0
$\begingroup$

Is there any way of computing analytically the exponential (or finding the eigenvalues, or determining if it's positive definite) of a $n\times n$ symmetric 2-tridiagonal matrix: $$ T_n^{(2)}= \begin{bmatrix} a_1 & 0 & b_1 & & & \\ 0 & a_2 & 0 & b_2 & & \\ b_1 & 0 & a_3 & 0 & b_3 & \\ & b_2 & 0 & a_4 & 0 & \ddots &\\ & & b_3 & 0 & a_5 & \ddots & b_{n-2}\\ & & & \ddots & \ddots & \ddots & 0 \\ & & & & b_{n-2} & 0 & a_n \end{bmatrix} $$

Actually I am interested only on the first and the second row of the $\exp[T_n^{(2)}]$, if that helps.

$\endgroup$
2
$\begingroup$

A partial result: If you permute the indices of this matrix as $$\{1,2,3,4,\cdots\}\mapsto\{1,3,5,\cdots,2,4,6,\cdots\}$$ then the matrix becomes block diagonal

$$T_n^{(2)}=\left( \begin{array}{cccc|cccc} a_1 & b_1 & 0&\cdots & \\ b_1 & a_3 & b_3 & \cdots & & \ddots \\ 0 & b_3 & a_5 & \cdots & & & \ddots\\ \vdots & \vdots &\vdots & \ddots \\ \hline & &&& a_2 & b_2 & 0&\cdots \\ &\ddots&&& b_2 & a_4 & b_4 & \cdots & & \\ &&\ddots&& 0 & b_4 & a_6 & \cdots & & \\ &&&& \vdots & \vdots &\vdots & \ddots \end{array}\right)$$ i.e. $T_n^{(2)}$ is the direct sum of two submatrices $\mathcal{O}_{n},$ $\mathcal{E}_n$. This converts the question from being about a single 2-tridiagonal matrix to a pair of standard 1-tridiagonal matrices. For instance, since $T_n^{(2)}=\mathcal{O}_{n}\oplus\mathcal{E}_n$ we have $\exp T_n^{(2)}=(\exp \mathcal{O}_{n})\oplus (\exp \mathcal{E}_{n}).$ There's an extensive amount of literature on such tridagonal matrices, so I won't attempt to cover it here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.