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The inverse of the symmetric tridiagonal matrix (Toeplitz) $$ t_{ij}=\begin{align} \begin{cases} -2 &\quad \text{if} \,\, i=j \\ 1 &\quad\text{if} \,\, \vert i-j\vert = 1 \end{cases} \end{align} $$

does not have any non zero entries according to online inverse calculators (I tried up to $5 \times 5-$matrices). Why is it so?

Rodrigo de Azevedo
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Tesla
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1 Answers1

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The expression for the $i,j$ element of the inverse of this $n\times n$ matrix has a nice form \begin{align} w_{ij}&=\frac{i\,j}{n+1}-\min(i,\,j) \tag{1}\label{1} , \end{align}

so $w_{ij}$ could be zero only if \begin{align} n&=\frac{i\,j}{\min(i,\,j)}-1 =\max(i,j)-1=n-1 , \end{align}

and the answer follows.

g.kov
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  • How did you obtain the expression for the $i,j$ element of the inverse? just looking at it and recognizing the pattern? – Tesla Nov 10 '19 at 08:49
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    @Tesla: I get it quite a while ago, via the inverse of the companion circulant matrix. And also, as far as I remember, this inverse was published is some old paper (not sure which one) with many similar results. But the pattern is indeed simple and you are right, it can be determined. – g.kov Nov 10 '19 at 09:16