2
$\begingroup$

Assume the following matrix $$ C_p^{(a,b)}:=\left( \begin{array}{cccccc} a &a &0 &\cdots &\cdots &0 \\ 0 &0 &a &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &0 \\ 0 &\cdots &\cdots &0 &0 &a \\ b &b &\cdots &\cdots &b &b \\ \end{array} \right)_{p \times p}\, . $$ Where $a$ and $b$ are any integer number. With the numerical simulation, i found that the $n$th power of the matrix $C_p^{(a,b)}$, has the following form $$ {(C_p^{(a,b)})}^n:=\left( \begin{array}{cccccc} {g_1^{a,b}}(n) &{g_1^{a,b}}(n) &\cdots &\cdots &{g_1^{a,b}}(n) \\ \\ {g_2^{a,b}}(n) &{g_2^{a,b}}(n) &\cdots &\cdots &{g_2^{a,b}}(n) \\ \\ \vdots &\cdots &\cdots &\cdots &\vdots \\ \vdots &\cdots &\cdots &\cdots &\vdots \\ \\ {g_p^{a,b}}(n) &{g_p^{a,b}}(n) &\cdots &\cdots &{g_p^{a,b}}(n) \\ \end{array} \right)_{p \times p}\, . $$ Where ${g_i^{a,b}}(n)$, $1\leq i \leq p$, are expressions based on the parameters $a$, $b$ and $n$. For example, two consecutive power of the matrix $C_7^{(2,3)}$, are as follows $$ {(C_7^{(2,3)})}^9:= \left( \begin {array}{ccccccc} 8000&8000&8000&8000&8000&8000&8000 \\ 12000&12000&12000&12000&12000&12000&12000 \\ 30000&30000&30000&30000&30000&30000&30000 \\ 75000&75000&75000&75000&75000&75000&75000 \\ 187500&187500&187500&187500&187500&187500&187500 \\ 468750&468750&468750&468750&468750&468750&468750 \\ 1171875&1171875&1171875&1171875&1171875&1171875& 1171875\end {array} \right)\, . $$

$$ {(C_7^{(2,3)})}^{10}:= \left( \begin {array}{ccccccc} 40000&40000&40000&40000&40000&40000& 40000\\ 60000&60000&60000&60000&60000&60000&60000 \\ 150000&150000&150000&150000&150000&150000&150000 \\375000&375000&375000&375000&375000&375000&375000 \\ 937500&937500&937500&937500&937500&937500&937500 \\ 2343750&2343750&2343750&2343750&2343750&2343750& 2343750\\ 5859375&5859375&5859375&5859375&5859375& 5859375&5859375\end {array} \right)\, . $$ Is there a way to find an explicit formula for ${g_i^{a,b}}(n)$, $1\leq i \leq p$ in general. The matrix $C_p^{(a,b)}$ is so interesting. If $a=-b$ then $$ \forall n\geq p \qquad {(C_p^{(a,b)})}^n=O_p\, . $$ Where $O_p$ is a zero matrix of order $p$. In some cases, ${g_i^{a,b}}(n)$, $1\leq i \leq p$, are fixed. For example, if $[a=-(d\pm1) \, \& \, b=d]$ or $[b=-(d\pm1) \, \& \, a=d]$ where $d$ is an integer number, then we have $$ \forall n\geq p-1 \qquad {(C_p^{(a,b)})}^n=\pm F_p\, . $$ Where $F_p$ is a fixed matrix of order $p$. For example, by using $C_5^{(-3,2)}$ and $C_4^{(3,-4)}$, we can see that $$ C_5^{(-3,2)}= \left( \begin {array}{ccccc} -3&-3&0&0&0\\0&0&-3&0 &0\\ 0&0&0&-3&0\\ 0&0&0&0&-3 \\ 2&2&2&2&2\end {array} \right) \Rightarrow \forall n\geq 4 \quad {(C_5^{(-3,2)})}^n= \left( \begin {array}{ccccc} 81&81&81&81&81\\ -54&- 54&-54&-54&-54\\ -18&-18&-18&-18&-18 \\ -6&-6&-6&-6&-6\\ -2&-2&-2&-2&-2 \end {array} \right)\, . $$ $$ C_4^{(3,-4)}= \left( \begin {array}{cccc} 3&3&0&0\\0&0&3&0 \\ 0&0&0&3\\ -4&-4&-4&-4 \end {array} \right) \Rightarrow \forall n\geq 3 \quad {(C_4^{(3,-4)})}^n=\pm \left( \begin {array}{cccc} 27&27&27&27\\ -36&-36&- 36&-36\\ 12&12&12&12\\ -4&-4&-4&-4 \end {array} \right)\, . $$ In some especial cases, i found an expression for ${g_i^{a,b}}(n)$, $1\leq i \leq p$. Assume $C_p^{(a,b)}$, for $a=b=1$, as follows $$ C_p^{(1,1)}:=\left( \begin{array}{cccccc} 1 &1 &0 &\cdots &\cdots &0 \\ 0 &0 &1&\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &0 \\ 0 &\cdots &\cdots &0 &0 &1 \\ 1 &1 &\cdots &\cdots &1 &1 \\ \end{array} \right)_{p \times p}\, . $$ With the induction on $n$, we can prove that for $n\geq p-1$, we have

