b

)i−n
, i = 1, 2, . . . , n if |a| = 2|b|, (41)

where
r ≡

√
a2 − 4|b|2.



16 Analytical Inversion of Tridiagonal Hermitian Matrices

In a similar way, we find expressions for the quantities νi determined in Item 4 of the
procedure. These quantities satisfy the following second-order recurrent relations:

bνi−2 + aνi−1 + bνi = 0 , i = 3, 4, . . . , n,

where ν1 = 1, ν2 = −a/b. Making calculations, we find that

νi = (−1)i−1
b

r

[(
a + r

2b

)i
−

(
a − r
2b

)i]
, i = 1, 2, . . . , n if |a| > 2|b| (42)

and

νi = (−1)i−1 i
(
a

2b

)i−1
, i = 1, 2, . . . , n if |a| = 2|b|. (43)

In Item 5 of the procedure, the quantity t is calculated. Using the expressions (40) and
(41), we get

t = (−1)n−1
r

b
2

[(
a + r

2b

)n+1
−

(
a − r
2b

)n+1]−1
if |a| > 2|b| (44)

and

t =
(−1)n−1

n + 1

2

a

(
a

2b

)n−1
if |a| = 2|b|. (45)

Finally, in Items 6 and 7 of the procedure, the elements xij of the inverse matrix A
−1

are calculated. If |a| > 2|b|, then we use the formulas (40), (42) and (44). For the values
j = 1, 2, . . . , n, we obtain that

xij =
(−1)j−i

r

[(
a + r

2b

)i
−

(
a − r
2b

)i] [(a + r
2b

)n+1−j
−

(
a − r
2b

)n+1−j]
[(

a + r

2b

)n+1
−

(
a − r
2b

)n+1] (46)
if i = 1, 2, . . . , j − 1 and

xij =
(−1)i−j

r

[(
a + r

2b

)n+1−i
−

(
a − r
2b

)n+1−i] [(a + r
2b

)j
−

(
a − r
2b

)j]
[(

a + r

2b

)n+1
−

(
a − r
2b

)n+1] (47)
if i = j, j + 1, . . . , n. As an example, consider the matrix

A =




5 2i
−2i 5 2i 0

...
...

...

0 −2i 5 2i
−2i 5



.

According to the expressions (46) and (47) we find that

xij =




(2i − 2−i)(2n+1−j − 2−n−1+j)
3(2n+1 − 2−n−1)

ii−j, i = 1, 2, . . . , j − 1,

(2n+1−i − 2−n−1+i)(2j − 2−j)
3(2n+1 − 2−n−1)

ii−j, i = j, j + 1, . . . , n,



Yu. Hakopian and A. Manukyan 17

where the symbol i stands for the imaginary unit.
Now consider the case |a| = 2|b|. For the values j = 1, 2, . . . , n, using the formulas (41),

(43) and (45), we find that

xij =




(−1)j−i
(n + 1 − j)i

n + 1

2

a

(
a

2b

)i−1 ( a
2b

)j−1
, i = 1, 2, . . . , j − 1,

(−1)i−j
(n + 1 − i)j

n + 1

2

a

(
a

2b

)i−1 ( a
2b

)j−1
, i = j, j + 1, . . . , n.

(48)

For the matrix

A =




2 i
−i 2 i 0

...
...

...

0 −i 2 i
−i 2



,

the expressions (48) take the following form:

xij =




(−1)j
(n − j + 1)i

n + 1
ii+j, i = 1, 2, . . . , j − 1,

(−1)j
(n − i + 1)j

n + 1
ii+j, i = j, j + 1, . . . , n,

j = 1, 2, . . . , n.

5. Conclusion

In this paper, we have constructed the computational procedure Inv 3d Hermitian for inver-
sion of tridiagonal Hermitian matrices. This procedure can be used as a numerical algorithm
with an optimal number of arithmetic operations (see the comment on the procedure at the
end of Section 3). In certain cases, the procedure can also be used to derive closed-form ex-
pressions for the elements of inverse matrices. In this regard, Toeplitz tridiagonal Hermitian
matrices in Section 4 were considered.

References

[1] G. H. Golub and Ch. F. van Loan, Matrix Computations, The John Hopkins University
Press, 1996.

[2] D. Kincaid and W. Cheney, Numerical Analysis, Brooks/Cole, Pacific Grove, CA,
1991.

[3] D. S. Watkins, Fundamentals of matrix computations, A Wiley Intercience Publ., 2010.

[4] B. Buchberger and G. A. Yemel’yanenko, ”Methods for inverting tridiagonal matrices”,
J. Comput. Math. and Math. Physics, vol. 13, No.3, pp. 546-554, 1973 (in Russian).

[5] M. El-Mikkawy and A. Karawia, ”Inversion of general tridiagonal matrices”, Applied
Math. Letters, vol. 19, pp. 712-720, 2006.

[6] V.P. Il’in and Yu. I. Kuznetsov, Tridiagonal Matrices and Their Applications, (in
Russian), Nauka, 1985.



1 8 Analytical Inversion of Tridiagonal Hermitian Matrices

[7 ] G.Y . H u a n d R .F. O'Co n n e ll, " A n a lyt ic a l in ve r s io n o f s ym m e t r ic t r id ia g o n a l m a t r ic e s " ,
J . P hys. A: M ath. Gen., vo l. 2 9 , p p . 1 5 1 1 -1 5 1 3 , 1 9 9 6 .

