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Summary
- Matrix
A matrix over a field $F$ is a rectangular array of elements from $F$. The symbol $M_{m\times n}(F)$ denotes the collection of all $m\times n$ matrices over $F$.
Matrices will usually be denoted by capital letters and the equation $A=[a_{ij}]$ means that the element in the $i$-th row and $j$-th column of the matrix $A$ equals $a_{ij}$. - Equality of matrices
Matrices $A$, $B$ are said to be equal if $A$ and $B$ have the same size and corresponding elements are equal; i.e. $A$ and $B$ $\in M_{m\times n}(F)$ and $A=[a_{ij}]$, $B=[b_{ij}]$, with $a_{ij}=b_{ij}$ for $1\leq i\leq m$, $1\leq j\leq n$. - Addition (subtraction) of matrices
Let $A=[a_{ij}]$ and $B=[b_{ij}]$ be of the same size. Then $A\pm B$ is the matrix obtained by adding (subtracting) corresponding elements of $A$ and $B$; that is $$A\pm B=[a_{ij}]\pm[b_{ij}]=[a_{ij}\pm b_{ij}].$$ - Scalar multiple of a matrix
Let $A=[a_{ij}]$ and $t\in F$ ($t$ is a scalar). Then $tA$ is the matrix obtained by multiplying all elements of $A$ by $t$; that is $$tA=t[a_{ij}]=[ta_{ij}].$$ - Additive inverse of a matrix
Let $A=[a_{ij}]$. Then $-A$ is the matrix obtained by replacing the elements of $A$ by their additive inverses; that is $$-A=-[a_{ij}]=[-a_{ij}].$$ - The zero matrix
For each $m$, $n$ the matrix in $M_{m\times n}(F)$, all of whose elements are zero, is called the zero matrix (of size $m\times n$) and is denoted by the symbol 0. - Matrix product
- Definition
Let $A=[a_{ij}]$ be a matrix of size $m\times n$ and $B=[b_{jk}]$ be a matrix of size $n\times p$; that is the number of columns of $A$ equals the number of rows of $B$. Then $AB$ is the $m\times p$ matrix $C=[c_{ik}]$ whose $(i, k)$-th element is defined by the formula $$c_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk}=a_{i1}b_{1k} +\cdots+a_{in}b_{nk}.$$ - Basic laws
$$(AB)C=A(BC)$$ $$t(AB)=(tA)B=A(tB)$$ $$(A+B)C=AC+BC$$ $$D(A+B)=DA+DB$$ where the size of $A$, $B$, $C$, $D$ should satisfy the definition of matrix product, and $t$ is a scalar in the field $F$.
- Definition
- The identity matrix
The $n\times n$ matrix $I_n=[\delta_{ij}]$, defined by $\delta_{ij}=1$ if $i=j$, $\delta_{ij}=0$ if $i\neq j$, is called the $n\times n$ identity matrix of order $n$. For example, $I_2=\begin{bmatrix}1&0\\0&1 \end{bmatrix}$.
If $A$ is $m\times n$, then $I_mA=A=AI_n$. - $k$-th power of a matrix
If $A$ is an $n\times n$ matrix, $A^k$ is defined recursively as follows: $A_0=I_n$, and $A^{k+1}=A^kA$ for $k\geq0$. For example, $A^1=A^0A=I_nA=A$ and $A^2=A^1A=AA$. - Mathematical Induction
If $P_n$ denotes a mathematical statement for each $n\geq1$, satisfying
(1) $P_1$ is true,
(2) the truth of $P_n$ implies that of $P_{n+1}$ for each $n\geq1$,
then $P_n$ is true for all $n\geq1$.
