Determinant

The determinant of a tridiagonal matrix A of order n can be computed from a three-term recurrence relation.^[4] Write f₁ = |a₁| = a₁ (i.e., f₁ is the determinant of the 1 by 1 matrix consisting only of a₁), and let

${\displaystyle f_{n}={\begin{vmatrix}a_{1}&b_{1}\\c_{1}&a_{2}&b_{2}\\&c_{2}&\ddots &\ddots \\&&\ddots &\ddots &b_{n-1}\\&&&c_{n-1}&a_{n}\end{vmatrix}}.}$

The sequence (f_i) is called the continuant and satisfies the recurrence relation

${\displaystyle f_{n}=a_{n}f_{n-1}-c_{n-1}b_{n-1}f_{n-2}}$

with initial values f₀ = 1 and f₋₁ = 0. The cost of computing the determinant of a tridiagonal matrix using this formula is linear in n, while the cost is cubic for a general matrix.

Inversion

The inverse of a non-singular tridiagonal matrix T

${\displaystyle T={\begin{pmatrix}a_{1}&b_{1}\\c_{1}&a_{2}&b_{2}\\&c_{2}&\ddots &\ddots \\&&\ddots &\ddots &b_{n-1}\\&&&c_{n-1}&a_{n}\end{pmatrix}}}$

is given by

${\displaystyle (T^{-1})_{ij}={\begin{cases}(-1)^{i+j}b_{i}\cdots b_{j-1}\theta _{i-1}\phi _{j+1}/\theta _{n}&{\text{ if }}i<j\\\theta _{i-1}\phi _{j+1}/\theta _{n}&{\text{ if }}i=j\\(-1)^{i+j}c_{j}\cdots c_{i-1}\theta _{j-1}\phi _{i+1}/\theta _{n}&{\text{ if }}i>j\\\end{cases}}}$

where the θ_i satisfy the recurrence relation

${\displaystyle \theta _{i}=a_{i}\theta _{i-1}-b_{i-1}c_{i-1}\theta _{i-2}\qquad i=2,3,\ldots ,n}$

with initial conditions θ₀ = 1, θ₁ = a₁ and the ϕ_i satisfy

${\displaystyle \phi _{i}=a_{i}\phi _{i+1}-b_{i}c_{i}\phi _{i+2}\qquad i=n-1,\ldots ,1}$

with initial conditions ϕ_n+1 = 1 and ϕ_n = a_n.^[5][6]

Closed form solutions can be computed for special cases such as symmetric matrices with all diagonal and off-diagonal elements equal^[7] or Toeplitz matrices^[8] and for the general case as well.^[9][10]

In general, the inverse of a tridiagonal matrix is a semiseparable matrix and vice versa.^[11]

Solution of linear system

A system of equations Ax = b for ${\displaystyle b\in \mathbb {R} ^{n}}$ can be solved by an efficient form of Gaussian elimination when A is tridiagonal called tridiagonal matrix algorithm, requiring O(n) operations.^[12]

Eigenvalues

When a tridiagonal matrix is also Toeplitz, there is a simple closed-form solution for its eigenvalues, namely:^[13][14]

${\displaystyle a+2{\sqrt {bc}}\cos \left({\frac {k\pi }{n+1}}\right),\qquad k=1,\ldots ,n.}$

A real symmetric tridiagonal matrix has real eigenvalues, and all the eigenvalues are distinct (simple) if all off-diagonal elements are nonzero.^[15] Numerous methods exist for the numerical computation of the eigenvalues of a real symmetric tridiagonal matrix to arbitrary finite precision, typically requiring ${\displaystyle O(n^{2})}$ operations for a matrix of size ${\displaystyle n\times n}$, although fast algorithms exist which (without parallel computation) require only ${\displaystyle O(n\log n)}$.^[16]

As a side note, an unreduced symmetric tridiagonal matrix is a matrix containing non-zero off-diagonal elements of the tridiagonal, where the eigenvalues are distinct while the eigenvectors are unique up to a scale factor and are mutually orthogonal.^[17]

Similarity to symmetric tridiagonal matrix

For unsymmetric or nonsymmetric tridiagonal matrices one can compute the eigendecomposition using a similarity transformation. Given a real tridiagonal, nonsymmetric matrix

${\displaystyle T={\begin{pmatrix}a_{1}&b_{1}\\c_{1}&a_{2}&b_{2}\\&c_{2}&\ddots &\ddots \\&&\ddots &\ddots &b_{n-1}\\&&&c_{n-1}&a_{n}\end{pmatrix}}}$

where ${\displaystyle b_{i}\neq c_{i}}$. Assume that each product of off-diagonal entries is strictly positive ${\displaystyle b_{i}c_{i}>0}$ and define a transformation matrix ${\displaystyle D}$ by

${\displaystyle D:=\operatorname {diag} (\delta _{1},\dots ,\delta _{n})\quad {\text{for}}\quad \delta _{i}:={\begin{cases}1&,\,i=1\\{\sqrt {\frac {c_{i-1}\dots c_{1}}{b_{i-1}\dots b_{1}}}}&,\,i=2,\dots ,n\,.\end{cases}}}$

The similarity transformation ${\displaystyle D^{-1}TD}$ yields a symmetric tridiagonal matrix ${\displaystyle J}$ by:^[18]

${\displaystyle J:=D^{-1}TD={\begin{pmatrix}a_{1}&\operatorname {sgn} b_{1}\,{\sqrt {b_{1}c_{1}}}\\\operatorname {sgn} b_{1}\,{\sqrt {b_{1}c_{1}}}&a_{2}&\operatorname {sgn} b_{2}\,{\sqrt {b_{2}c_{2}}}\\&\operatorname {sgn} b_{2}\,{\sqrt {b_{2}c_{2}}}&\ddots &\ddots \\&&\ddots &\ddots &\operatorname {sgn} b_{n-1}\,{\sqrt {b_{n-1}c_{n-1}}}\\&&&\operatorname {sgn} b_{n-1}\,{\sqrt {b_{n-1}c_{n-1}}}&a_{n}\end{pmatrix}}\,.}$

Note that ${\displaystyle T}$ and ${\displaystyle J}$ have the same eigenvalues.