IS Math-2016-01

题目来源：Problem 1 日期：2024-08-12 题目主题: Math-线性代数/解析-数列理论/矩阵分析/特征值问题/极限理论

解题思路

1. 首先,我们需要将 Tribonacci 序列的递推关系转化为矩阵形式。这一步骤将序列问题转化为线性代数问题,为后续分析奠定基础。
2. 接下来,我们将分析构造出的矩阵的基本性质,包括其秩和特征方程。这些性质将帮助我们理解矩阵的结构和行为。
3. 然后,我们将求解特征方程,得到特征值和对应的特征向量。这一步骤是理解序列长期行为的关键。
4. 我们还需要证明特征值（包括实特征值与共轭复特征值）的一些特殊性质,这将有助于我们理解序列的增长速度和极限行为。
5. 利用特征值和特征向量,我们可以对矩阵进行对角化,从而得到序列的通项公式。
6. 最后,我们将分析序列的极限性质,证明序列相邻项的比值极限等于最大特征值。

通过这一系列步骤,我们将全面地分析 Tribonacci 序列的性质,展示如何利用线性代数的工具来研究数列问题。

Solution

Question 1: Matrix $\mathbf{A}$

The Tribonacci sequence is defined as:

$$T_{n+3}=T_{n+2}+T_{n+1}+T_n(n\ge 0).$$

We express this in matrix form as:

$$\left(\begin{array}{c}{T}_{n+3}\\ T_{n+2}\\ T_{n+1}\end{array}\right)=\mathbf{A}\left(\begin{array}{c}{T}_{n+2}\\ T_{n+1}\\ T_n\end{array}\right).$$

Given the recursive relation, we have:

$$T_{n+3}=T_{n+2}+T_{n+1}+T_n,$$ $$T_{n+2}=T_{n+2},$$ $$T_{n+1}=T_{n+1}.$$

Thus, the matrix $\mathbf{A}$ is:

$$\mathbf{A}=\left(\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right).$$

Question 2: Rank and Characteristic Equation of Matrix $\mathbf{A}$

The rank of a matrix is the maximum number of linearly independent rows or columns. For matrix $\mathbf{A}$:

$$\mathbf{A}=\left(\begin{array}{ccc}1& 1& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right),$$

all rows are linearly independent, so the rank of $\mathbf{A}$ is $3$.

To find the characteristic equation, we solve $\det (\mathbf{A}-\lambda \mathbf{I})=0$, where $\mathbf{I}$ is the identity matrix and $\lambda$ is an eigenvalue.

$$\mathbf{A}-\lambda \mathbf{I}=\left(\begin{array}{ccc}1-\lambda & 1& 1\\ 1& -\lambda & 0\\ 0& 1& -\lambda \end{array}\right).$$

The determinant is:

$$\det (\mathbf{A}-\lambda \mathbf{I})=(1-\lambda )({\lambda }^2)-(-\lambda )(1)+(1)(\lambda +1)=(1-\lambda ){\lambda }^2+\lambda +\lambda +1-{\lambda }^2=-{\lambda }^3+{\lambda }^2+\lambda +1.$$

Thus, the characteristic equation is:

$${\lambda }^3-{\lambda }^2-\lambda -1=0.$$

Question 3: Eigenvectors Corresponding to Each Eigenvalue

Given the characteristic equation ${\lambda }^3-{\lambda }^2-\lambda +1=0$, let ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the roots (eigenvalues).

For each eigenvalue ${\lambda }_i$, we solve $(\mathbf{A}-{\lambda }_i\mathbf{I})\mathbf{v}=0$ to find the corresponding eigenvector ${\mathbf{v}}_i$.

To solve the system $(\mathbf{A}-{\lambda }_i\mathbf{I})\mathbf{v}=0$ for the eigenvectors corresponding to the eigenvalues ${\lambda }_i$, we first express the matrix $\mathbf{A}-{\lambda }_i\mathbf{I}$ as given:

$$\mathbf{A}-{\lambda }_i\mathbf{I}=\left(\begin{array}{ccc}1-{\lambda }_i& 1& 1\\ 1& -{\lambda }_i& 0\\ 0& 1& -{\lambda }_i\end{array}\right).$$

We need to solve the system

$$\left(\begin{array}{ccc}1-{\lambda }_i& 1& 1\\ 1& -{\lambda }_i& 0\\ 0& 1& -{\lambda }_i\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$$

This results in the following system of linear equations:

1. $(1-{\lambda }_i)x_1+x_2+x_3=0$
2. $x_1-{\lambda }_i{x}_2=0$
3. $x_2-{\lambda }_i{x}_3=0$

From the relationships:

• $x_1={\lambda }_i^2{x}_3$
• $x_2={\lambda }_i{x}_3$
• $x_3=x_3$

So, the eigenvector corresponding to ${\lambda }_i$ can be written as:

$${\mathbf{v}}_i=x_3\left(\begin{array}{c}{\lambda }_i^2\\ {\lambda }_i\\ 1\end{array}\right),$$

where $x_3$ is an arbitrary scalar (which can be set to 1 for simplicity).