Problems 2.4
1. Let $A$, $B$, $C$, $D$ be matrices defined by $$A=\begin{bmatrix}3 & 0\\ -1 & 2\\ 1 & 1\end{bmatrix}, \, B=\begin{bmatrix}1& 5& 2\\-1& 1& 0\\ -4& 1& 3 \end{bmatrix},\, C=\begin{bmatrix}-3& -1\\ 2& 1\\ 4& 3 \end{bmatrix}, \, D=\begin{bmatrix}4& -1\\ 2& 0 \end{bmatrix}.$$ Which of the following matrices are defined? Compute those matrices which are defined. $$A+B,\ A+C,\ AB,\ BA,\ CD,\ DC,\ D^2.$$
Solution:
$$A+C=\begin{bmatrix}3-3& 0-1\\-1+2& 2+1\\ 1+4& 1+3 \end{bmatrix}=\begin{bmatrix}0& -1\\1& 3\\ 5& 4 \end{bmatrix}$$
$$BA=\begin{bmatrix}1\times3-5\times1+2\times1& 1\times0+5\times2+2\times1\\-1\times3-1\times1+0\times1& -1\times0+1\times2+0\times1\\ -4\times3-1\times1+3\times1&-4\times0+1\times2+3\times1 \end{bmatrix}=\begin{bmatrix}0 &12\\-4&2\\-10&5 \end{bmatrix}$$ $$CD=\begin{bmatrix}-14 &3\\10& -2\\22& -4 \end{bmatrix}$$ $$D^2=\begin{bmatrix}14& -4\\8& -2 \end{bmatrix}$$
2. Let $A=\begin{bmatrix}-1& 0& 1\\ 0& 1& 1 \end{bmatrix}.$ Show that if $B$ is a $3\times2$ such that $AB=I_2$, then $$B=\begin{bmatrix}a& b\\ -a-1& 1-b\\a+1& b \end{bmatrix}$$ for suitable numbers $a$ and $b$. Use the associative law to show that $(BA)^2B=B$.
Solution:
Let $B=\begin{bmatrix}a& b\\c& d\\e& f \end{bmatrix}$, and we have $$AB=\begin{bmatrix}-1& 0& 1\\ 0& 1& 1 \end{bmatrix}\cdot\begin{bmatrix}a& b\\c& d\\e& f \end{bmatrix}=\begin{bmatrix}-a+e &-b+f\\c+e&d+f \end{bmatrix}=I_2=\begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix}$$ Thus $$\begin{cases}-a+e=1\\-b+f=0\\c+e=0\\d+f=1 \end{cases}\Rightarrow\begin{cases}c=-e=-(a+1)=-a-1\\ d=1-f=1-b\\e=a+1\\f=b \end{cases}$$ That is $$B=\begin{bmatrix}a& b\\ -a-1& 1-b\\a+1& b \end{bmatrix}$$ Next $$(BA)^2B=BA\cdot BA\cdot B=B\cdot(AB)\cdot(AB)=B\cdot I_2\cdot I_2=B\cdot I_2=B$$
3. If $A=\begin{bmatrix}a& b\\c& d \end{bmatrix}$, prove that $A^2-(a+d)A+(ad-bc)I_2=0$
Solution:
Since
$$A^2=\begin{bmatrix}a& b\\c& d \end{bmatrix}\cdot\begin{bmatrix}a& b\\c& d \end{bmatrix}=\begin{bmatrix}a^2+bc& ab+bd\\ac+cd& bc+d^2 \end{bmatrix}$$ $$(a+d)\cdot A=\begin{bmatrix}a^2+ad& ab+bd\\ac+cd& ad+d^2 \end{bmatrix}$$ $$(ad-bc)I_2=(ad-bc)\cdot\begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix}=\begin{bmatrix}ad-bc& 0\\ 0& ad-bc \end{bmatrix}$$ Thus $$A^2-(a+d)A+(ad-bc)I_2$$ $$ =\begin{bmatrix}a^2+bc& ab+bd\\ac+cd& bc+d^2 \end{bmatrix}- \begin{bmatrix}a^2+ad& ab+bd\\ac+cd& ad+d^2 \end{bmatrix} +\begin{bmatrix}ad-bc& 0\\ 0& ad-bc \end{bmatrix} $$ $$=\begin{bmatrix}0& 0\\0& 0 \end{bmatrix}=0$$
4. If $A=\begin{bmatrix}4& -3\\ 1& 0 \end{bmatrix}$, use the fact $A^2=4A-3I_2$ and the mathematical induction, to prove that $$A^n={3^n-1\over2}A+{3-3^n\over2}I_2,\ \text{if}\ n\geq1.$$
Solution:
Note that according to the previous problem we have $a=4,\ b=-3,\ c=1,\ d=0$ in this problem, and hence we have $A^2-4A+3I_2=0$ which provided by the condition. By using mathematical induction, we have $$P_1: A^1={3-1\over2}A+{3-3\over2}I_2=A$$ is true; And suppose $$P_n: A^n={3^n-1\over2}A+{3-3^n\over2}I_2$$ is true, then we have $$P_{n+1}: A^{n+1}=A\cdot A^n={3^n-1\over2}A^2+{3-3^n\over2}AI_2$$ $$={3^n-1\over2}\cdot(4A-3I_2)+{3-3^n\over2}A$$ $$={4\cdot3^n-4+3-3^n\over2}A- {3^{n+1}-3\over2}I_2$$ $$={3^{n+1}-1\over2}A+{3-3^{n+1}\over2}I_2$$ which is true. Thus the original assumption $P_n$ is true.
