We know the formula of determinant of every tridiagonal matrix in continued fraction form via LU factorization of the matrix. Does anyone know about the explicit formula of determinant of tridiagonal matrix in another difference form?
Dear Wiwat,
In one old Russian textbook, I've found the following result: the determinant of the tridiagonal matrix of the form
a_1 b_1 0 0 0 ...
-c_2 a_2 b_2 0 0 ...
0 - c_3 a_3 b_3 0 ...
...
of the order n equals the sum of its leading term a_1a_2...a_n and all the terms obtained from this product by replacement one or several pairs of neighbour product a_ja_{j+1} by b_j c_{j+1}. So, for the order 5 it has the form:
a_1a_2a_3a_4a_5+b_1c_2a_3a_4a_5+a_1b_2c_3a_4a_5+a_1a_2b_3c_4a_5
+a_1a_2a_3b_4c_5+b_1c_2b_3c_4a_5+b_1c_2a_3b_4c_5+a_1b_2c_3b_4c_5
Dear Wiwat, if I "good" understand your question, this is an answer on Wikipedia:
The determinant of a tridiagonal matrix A of order n can be computed from a three-term recurrence relation. Write f1 = |a1| = a1 and fn=det A=an fn-1 -cn-1 bn-1 fn-2, where A is the tridiagonal matrix with ( a1,. . . ,an ) the main diagonal, ( b1 ,. . .. , bn-1 ) the upper one and ( c1 , . . . ,cn-1 ) the lower one
https://en.wikipedia.org/wiki/Tridiagonal_matrix
Dear Hanifa,
Thanks for your answer.
I would like to have the explicit formula of the determinant, that is, for given a tridiagonal matrix, we can compute the determinate by substitute ai, bj, and ck in the formula only.
Dear Wiwat,
In one old Russian textbook, I've found the following result: the determinant of the tridiagonal matrix of the form
a_1 b_1 0 0 0 ...
-c_2 a_2 b_2 0 0 ...
0 - c_3 a_3 b_3 0 ...
...
of the order n equals the sum of its leading term a_1a_2...a_n and all the terms obtained from this product by replacement one or several pairs of neighbour product a_ja_{j+1} by b_j c_{j+1}. So, for the order 5 it has the form:
a_1a_2a_3a_4a_5+b_1c_2a_3a_4a_5+a_1b_2c_3a_4a_5+a_1a_2b_3c_4a_5
+a_1a_2a_3b_4c_5+b_1c_2b_3c_4a_5+b_1c_2a_3b_4c_5+a_1b_2c_3b_4c_5
Dear Alexei,
Thank you for your valuable answer. May I would like to have the name of the Russian textbook and the authors.
Regards,
Wiwat Wanicharpichat
Surprise for me: it looks like be translated into English
Higher Algebra: Linear Algebra, Polynomials, General Algebra (International Series of Monographs in Pure and Applied Mathematics
Mishina, A.P., and I.V. Proskuryakov
Pergamon Press. 1965.
http://www.biblio.com/book/higher-algebra-linear-algebra-polynomials-general/d/707682709
In my (Russian) edition the result you need is in p.28 (section "Jacobi's determinants. Continuant and its relationship to Fibonacci's numbers")
AU
Dear Wiwat,
What gave Alexei Uteshev is exactly what is in my answer. by computing the recurrence relation, we obtain the determinant in terms of ai, bj , and ck only. i verified it on 5 by 5 matrix. Just i have to add f0 =1 for the recurrence relation above, which also called continuant.
https://en.wikipedia.org/wiki/Continuant_(mathematics)
Dear Alexei and Hanifa,
Thank you so much for your kindly comments and advice.
Regards,
Wiwat Wanicharpichat
The explicit expression for the determinant exists though it is complicated. One possible formula in stated in Proposition 4 in the paper: http://arxiv.org/abs/1201.1743. The formula is written in terms of a function called \mathfrak{F} which is defined in Definition 1 therein.
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