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I am working on the inverse of the sum of an identity matrix and a Toeplitz matrix, and trying to find the formula for the $(1,1)$ element of the inverse. For example, Assume $c \neq 0$ is a constant, and let $$A_{t}=cI_{t}+\Omega_{t},$$ where, for $t=4$,

$$ \Omega _{t}=\left( \begin{array}{cccc} 1 & \rho & \rho ^{2} & \rho ^{3} \\ \rho & 1 & \rho & \rho ^{2} \\ \rho ^{2} & \rho & 1 & \rho \\ \rho ^{3} & \rho ^{2} & \rho & 1% \end{array}% \right), \quad \text{with} \quad \left\vert \rho \right\vert <1. $$

In general, the $(i,j)$-th element of $\Omega _{t}$ is given by $\rho ^{|i-j|}$. Is there any way to find a general formula for the $(1,1)$-th element of $A_{t}^{-1}$ for different $t,$ say, $t=2,3,4,\dots$?

Rodrigo de Azevedo
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Charles Chou
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2 Answers2

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By Cramer's rule, $U_t=A_t^{-1}(1,1)$ is given by the ratios of two determinants. Let:

$$A_3=\left(\begin{array}{ccc}c+1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right),\qquad B_3=\left(\begin{array}{ccc}1&\rho&\rho^2 \\ \rho&c+1&\rho\\ \rho^2&\rho&c+1\end{array}\right)$$ and define $A_t,B_t$ in a similar fashion. By replacing the first column with the first column minus $\rho$ times the second column, then expanding along the first column, we get: $$ \det A_{t+1} = (c+1-\rho^2) \det A_t + c\rho^2 \det B_t, $$ $$ \det B_{t+1} = (1-\rho^2) \det A_t + c\rho^2 \det B_t, $$ and by eliminating $B_t$, $$ \det A_{t+1} = (c+1+c\rho^2-\rho^2) \det A_t -c^2\rho^2\det A_{t-1} $$ then dividing the whole line by $\det A_t$, $$ \frac{1}{U_{t+1}} = (c+1+c\rho^2-\rho^2)-c^2\rho^2 U_{t} $$ or:

$$ U_{t+1} =\frac{1}{(c+1+c\rho^2-\rho^2)-(c^2\rho^2)U_t}$$

where obviously $U_1=\frac{1}{c+1}$. If we set $U_t=\frac{V_t}{c\rho},$

$$ V_{t+1} =\frac{1}{\frac{c+1+c\rho^2-\rho^2}{c\rho}-V_t},$$ so by setting $\beta=\frac{c+1+c\rho^2-\rho^2}{c\rho}$ and assuming $\beta \geq 2$ we have: $$ \lim_{t\to +\infty}U_t = \frac{-\beta+\sqrt{\beta^2-4}}{2c\rho}$$ by the usual convergence properties of the negative continued fractions.

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Jack D'Aurizio
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  • I very much appreciate for your input on this! Does Ut converge to a stable solution if t is large? Many thanks! – Charles Chou Jun 02 '15 at 20:12
  • @CharlesChou: I have just expanded and fixed my answer to reply to your last point, too. – Jack D'Aurizio Jun 02 '15 at 20:19
  • Thanks a lot for your time and help! – Charles Chou Jun 02 '15 at 20:23
  • @ Jack D'Aurizio: Based on your definition of Bt, I just noticed that your calculation doesn't go through. The second term for your first equation is the determinant of (\rho, \rho^2; \rho, 1+c) instead of B2 you defined. Thanks – Charles Chou Jun 02 '15 at 21:12
  • @CharlesChou: I am quite sure the recursion are right. Start from $A_{t+1}$, for instance. Combine the first two columns as written in order to have just two non-zero elements in the first column, $c+1-\rho^2$ and $-c\rho$. Expand along the first column. You will find $(c+1-\rho^2)$ times the determinant of $A_t$ plus $c\rho$ times the determinant of a matrix like $B_t$, in which the first row is multiplied by $\rho$. The same works for $B_{t+1}$. Combining the two recursions, you get a two-terms recursion for $\det A_{t+1}$. – Jack D'Aurizio Jun 02 '15 at 21:37
  • For short, the determinant of $(\rho,\rho^2;\rho,1+c)$ is just $\rho$ times the determinant of $B_2$, by factoring a $\rho$ out of the first row. – Jack D'Aurizio Jun 02 '15 at 21:39
  • @CharlesChou: yes, if multiply a whole row/column of a matrix by some amount the determinant is (quite trivially) multiplied by the same amount. – Jack D'Aurizio Jun 02 '15 at 21:41
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    Sure, my apology on that. :) – Charles Chou Jun 02 '15 at 21:44
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This is not an answer, but just to demonstrate what we're dealing with here, I've input WolframAlpha's answer for the inverses in the links below.

Click for t=2.

Click for t=3.

Click for t=4.

Xoque55
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  • If it's not an answer (but difficult to put as a comment), make this answer "community wiki". See the check-box under the question when you edit. – Ben Grossmann Jun 02 '15 at 17:31
  • Many thanks for your input, I have also tried to calculate the inverse for finite t. It turns out to be difficult to find a general form for the (1,1)-th element. – Charles Chou Jun 02 '15 at 18:17
  • @Omnomnomnom Thanks for the advice, I'll be more mindful of that in the future! – Xoque55 Jun 03 '15 at 02:10