5. A sequence of numbers $x_1,\ x_2,\ \cdots,\ x_n,\ \cdots$ satisfies the recurrence relation $x_{n+1}=ax_n+bx_{n-1}$ for $n\geq1$, where $a$ and $b$ are constant. Prove that $$\begin{bmatrix}x_{n+1}\\ x_n \end{bmatrix}=A\begin{bmatrix}x_n\\ x_{n-1} \end{bmatrix},$$ where $A=\begin{bmatrix}a& b\\1& 0 \end{bmatrix}$ and hence express $\begin{bmatrix}x_{n+1}\\x_n \end{bmatrix}$ in terms of $\begin{bmatrix}x_1\\x_0 \end{bmatrix}$.
If $a=4$ and $b=-3$, use the previous question to find a formula for $x_n$ in terms of $x_1$ and $x_0$.
Solution:
It is easy to see that $$A\begin{bmatrix}x_n\\ x_{n-1} \end{bmatrix}=\begin{bmatrix}a& b\\1& 0 \end{bmatrix}\cdot\begin{bmatrix}x_n\\ x_{n-1} \end{bmatrix}=\begin{bmatrix}ax_n+bx_{n-1}\\x_n \end{bmatrix}=\begin{bmatrix}x_{n+1}\\ x_n \end{bmatrix}$$ By recurrence relation we have $$\begin{bmatrix}x_{n+1}\\ x_n \end{bmatrix}=A\begin{bmatrix}x_n\\ x_{n-1} \end{bmatrix}=A\cdot A\begin{bmatrix}x_{n-1}\\ x_{n-2} \end{bmatrix}=A^2\begin{bmatrix}x_{n-1}\\ x_{n-2} \end{bmatrix}=\cdots=A^n\begin{bmatrix}x_{1}\\ x_{0} \end{bmatrix}$$ When $a=4$ and $b=-3$, $A=\begin{bmatrix}4& -3\\1& 0 \end{bmatrix}$, and according to the previous results we have $$\begin{bmatrix}x_{n+1}\\ x_n \end{bmatrix}=A^n\begin{bmatrix}x_{1}\\ x_{0} \end{bmatrix} =\left({3^n-1\over2}A+{3-3^n\over2}I_2\right)\cdot\begin{bmatrix}x_{1}\\ x_{0} \end{bmatrix}$$ $$=\left({3^n-1\over2}\cdot\begin{bmatrix}4& -3\\1& 0 \end{bmatrix}+{3-3^n\over2}\cdot\begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix}\right)\cdot\begin{bmatrix}x_{1}\\ x_{0} \end{bmatrix}$$ $$=\begin{bmatrix}2\cdot(3^n-1)+{3-3^n\over2}&-{3\over2}\cdot (3^n-1)\\ {3^n-1\over2}&{3-3^n\over2}\end{bmatrix}\cdot\begin{bmatrix}x_{1}\\ x_{0} \end{bmatrix}$$ Thus we only need to compute the entry of $(2, 1)$ which is $x_n$: $$x_n={3^n-1\over2}x_1+{3-3^n\over2}x_0.$$