[8 ] Y . H u a n g a n d W .F. Mc Co ll, " A n a lyt ic in ve r s io n o f g e n e r a l t r id ia g o n a l m a t r ic e s " , J .
P hys. A: M ath. Gen., vo l. 3 0 , p p . 7 9 1 9 -7 9 3 3 , 1 9 9 7 .

[9 ] J. W . L e wis , " In ve r s io n o f t r id ia g o n a l m a t r ic e s " , Numer. M ath., vo l. 3 8 , p p . 3 3 3 -3 4 5 ,
1 9 8 2 .

[1 0 ] R . A . U s m a n i, " In ve r s io n o f Ja c o b i's t r id ia g o n a l m a t r ix" , Computers M ath. Applic.,
vo l. 2 7 , n o . 8 , p p . 5 9 -6 6 , 1 9 9 4 .

[1 1 ] R . H o r n a n d Ch . Jo h n s o n , M atrix Analysis, Ca m b r id g e U n ive r s it y P r e s s , 1 9 8 6 .

ºñ»ù³ÝÏÛáõݳ·Í³ÛÇÝ Ñ»ñÙÇïÛ³Ý Ù³ïñÇóÝ»ñÇ ³Ý³ÉÇïÇÏ
ѳϳ¹³ñÓáõÙ

ÚáõñÇ è. гÏáµÛ³Ý ¨ ²í»ïÇù Ð. سÝáõÏÛ³Ý

ºñ¨³ÝÇ å»ï³Ï³Ý ѳٳÉë³ñ³Ý, ºñ¨³Ý, г۳ëï³Ý

e-mail: yuri.hakopian@ysu.am, avetiq.manukyan1@ysumail.am

²Ù÷á÷áõÙ

Ðá¹í³ÍáõÙ ïñíáõÙ ¿ »ñ»ù³ÝÏÛáõݳ·Í³ÛÇÝ Ñ»ñÙÇïÛ³Ý Ù³ïñÇóÝ»ñÇ Ñ³Ï³¹³ñÓÙ³Ý
³É·áñÇÃÙÁ, áñÇ Ãí³ÛÇÝ Çñ³Ï³Ý³óáõÙÁ å³Ñ³ÝçáõÙ ¿ ûåïÇÙ³É Ãíáí Ãí³µ³Ý³Ï³Ý
·áñÍáÕáõÃÛáõÝÝ»ñ: гßíáÕ³Ï³Ý åñáó»¹áõñ³Ý Çñ»ÝÇó Ý»ñϳ۳óÝáõÙ ¿ ѳϳ¹³ñÓ
Ù³ïñÇóÇ ï³ññ»ñÇ Ñ³ßíÙ³ÝÁ ѳݷ»óÝáÕ ³Ý¹ñ³¹³ñÓ ³éÝãáõÃÛáõÝÝ»ñÇ Ñ³çáñ¹³-
ϳÝáõÃÛ³Ý: гïáõÏ ïÇåÇ Ù³ïñÇóÝ»ñÇ Ñ³Ù³ñ ¨, Ù³ëݳíáñ³å»ë, ïÛáåÉÇóÛ³Ý
»ñ»ù³ÝÏÛáõݳ·Í³ÛÇÝ Ñ»ñÙÇïÛ³Ý Ù³ïñÇóÝ»ñÇ Ñ³Ù³ñ, ëï³óí³Í ³éÝãáõÃÛáõÝÝ»ñÁ
ѳݷ»óÝáõÙ »Ý ѳϳ¹³ñÓ Ù³ïñÇóÇ ï³ññ»ñÇ Ñ³Ù³ñ µ³ó³Ñ³Ûï µ³Ý³Ó¨»ñÇ:

Àíàëèòè÷åñêîå îáðàùåíèå òðåõäèàãîíàëüíûõ
èçîáðàæåíè

Þðèé Ð. Àêîïÿí è Àâåòèê À. Ìàíóêÿí

Åðåâàíñêèé ãîñóäàðñòâåííûé óíèâåðñèòåò, Åðåâàí, Àðìåíèÿ
e-mail: yuri.hakopian@ysu.am, avetiq.manukyan1@ysumail.am

Àííîòàöèÿ

 ñòàòüå äàåòñÿ àëãîðèòì îáðàùåíèÿ òðåõäèàãîíàëüíûõ ýðìèòîâûõ ìàòðèö,
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´³Ý³ÉÇ µ³é»ñ` ѳϳ¹³ñÓ Ù³ïñÇó, »ñ»ù³ÝÏÛáõݳ·Í³ÛÇÝ Ù³ïñÇó, Ñ»ñÙÇïÛ³Ý
Ù³ïñÇó, ïÛáåÉÇóÛ³Ý Ù³ïñÇó:



Yu. Hakopian and A. Manukyan 1 9

ïîñëåäîâàòåëüíîñòü ðåêóððåíòíûõ ñîîòíîøåíèé, ïðèâîäÿùèõ ê âû÷èñëåíèþ
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Êëþ÷åâûå ñëîâà: îáðàòíàÿ ìàòðèöà, òðåõäèàãîíàëüíàÿ ìàòðèöà, ýðìèòîâà
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