Explicit inverses of some tridiagonal matrices - Estudo Geral ...

14 C.M. da Fonseca, J. Petronilho / Linear Algebra and its Applications 325 (2001) 7–21
i.e.,
φ i = (−1) i+1 Q n+1−i (0),
where
Q 2i (x) = P i [π 2 (x)] , Q 2i+1 (x) = (x − a 2 )Pi ∗ [π 2 (x)] ,
with π 2 (x), P i (x) and Pi ∗ (x) the same as in (11), (13) and (12), respectively. Observe
that if a 1 = a 2 , then θ i = φ n+1−i .
We have determined completely the θ i ’s and φ i ’s of (7 ) – thus the inverse of the
matrix – in the case of a tridiagonal 2-Toeplitz matrix (1).
Theorem 4.1. Let A be the tridiagonal matrix (1), with a 1 a 2 /= 0 and b 1 b 2 c 1 c 2 > 0.
Put
π 2 (x):=(x − a 1 )(x − a 2 ), β := √ b 2 c 2 /b 1 c 1
and let {Q i (·; α, γ )} be the sequence of polynomials defined by
(√ ) [ ( )
i π2 (x) − b 1 c 1 − b 2 c 2
Q 2i (x; α, γ )= b1 b 2 c 1 c 2 U i
2 √ b 1 b 2 c 1 c
( )] 2
π2 (x) − b 1 c 1 − b 2 c 2
+γU i−1
2 √ b 1 b 2 c 1 c 2
(√ ) ( )
i π2 (x) − b 1 c 1 − b 2 c 2
Q 2i+1 (x; α, γ ) = (x − α) b1 b 2 c 1 c 2 Ui
2 √ ,
b 1 b 2 c 1 c 2
where α and γ are some parameters. Under these conditions,
⎧
⎪⎨ (−1) i+j b
(A −1 p ⌊(j−i)/2⌋
i
) ij =
⎪⎩
(−1) i+j c ⌊(j−i)/2⌋
p j
b ⌊(j−i+1)/2⌋
q i
θ i−1 φ j+1 /θ n if i j,
c ⌊(j−i+1)/2⌋
q j
θ j−1 φ i+1 /θ n if i>j,
(14)
where p l = (3 − (−1) l )/2, q l = (3 + (−1) l )/2, ⌊z⌋ denotes the greater integer
less or equal to the real number z,
θ i = (−1) i Q i (0; a 1 ,β) , (15)
and
⎧
⎨(−1) i Q n+1−i (0; a 1 , 1/β)
φ i =
⎩
(−1) i+1 Q n+1−i (0; a 2 ,β)
if n is odd,
if n is even.
Remark. As we already noticed, under the conditions of Theorem 4.1, if n is odd
and b 1 c 1 = b 2 c 2 , then θ i = φ n+1−i ;andifn is even and a 1 = a 2 , then also θ i =
φ n+1−i .