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What is the trick for evaluating the determinant of this matrix?

$$\begin{bmatrix} 2 & -1 \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ && -1 & 2 & -1 \\ &&& -1 & 2 & -1 \\ &&&& -1 & 2 \end{bmatrix}$$

Rodrigo de Azevedo
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sailor
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    What do you mean to evaluate it? – Amzoti Feb 15 '15 at 01:51
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    Determinate? What do you mean by evaluate? – Chinny84 Feb 15 '15 at 01:52
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    there is a three term recursive formula. you can look up the determinant of a tridiagonal matrix. – abel Feb 15 '15 at 01:57
  • Apologies. Question updated. – sailor Feb 15 '15 at 01:57
  • This is not just any tridiagonal matrix, it's a very special one. I think we can find a basis of eigenvectors analytically. The $j$th component of eigenvector $m$ is something like $\sin(m \pi j/N) $. Knowing the eigenvalues, we get the determinant. – littleO Feb 15 '15 at 02:44
  • See also: http://math.stackexchange.com/questions/995779/proving-the-determinant-of-a-tridiagonal-matrix – Martin Sleziak Feb 15 '15 at 09:08
  • Elaborating on my previous comment: it's useful to know that the $m$th eigenvector of this $N \times N$ matrix is the vector whose $j$th component is $\sin(m \pi j/(N+1))$. This can be checked by a direct calculation, which reveals that the corresponding eigenvalue is $2 - 2 \cos(m \pi/(N+1))$. One could then show that the product of the eigenvalues is $N+1$. This isn't as simple as the answer given, but sometimes it's useful to know the eigenvalues and eigenvectors. – littleO Feb 15 '15 at 09:55
  • For a more general version, see this answer. – Martin Sleziak Feb 15 '15 at 10:44

1 Answers1

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let $f_n$ be the the determinant of the $n \times n$ tridiagonal matrix with diagonal elements all equal to $2$ and the sub and super diagonal has $-1$ on them. then by expanding the determinant by the first row, you get the recursive relation $$f_n = 2f_{n-1} - f_{n-2},\quad f_1 = 2,\ f_2 = 3.$$ you can verify that $$f_n = n+1$$ is in fact the solution.

Michael Hardy
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abel
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