$$ {(C_p^{(1,1)})}^n:=\left( \begin{array}{cccccc} 2^{n-(p-1)} &2^{n-(p-1)} &\cdots &\cdots &2^{n-(p-1)} \\ \\ 2^{n-(p-1)} &2^{n-(p-1)} &\cdots & \cdots & 2^{n-(p-1)} \\ \\ 2^{n-(p-2)} &2^{n-(p-2)} &\cdots & \cdots & 2^{n-(p-2)} \\ \\ 2^{n-(p-3)} &2^{n-(p-3)} &\cdots & \cdots & 2^{n-(p-3)} \\ \\ \vdots &\cdots &\cdots &\cdots &\vdots \\ \vdots &\cdots &\cdots &\cdots &\vdots \\ \\ 2^{n-1} &2^{n-1} &\cdots &\cdots & 2^{n-1} \\ \end{array} \right)_{p \times p}\, . $$ Is there a method to find a general expression for ${g_i^{a,b}}(n)$, $1\leq i \leq p$? I would greatly appreciate for any suggestions.

$\endgroup$
5
$\begingroup$

The characteristic polynomial of $C_p^{(a,b)}$ is $\lambda^p - (a+b) \lambda^{p-1}$. Therefore, for $m \ge p$ we have $$(C_p^{(a,b)})^m = (a+b)^{m-p} (C_p^{(a,b)})^{p-1}$$ It appears that $B = (C_p^{(a,b)})^{p-1}$ has entries $$ \eqalign{b_{1j} &= a^{p-1}\cr b_{ij} &= a^{p-i} b (a+b)^{i-2}\ \text{for}\ i \ge 2\cr}$$

EDIT: We can exhibit the Jordan form of $C_p^{(a,b)}$ explicitly: $C_p^{(a,b)} = S J S^{-1}$ where

$$ J = \pmatrix{0 & 1 & 0 & \ldots & 0 & 0\cr 0 & 0 & 1 & \ldots & 0 & 0\cr 0 & 0 & 0 & \ldots & 0 & 0\cr \ldots &\ldots &\ldots &\ldots & \ldots & \ldots\cr 0 & 0 & 0 & \ldots &1 & 0\cr 0 & 0 & 0 & \ldots & 0 & 0\cr 0 & 0 & 0 & \ldots & 0 & a+b\cr} $$ $$ S = \pmatrix{\frac{a^{p-2} b}{a+b} & \frac{a^{p-3}(a+b)^2 - a^{p-1}}{(a+b)^2} & \frac{a^{p-4}(a+b)^3 - a^{p-1}}{(a+b)^3} & \ldots & \frac{(a+b)^{p-1} - a^{p-1}}{(a+b)^{p-1}} & \frac{a^{p-1}}{(a+b)^{p-1}}\cr -\frac{a^{p-2} b}{a+b} & -\frac{a^{p-2} b}{(a+b)^2} & -\frac{a^{p-2} b}{(a+b)^3} & \ldots & -\frac{a^{p-2} b}{(a+b)^{p-1}} & \frac{a^{p-2} b}{(a+b)^{p-1}}\cr 0 & -\frac{a^{p-3} b}{a+b} & -\frac{a^{p-3} b}{(a+b)^2} & \ldots & -\frac{a^{p-3} b}{(a+b)^{p-2}} & \frac{a^{p-3} b}{(a+b)^{p-2}}\cr \ldots &\ldots &\ldots &\ldots & \ldots & \ldots\cr 0 & 0 & 0 & \ldots & -\frac{b}{a+b} & \frac{b}{a+b}}$$ $$ S^{-1} = \pmatrix{0 & -\frac{a+b}{a^{p-2} b} & \frac{1}{a^{p-3} b} & 0 & \ldots & 0 & 0\cr 0 & 0 & -\frac{a+b}{a^{p-3} b} & \frac{1}{a^{p-4} b} & \ldots & 0 & 0\cr 0 & 0 & 0 & -\frac{a+b}{a^{p-4} b} & \ldots & 0 & 0\cr \ldots &\ldots &\ldots &\ldots & \ldots & \ldots & \ldots \cr 0 & 0 & 0 & 0 & \ldots & -\frac{a+b}{ab} & \frac{1}{b}\cr 1 & 1 & 1 & 1 & \ldots & 1 & -\frac{a}{b}\cr 1 & 1 & 1 & 1 & \ldots & 1 & 1\cr}$$

$\endgroup$
1
$\begingroup$

Let $\left( \begin{array}{c} {g_1^{(a,b)}}(1) \\ \\ {g_2^{(a,b)}}(1) \\ \\ \vdots \\ \vdots \\ \\ {g_p^{(a,b)}}(1) \\ \end{array} \right)$ be the first column of $C_p^{(a,b)}$. From equation ${(C_p^{(a,b)})}^n=C_p^{(a,b)}\cdot {(C_p^{(a,b)})}^{n-1}$ we conclude:

$$\left\{ \begin{array}{c} g_1^{(a,b)}(n)=a\cdot (g_1^{(a,b)}(n-1)+g_2^{(a,b)}(n-1)) \\g_i^{(a,b)}(n)=a\cdot g_{i+1}^{(a,b)}(n-1) , 2\le i\le p-1\\g_p^{(a,b)}(n)=b \cdot \sum_1^p g_i^{(a,b)}(n-1) \\ \end{array} \right.$$

$\endgroup$

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.