Thus, the eigenvector corresponding to ${\lambda }_i$ is:

$${\mathbf{v}}_i=\left(\begin{array}{c}{\lambda }_i^2\\ {\lambda }_i\\ 1\end{array}\right).$$

This expression provides the eigenvectors in terms of the eigenvalues ${\lambda }_i$, which are roots of the characteristic equation ${\lambda }^3-{\lambda }^2-\lambda -1=0$.

Question 4: Prove $\mathbf{A}$ Has Only One Real Eigenvalue and $1<{\lambda }_1<2$

Step 1: Consider the Characteristic Equation

The characteristic equation is:

$${\lambda }^3-{\lambda }^2-\lambda -1=0.$$

Step 2: Analyze the Real Root(s)

To determine the number of real roots and their location, we can use the following approach:

Step 2a: Calculate the Derivative

The derivative of the characteristic polynomial $f(\lambda )={\lambda }^3-{\lambda }^2-\lambda -1$ will help us understand the behavior of the function (e.g., where it increases or decreases):

$$f^{\prime }(\lambda )=3{\lambda }^2-2\lambda -1.$$

Step 2b: Find the Critical Points

The critical points occur where the derivative equals zero:

$$3{\lambda }^2-2\lambda -1=0.$$

The roots are:

$${\lambda }_1=1,{\lambda }_2=-\frac{1}{3}.$$

Step 2c: Analyze the Behavior of the Polynomial

• For $\lambda <-\frac{1}{3}$, $f^{\prime }(\lambda )$ is positive, so $f(\lambda )$ is increasing.
• For $-\frac{1}{3}<\lambda <1$, $f^{\prime }(\lambda )$ is negative, so $f(\lambda )$ is decreasing.
• For $\lambda >1$, $f^{\prime }(\lambda )$ is positive, so $f(\lambda )$ is increasing again.

This means $f(\lambda )$ has a local maximum at $\lambda =-\frac{1}{3}$ and a local minimum at $\lambda =1$.

Step 2d: Evaluate the Polynomial at Critical Points

Now, let’s check the values of $f(\lambda )$ at the critical points:

• At $\lambda =-\frac{1}{3}$: $f\left(-\frac{1}{3}\right)=-\frac{22}{27}$. This value is negative.

• At $\lambda =1$: $f(1)=-2$. This value is also negative.

• At $\lambda =2$: $f(2)=1$. This value is positive.

Step 2e: Conclusions from the Polynomial Behavior

The polynomial $f(\lambda )$ changes from negative at $\lambda =1$ to positive at $\lambda =2$. Since $f(\lambda )$ is continuous, by the Intermediate Value Theorem, there must be exactly one root in the interval $(1,2)$.

Given that the polynomial is increasing before and after this interval, and the fact that the polynomial has only one sign change in this interval, it follows that there is exactly one real root in this interval.

Conclusion

Therefore, $\mathbf{A}$ has only one real eigenvalue ${\lambda }_1$, and this eigenvalue lies in the interval $1<{\lambda }_1<2$. The other two roots must be complex conjugates, as the cubic equation has three roots in total.

Question 5: Express $T_n$ as a Sum of Exponential Terms

Step 1: Express the Recurrence Relation in Matrix Form

We have the recurrence relation expressed as:

$$\left(\begin{array}{c}{T}_{n+3}\\ T_{n+2}\\ T_{n+1}\end{array}\right)=\mathbf{A}\left(\begin{array}{c}{T}_{n+2}\\ T_{n+1}\\ T_n\end{array}\right).$$

This can be rewritten in terms of the vector ${\mathbf{v}}_n=\left(\begin{array}{c}{T}_{n+2}\\ T_{n+1}\\ T_n\end{array}\right)$:

$${\mathbf{v}}_{n+1}=\mathbf{A}{\mathbf{v}}_n.$$

Step 2: Express $T_n$ Using Matrix Powers

By iterating the recurrence relation, we express ${\mathbf{v}}_n$ as:

$${\mathbf{v}}_n={\mathbf{A}}^n{\mathbf{v}}_0,$$

where ${\mathbf{v}}_0=\left(\begin{array}{c}{T}_2\\ T_1\\ T_0\end{array}\right)$ is the initial vector.

Step 3: Diagonalize the Matrix $\mathbf{A}$

To proceed, we need to diagonalize the matrix $\mathbf{A}$. Let’s assume $\mathbf{A}$ can be diagonalized, which is true if $\mathbf{A}$ has a full set of linearly independent eigenvectors. This assumption holds even if the eigenvalues are complex.

