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Suppose we consider the tridiagonal matrix of size $n \times n$ $$A = \begin{pmatrix}1&1&&&\\ 1&1&1&\ddots\\ &\ddots&\ddots&\ddots\\ &&1&1&1\\ &&&1&1\end{pmatrix}$$

I am thinking if there is a general form of the matrix $A^{-1}$?

I have observed numerically that when $n$ is of the form $3k$ or $3k-2$, then the determinant of $A$ is $-1$ and $1$ respectively. If $n$ is of the form $3k-1$, then $det(A) = 0$, where $k \in \Bbb{N}$ and $k \geq 2$.

BAYMAX
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    See a slightly more general case here – Jean Marie Sep 01 '22 at 10:32
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    In the last line of the answer, should it be $A^{-1} = V D^{-1} V^{T}$ ? – BAYMAX Sep 02 '22 at 02:44
  • You are perfectly right. – Jean Marie Sep 02 '22 at 04:59
  • Oh ok, i have edited it. But is it immidiate why $\det(A) = 0$ when $n = 3k-1$ ? i tried to plug it in but could not see that through eyeballing.? – BAYMAX Sep 02 '22 at 08:21
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    This is a Toeplitz matrix and as any such matrix it will be diagonalizable with complex exponential eigenfunctions. The inverse will be readily found from the diagonalization iff diagonal elements are non-zero. Jean Marie seems to have already derived it. – mathreadler Sep 02 '22 at 12:18

1 Answers1

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Here is an answer inspired (in particular) by the answer by Calvin Lin to this question.

Let us give the indexed name $A_n$ in the case $A$ is $n \times n$. Let $D_n$ denote its determinant. We have in particular

$$D_1=1, \ \ D_2=0 \tag{1}$$

Using two successive Laplace expansions of $D_n$ with respect to columns, one obtains the recurrence relationship:

$$D_n=D_{n-1}-D_{n-2}\tag{2}$$

(2) is a linear recurrence relationship with constant coefficients and initial conditions (1) ; it has a unique solution which cannot be "eyeballed".

The characteristic equation of (2) is $x^2-x+1=0$.

Its two roots are

$$\frac{1 \pm \sqrt{3}i}{2}= w \ or \ \bar{w} \ \text{with} \ w:=e^{i \pi/3}.$$

Therefore, the solution of (2) has the form:

$$D_n=aw^n+b\bar{w}^n\tag{3}$$

for certain constants $a$ and $b$.

Taking into account the "initial conditions" (1), we must have:

$$\begin{cases}aw+b \bar{w}&=&1\\aw^2+b \bar{w}^2&=&0\end{cases},$$

a linear system whose solution is:

$$a=\frac{-iw}{\sqrt{3}}, \ \ b=\frac{i \bar{w}}{\sqrt{3}}\tag{4}$$

Plugging (4) into (3), we get:

$$D_n=\frac{i}{\sqrt{3}}\left(-w w^n+\bar{w}\bar{w}^n\right)=\frac{i}{\sqrt{3}}\left(- w^{n+1}+\bar{w}^{n+1}\right)$$

$$D_n=\frac{2}{\sqrt{3}}\sin \left((n+1)\frac{\pi}{3}\right)$$

Please note that the final expression is real.

Jean Marie
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  • Awesome! The characteristic equation of (2) ? the characteristic equation of recurrence relation? – BAYMAX Sep 03 '22 at 07:40