6. Let $A=\begin{bmatrix}2a&-a^2\\1&0 \end{bmatrix}$.
(a) Prove that $$A^n=\begin{bmatrix}(n+1)a^n&-na^{n+1}\\na^{n-1} & (1-n)a^n \end{bmatrix},\ \text{if}\ n\geq1.$$
(b) A sequence $x_0,\ x_1,\ \cdots,\ x_n,\ \cdots$ satisfies $x_{n+1}=2ax_n-a^2x_{n-1}$ for $n\geq1$. Use part (a) and the previous question to prove that $x_n=na^{n-1}x_1+(1-n)a^nx_0$ for $n\geq1$.
Solution:
(a) Using mathematical induction: $$P_1: A^1=\begin{bmatrix}(1+1)a&-a^2\\1&0 \end{bmatrix}=A$$ is true; Suppose that $$P_n: A^n=\begin{bmatrix}(n+1)a^n&-na^{n+1}\\na^{n-1} & (1-n)a^n \end{bmatrix}$$ is true, and we have $$P_{n+1}: A^{n+1}=A\cdot A^n=\begin{bmatrix}2a&-a^2\\1&0 \end{bmatrix}\cdot\begin{bmatrix}(n+1)a^n&-na^{n+1}\\na^{n-1} & (1-n)a^n \end{bmatrix}$$ $$=\begin{bmatrix}(2n+2)a^{n+1}-na^{n+1}&-2na^{n+2}-(1-n)a^{n+2}\\(n+1)a^{n} & -na^{n+1} \end{bmatrix}$$ $$=\begin{bmatrix}(n+2)a^{n+1}&-(n+1)a^{n+2}\\(n+1)a^{n} & -na^{n+1} \end{bmatrix}$$ follows the form of $A^{n}$, and hence the original assumption $P_n$ is true.
(b) Since $x_{n+1}=2ax_n-a^2x_{n-1}$, we have $$\begin{bmatrix}x_{n+1}\\x_n \end{bmatrix}=\begin{bmatrix}2a&-a^2\\1&0\end{bmatrix}\cdot\begin{bmatrix}x_n\\x_{n-1} \end{bmatrix} = A^{n}\cdot\begin{bmatrix}x_1\\x_0 \end{bmatrix}$$ $$=\begin{bmatrix}(n+1)a^n&-na^{n+1}\\na^{n-1} & (1-n)a^n \end{bmatrix}\cdot\begin{bmatrix}x_1\\x_0 \end{bmatrix}$$ Thus the entry of $(2, 1)$ is $x_n$: $$x_n=na^{n-1}x_1+(1-n)a^nx_0.$$
7. Let $A=\begin{bmatrix}a&b\\c&d \end{bmatrix}$ and suppose that $\lambda_1$ and $\lambda_2$ are the roots of the quadratic polynomial $x^2-(a+d)x+(ad-bc).$ ($\lambda_1$ and $\lambda_2$ may be equal.)
Let $k_n$ be defined by $k_0=0$, $k_1=1$ and for $n\geq2$ $$k_n=\sum_{i=1}^{n}\lambda_1^{n-i}\lambda_2^{i-1}.$$ Prove that $$k_{n+1}=(\lambda_1+\lambda_2)k_n-\lambda_1\lambda_2k_{n-1},$$ if $n\geq1$. Also prove that $$k_n=\begin{cases}\displaystyle{(\lambda_1^{n}-\lambda_2^{n}) \over (\lambda_1-\lambda_2)}& \mbox{if} \lambda_1\neq\lambda_2,\\ n\lambda_1^{n-1} & \mbox{if} \lambda_1=\lambda_2.\end{cases}$$ Use mathematical induction to prove that if $n\geq1$, $$A^n=k_nA-\lambda_1\lambda_2k_{n-1}I_2.$$
Solution:
First, we prove $k_{n+1}=(\lambda_1+\lambda_2)k_n-\lambda_1\lambda_2k_{n-1}$.