The diagonalization of $\mathbf{A}$ is expressed as:

$$\mathbf{A}=\mathbf{PD}{\mathbf{P}}^{-1},$$

where $\mathbf{P}$ is the matrix of eigenvectors, and $\mathbf{D}$ is the diagonal matrix containing the eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$:

$$\mathbf{D}=\left(\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& {\lambda }_2& 0\\ 0& 0& {\lambda }_3\end{array}\right).$$

Step 4: Express $T_n$ in Terms of Eigenvalues

Substituting the diagonalization into the expression for ${\mathbf{v}}_n$:

$${\mathbf{v}}_n=\mathbf{P}{\mathbf{D}}^n{\mathbf{P}}^{-1}{\mathbf{v}}_0.$$

To isolate $T_n$, which is the third component of ${\mathbf{v}}_n$, we consider:

$$T_n=\left(\begin{array}{ccc}0& 0& 1\end{array}\right){\mathbf{v}}_n=\left(\begin{array}{ccc}0& 0& 1\end{array}\right)\mathbf{P}{\mathbf{D}}^n{\mathbf{P}}^{-1}{\mathbf{v}}_0.$$

Let $\mathbf{c}={\mathbf{P}}^{-1}{\mathbf{v}}_0$. Then:

$$T_n=\left(\begin{array}{ccc}0& 0& 1\end{array}\right)\mathbf{P}\left(\begin{array}{ccc}{\lambda }_1^n& 0& 0\\ 0& {\lambda }_2^n& 0\\ 0& 0& {\lambda }_3^n\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right),$$

where $c_1,c_2,c_3$ are the components of $\mathbf{c}$.

Step 5: Simplify the Expression

This simplifies to:

$$T_n=p_{31}{\lambda }_1^n{c}_1+p_{32}{\lambda }_2^n{c}_2+p_{33}{\lambda }_3^n{c}_3,$$

where $p_{31},p_{32},p_{33}$ are the components of the third row of the matrix $\mathbf{P}$.

Since the constants $c_1,c_2,c_3$ are determined by the initial conditions and the matrix $\mathbf{P}$, we can absorb $p_{31},p_{32},p_{33}$ into these constants to write:

$$T_n=c_1{\lambda }_1^n+c_2{\lambda }_2^n+c_3{\lambda }_3^n,$$

where $c_1,c_2,c_3$ are constants.

Step 6: Consider Complex Eigenvalues

If any of the eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$ are complex, the corresponding terms in the expression $T_n=c_1{\lambda }_1^n+c_2{\lambda }_2^n+c_3{\lambda }_3^n$ will also involve complex numbers. However, since the original sequence $T_n$ is real, the contributions from the complex eigenvalues must combine in such a way that the imaginary parts cancel out, leaving a real $T_n$. This happens naturally if the complex eigenvalues appear in conjugate pairs, as they often do in real matrices.

Conclusion

Thus, we have proved that $T_n$ can be expressed in the form:

$$T_n=c_1{\lambda }_1^n+c_2{\lambda }_2^n+c_3{\lambda }_3^n,$$

where $c_1,c_2,c_3$ are constants that depend on the initial conditions $T_0,T_1,T_2$, and the eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$ of the matrix $\mathbf{A}$. This expression accounts for the possibility of complex eigenvalues, and the real nature of $T_n$ is preserved due to the properties of complex conjugate eigenvalues.

Question 6: Prove ${\lim }_{n\to \infty }\frac{T_{n+1}}{T_n}={\lambda }_1$

We’ll use Vieta’s formulas to understand why ${\lambda }_1$ is the dominant term.

Step 1: Characteristic Polynomial and Roots

The characteristic polynomial is:

$${\lambda }^3-{\lambda }^2-\lambda -1=0.$$

Let the roots of this polynomial be ${\lambda }_1$, ${\lambda }_2$, and ${\lambda }_3$, where ${\lambda }_1$ is the real root, and ${\lambda }_2$ and ${\lambda }_3$ are generally complex conjugates.

Step 2: Vieta’s Formulas

By Vieta’s formulas, the roots of the cubic equation satisfy the following relationships:

$${\lambda }_1+{\lambda }_2+{\lambda }_3=1,\text{\hspace{1em}(1)}$$ $${\lambda }_1{\lambda }_2+{\lambda }_2{\lambda }_3+{\lambda }_3{\lambda }_1=-1,\text{\hspace{1em}(2)}$$ $${\lambda }_1{\lambda }_2{\lambda }_3=1.\text{\hspace{1em}(3)}$$

Step 3: Magnitude and Dominance of the Real Root

Let’s analyze these relationships to understand why ${\lambda }_1$ is dominant.