$$(\lambda_1+\lambda_2)k_n-\lambda_1\lambda_2k_{n-1} =(\lambda_1+\lambda_2(\lambda_1^{n-1}+\lambda_1^{n-2} \lambda_2 +\cdots+\lambda_1\lambda_2^{n-2}+\lambda_2^{n-1})$$ $$-\lambda_1\lambda_2(\lambda_1^{n-2}+\lambda_1^{n-3}\lambda_2 +\cdots+\lambda_1\lambda_2^{n-3}+\lambda_2^{n-2})$$ $$=(\lambda_1^{n}+\lambda_1^{n-1}\lambda_2+\cdots+\lambda_1^2 \lambda_2^{n-2}+\lambda_1\lambda_2^{n-1}) + (\lambda_1^{n-1}\lambda_2+\lambda_1^{n-2}\lambda_2^2+\cdots +\lambda_1\lambda_2^{n-1}+\lambda_2^{n})$$ $$-(\lambda_1^{n-1}\lambda_2+\lambda_1^{n-2}\lambda_2^2+\cdots +\lambda_1\lambda_2^{n-1}+\lambda_2^{n})$$ $$=\lambda_1^{n}+\lambda_1^{n-1}\lambda_2+\cdots+\lambda_1^2 \lambda_2^{n-2}+\lambda_1\lambda_2^{n-1} =\sum_{i=1}^{n}\lambda_1^{n+1-i}\lambda_2^{i-1}=k_{n+1}$$
Next, we prove $$k_n=\begin{cases}\displaystyle{(\lambda_1^{n}-\lambda_2^{n}) \over (\lambda_1-\lambda_2)}& \mbox{if}\ \lambda_1\neq\lambda_2,\\ n\lambda_1^{n-1} & \mbox{if}\ \lambda_1=\lambda_2.\end{cases}$$
- If $\lambda_1=\lambda_2$
$$k_n=\sum_{i=1}^{n}\lambda_1^{n-i}\lambda_1^{i-1}= \sum_{i=1}^{n}\lambda_1^{n-1} = n\cdot\lambda_1^{n-1}$$ - If $\lambda_1\neq\lambda_2$
$$(\lambda_1-\lambda_2)k_n=(\lambda_1-\lambda_2)(\lambda_1^{n-1}+\lambda_1^{n-2} \lambda_2 +\cdots+\lambda_1\lambda_2^{n-2}+\lambda_2^{n-1})$$ $$= (\lambda_1^{n}+\lambda_1^{n-1}\lambda_2+\cdots+\lambda_1^2 \lambda_2^{n-2}+\lambda_1\lambda_2^{n-1}) - (\lambda_1^{n-1}\lambda_2+\lambda_1^{n-2}\lambda_2^2+\cdots +\lambda_1\lambda_2^{n-1}+\lambda_2^{n}) =\lambda_1^{n}-\lambda_2^{n}$$ $$\Rightarrow k_n= {(\lambda_1^{n}-\lambda_2^{n}) \over (\lambda_1-\lambda_2)}$$
8. Use Question 7 to prove that if $A=\begin{bmatrix}1&2\\2&1 \end{bmatrix}$, then $$A^{n}={3^n\over2}\begin{bmatrix}1&1\\1&1 \end{bmatrix}+{(-1)^{n-1}\over2}\begin{bmatrix}-1&1\\1&-1 \end{bmatrix}$$ if $n\geq1$.
Solution:
$\lambda_1$ and $\lambda_2$ are the roots of $$x^2-x-3=(x-3)(x+1) \Rightarrow \lambda_1=3,\ \lambda_2=1$$ Note that $\lambda_1\neq\lambda_2$ and use the formula from the previous problem we have $$k_n={3^n-(-1)^n\over3-(-1)}={3^n-(-1)^n\over4},\ k_{n-1}={3^{n-1}- (-1)^{n-1}\over4}$$ Hence $$A^{n}=k_nA-\lambda_1\lambda_2k_{n-1}I_2$$ $$={3^n-(-1)^n\over4}\cdot\begin{bmatrix}1&2\\2&1 \end{bmatrix} +3\cdot{3^{n-1}- (-1)^{n-1}\over4}\cdot\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}$$ $$={3^n + (-1)^{n-1} \over4} \cdot \begin{bmatrix}1&2\\2&1 \end{bmatrix} +{3^{n} - 3\cdot(-1)^{n-1}\over4}\cdot\begin{bmatrix}1 & 0\\ 0 & 1 \end{bmatrix}$$ $$=\begin{bmatrix}\displaystyle{3^n+(-1)^{n-1} +3^n -3\cdot(-1)^{n-1} \over 4} & \displaystyle{3^n+(-1)^{n-1} \over 2}\\ \displaystyle{3^n+(-1)^{n-1} \over 2} & \displaystyle{3^n+(-1)^{n-1} +3^n -3\cdot(-1)^{n-1} \over 4}\end{bmatrix}$$ $$=\begin{bmatrix}\displaystyle{3^n\over 2} - {(-1)^{n-1}\over 2} & \displaystyle{3^n\over 2} + {(-1)^{n-1}\over 2}\\ \displaystyle{3^n\over 2} + {(-1)^{n-1}\over 2} & \displaystyle{3^n\over 2} - {(-1)^{n-1}\over 2}\end{bmatrix}$$ $$ = {3^n\over2} \begin{bmatrix}1&1\\1&1 \end{bmatrix} +{(-1)^{n-1}\over2} \begin{bmatrix}-1&1\\1&-1 \end{bmatrix}$$
9. The Fibonacci numbers are defined by the equations $F_0=0$, $F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ if $n\geq1$. Prove that $$F_n={1\over\sqrt{5}}\left(\left({1+\sqrt{5}\over2}\right)^n -\left({1-\sqrt{5}\over2}\right)^n\right)$$
Solution:
Since $F_{n+1}=F_n+F_{n-1}$, we have $$\begin{bmatrix}F_{n+1}\\ F_{n} \end{bmatrix}=\begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}\cdot\begin{bmatrix}F_{n}\\ F_{n-1} \end{bmatrix}= A\cdot\begin{bmatrix}F_{n}\\ F_{n-1} \end{bmatrix} = A^{n}\cdot\begin{bmatrix}F_{1}\\ F_{0} \end{bmatrix} = A^{n}\cdot \begin{bmatrix}1 \\ 0 \end{bmatrix}$$ where $A = \begin{bmatrix}1 & 1\\ 1 & 0 \end{bmatrix}$. From problem 7, we know $\lambda_1$ and $\lambda_2$ are the roots of $x^2-x-1\Rightarrow \lambda_1={1+\sqrt{5}\over2},\ \lambda_2={1-\sqrt{5}\over2}$. And $$k_n={\lambda_1^n-\lambda_2^n\over\lambda_1-\lambda_2} = {\left({1+\sqrt{5}\over2}\right)^n-\left({1-\sqrt{5}\over2}\right)^n \over\sqrt{5}}$$ $$\Rightarrow A^{n}=k_nA-\lambda_1\lambda_2k_{n-1}I_2=k_nA+k_{n-1}I_2$$ Hence $$\begin{bmatrix}F_{n+1}\\ F_{n} \end{bmatrix} = A^{n}\cdot \begin{bmatrix}1 \\ 0 \end{bmatrix} = (k_nA+k_{n-1}I_2)\cdot \begin{bmatrix}1 \\ 0 \end{bmatrix}$$ $$ = \begin{bmatrix}k_n+k_{n-1} & k_n\\k_n& k_{n-1}\end{bmatrix} \cdot \begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}k_n+k_{n-1}\\ k_n\end{bmatrix} $$ That is, $$F_n=k_n = {1 \over\sqrt{5}}\cdot \left(\left({1+\sqrt{5}\over2}\right)^n - \left({1-\sqrt{5}\over2}\right)^n\right)$$
10. Let $r > 1$ be an integer. Let $a$ and $b$ be arbitrary positive integers. Sequences $x_n$ and $y_n$ of positive integers are defined in terms of $a$ and $b$ by the recurrence relations $$x_{n+1}=x_n+ry_n$$ $$y_{n+1} = x_n + y_n,$$ for $n\geq0$, where $x_0=a$ and $y_0=b$.