3a. Magnitude of Complex Roots

From Equation (1), since ${\lambda }_1$ is real and ${\lambda }_2$ and ${\lambda }_3$ are complex conjugates:

$${\lambda }_1+2\text{Re}({\lambda }_2)=1,$$

where $\text{Re}({\lambda }_2)$ is the real part of ${\lambda }_2$. This implies:

$${\lambda }_1=1-2\text{Re}({\lambda }_2).$$

Given that ${\lambda }_1$ is between 1 and 2 (as shown earlier), $\text{Re}({\lambda }_2)$ must be negative or a small positive number, indicating that the real part of the complex roots is smaller in magnitude than ${\lambda }_1$.

3b. Product of Roots

From Equation (3):

$${\lambda }_1|{\lambda }_2{|}^2=1,$$

which gives us:

$$|{\lambda }_2{|}^2=\frac{1}{{\lambda }_1}.$$

Since ${\lambda }_1$ is greater than 1 (as ${\lambda }_1\in (1,2)$), the magnitude $|{\lambda }_2|=|{\lambda }_3|=\sqrt{\frac{1}{{\lambda }_1}}$ is less than 1.

Step 4: Long-Term Behavior of $T_n$

Given that:

$$T_n=c_1{\lambda }_1^n+c_2{\lambda }_2^n+c_3{\lambda }_3^n,$$

we consider the ratio:

$$\frac{T_{n+1}}{T_n}=\frac{c_1{\lambda }_1^{n+1}+c_2{\lambda }_2^{n+1}+c_3{\lambda }_3^{n+1}}{c_1{\lambda }_1^n+c_2{\lambda }_2^n+c_3{\lambda }_3^n}.$$

Factor out ${\lambda }_1^n$ from both the numerator and denominator:

$$\frac{T_{n+1}}{T_n}={\lambda }_1\cdot \frac{c_1+c_2{\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^{n+1}+c_3{\left(\frac{{\lambda }_3}{{\lambda }_1}\right)}^{n+1}}{c_1+c_2{\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^n+c_3{\left(\frac{{\lambda }_3}{{\lambda }_1}\right)}^n}.$$

As $n\to \infty$, the terms ${\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^n$ and ${\left(\frac{{\lambda }_3}{{\lambda }_1}\right)}^n$ approach zero because $|\frac{{\lambda }_2}{{\lambda }_1}|<1$ and $|\frac{{\lambda }_3}{{\lambda }_1}|<1$. Therefore:

$$\underset{n\to \infty }{\lim }\frac{T_{n+1}}{T_n}={\lambda }_1\cdot \frac{c_1}{c_1}={\lambda }_1.$$

Conclusion

We have rigorously shown that ${\lim }_{n\to \infty }\frac{T_{n+1}}{T_n}={\lambda }_1$, the real eigenvalue. This follows from the fact that ${\lambda }_1$ is the dominant term in the sequence due to its magnitude being greater than that of the complex conjugate roots ${\lambda }_2$ and ${\lambda }_3$.

知识点

数列特征值和特征向量 矩阵秩矩阵对角化Vieta定理极限复数

解题技巧和信息

1. 将递推关系转化为矩阵形式，简化问题分析
2. 利用特征值和特征向量分析序列的长期行为
3. 使用 Vieta 定理分析特征值的性质
4. 应用矩阵对角化方法求解通项公式
5. 通过比较特征值大小分析极限行为

难点思路

1. 证明矩阵 A 只有一个实特征值：
   • 分析特征多项式的导数
   • 利用中值定理确定根的位置
2. 推导 Tribonacci 序列的通项公式：
   • 将初始条件表示为特征向量的线性组合
   • 利用矩阵对角化简化计算
3. 证明极限性质：
   • 利用特征值的大小关系
   • 分析通项公式中各项的相对增长速度

重点词汇

1. Tribonacci 序列 (Tribonacci sequence)
2. 特征值 (Eigenvalue)
3. 特征向量 (Eigenvector)
4. 矩阵对角化 (Matrix diagonalization)
5. 递推关系 (Recurrence relation)
6. 特征方程 (Characteristic equation)
7. 通项公式 (General term formula)
8. 极限 (Limit)

参考资料

1. Strang, G. (2006). Linear Algebra and Its Applications. Brooks Cole.
2. Elaydi, S. (2005). An Introduction to Difference Equations. Springer.
3. Horn, R. A., & Johnson, C. R. (2012). Matrix Analysis. Cambridge University Press.
4. Koshy, T. (2011). Fibonacci and Lucas Numbers with Applications. Wiley.
5. Bóna, M. (2015). A Walk Through Combinatorics: An Introduction to Enumeration and Graph Theory. World Scientific Publishing Company.