Use Question 7 to prove that $$\lim_{n\to\infty}{x_n\over y_n}=\sqrt{r}$$
Solution:
$$\begin{bmatrix}x_{n+1}\\ y_{n+1} \end{bmatrix} = \begin{bmatrix}1 & r\\1 & 1 \end{bmatrix} \cdot \begin{bmatrix}x_{n}\\ y_{n} \end{bmatrix} = A\cdot \begin{bmatrix}x_{n}\\ y_{n} \end{bmatrix} = A^{n+1}\cdot \begin{bmatrix}x_{0}\\ y_{0} \end{bmatrix} = A^{n+1}\cdot \begin{bmatrix}a\\ b \end{bmatrix}$$ where $A = \begin{bmatrix}1 & r\\1 & 1 \end{bmatrix}$.
$\lambda_1$ and $\lambda_2$ are the roots of $x^2-2x+(1-r)$, hence $$\lambda_1=1+\sqrt{r},\ \lambda_2 = 1-\sqrt{r}\Rightarrow k_n={\lambda_1^n-\lambda_2^n\over \lambda_1-\lambda_2} ={\lambda_1^n-\lambda_2^n\over 2\sqrt{r}}$$ And $$A^{n} = k_nA-\lambda_1\lambda_2k_{n-1}I_2 = k_nA-(1-r)k_{n-1}I_2$$ $$\Rightarrow \begin{bmatrix}x_{n}\\ y_{n} \end{bmatrix} = (k_nA-(1-r)k_{n-1}I_2)\cdot \begin{bmatrix}a\\ b \end{bmatrix}$$ $$=\left(\begin{bmatrix}k_n & rk_n\\k_n & k_n \end{bmatrix}-\begin{bmatrix}(1-r)k_{n-1} & 0\\0 & (1-r)k_{n-1} \end{bmatrix}\right) \cdot \begin{bmatrix}a\\ b \end{bmatrix}$$ $$= \begin{bmatrix}k_n-(1-r)k_{n-1} & rk_n\\k_n & k_n-(1-r)k_{n-1} \end{bmatrix}\cdot \begin{bmatrix}a\\ b \end{bmatrix}$$ $$=\begin{bmatrix}a(k_n-(1-r)k_{n-1}) + brk_n\\ ak_n + b(k_n-(1-r)k_{n-1}) \end{bmatrix}$$ Since $$\lim_{n\to\infty}{k_n\over k_{n-1}}=\lim_{n\to\infty}{\lambda_1^{n}-\lambda_2^{n}\over \lambda_1^{n-1}-\lambda_2^{n-1}} = \lim_{n\to\infty} {\lambda_1^n\left(1-\left(\displaystyle{\lambda_2\over\lambda_1} \right)^n\right) \over \lambda_1^{n-1}\left(1-\left(\displaystyle {\lambda_2\over\lambda_1} \right)^{n-1}\right)} = \lambda_1 = 1+\sqrt{r}$$ Thus $$\lim_{n\to\infty}{x_n\over y_n}=\lim_{n\to\infty} {a(k_n-(1-r)k_{n-1}) + brk_n \over ak_n + b(k_n-(1-r)k_{n-1})}$$ $$=\lim_{n\to\infty} {a(1+\sqrt{r}-(1-r))+br(1+\sqrt{r}) \over a(1+\sqrt{r}) + b(1+\sqrt{r}-(1-r))}$$ $$=\lim_{n\to\infty} {a(\sqrt{r} + r) + br(1+\sqrt{r}) \over a(1+\sqrt{r}) + b(\sqrt{r}+r)}$$ $$=\lim_{n\to\infty}{\sqrt{r}[a(1+\sqrt{r}) + b(\sqrt{r}+r)] \over a(1+\sqrt{r}) + b(\sqrt{r}+r)} = \sqrt{r}$$
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