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# 数学代写|线性代数代写Linear algebra代考|KMA254 Eigenvectors and Difference Equations

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## 数学代写|线性代数代写Linear algebra代考|Eigenvectors and Difference Equations

This section concludes by showing how to construct solutions of the first-order difference equation discussed in the chapter introductory example:
$$\mathbf{x}_{k+1}=A \mathbf{x}_k \quad(k=0,1,2, \ldots)$$
If $A$ is an $n \times n$ matrix, then (8) is a recursive description of a sequence $\left{\mathbf{x}_k\right}$ in $\mathbb{R}^n$. A solution of (8) is an explicit description of $\left{\mathbf{x}_k\right}$ whose formula for each $\mathbf{x}_k$ does not depend directly on $A$ or on the preceding terms in the sequence other than the initial term $\mathbf{x}_0$.

The simplest way to build a solution of (8) is to take an eigenvector $\mathbf{x}0$ and its corresponding eigenvalue $\lambda$ and let $$\mathbf{x}_k=\lambda^k \mathbf{x}_0 \quad(k=1,2, \ldots)$$ This sequence is a solution because $$A \mathbf{x}_k=A\left(\lambda^k \mathbf{x}_0\right)=\lambda^k\left(A \mathbf{x}_0\right)=\lambda^k\left(\lambda \mathbf{x}_0\right)=\lambda^{k+1} \mathbf{x}_0=\mathbf{x}{k+1}$$
Linear combinations of solutions in the form of equation (9) are solutions, too! See Exercise $33 .$

## 数学代写|线性代数代写Linear algebra代考|THE CHARACTERISTIC EQUATION

Useful information about the eigenvalues of a square matrix $A$ is encoded in a special scalar equation called the characteristic equation of $A$. A simple example will lead to the general case.
EXAMPLE 1 Find the eigenvalues of $A=\left[\begin{array}{rr}2 & 3 \ 3 & -6\end{array}\right]$.
SOLUTION We must find all scalars $\lambda$ such that the matrix equation
$$(A-\lambda I) \mathbf{x}=\mathbf{0}$$
has a nontrivial solution. By the Invertible Matrix Theorem in Section 2.3, this problem is equivalent to finding all $\lambda$ such that the matrix $A-\lambda I$ is not invertible, where
$$A-\lambda I=\left[\begin{array}{rr} 2 & 3 \ 3 & -6 \end{array}\right]-\left[\begin{array}{ll} \lambda & 0 \ 0 & \lambda \end{array}\right]=\left[\begin{array}{cc} 2-\lambda & 3 \ 3 & -6-\lambda \end{array}\right]$$
By Theorem 4 in Section 2.2, this matrix fails to be invertible precisely when its determinant is zero. So the eigenvalues of $A$ are the solutions of the equation
$$\operatorname{det}(A-\lambda I)=\operatorname{det}\left[\begin{array}{cc} 2-\lambda & 3 \ 3 & -6-\lambda \end{array}\right]=0$$
Recall that
$$\operatorname{det}\left[\begin{array}{ll} a & b \ c & d \end{array}\right]=a d-b c$$
So
\begin{aligned} \operatorname{det}(A-\lambda I) &=(2-\lambda)(-6-\lambda)-(3)(3) \ &=-12+6 \lambda-2 \lambda+\lambda^2-9 \ &=\lambda^2+4 \lambda-21 \ &=(\lambda-3)(\lambda+7) \end{aligned}
If $\operatorname{det}(A-\lambda I)=0$, then $\lambda=3$ or $\lambda=-7$. So the eigenvalues of $A$ are 3 and $-7$.
The determinant in Example 1 transformed the matrix equation $(A-\lambda I) \mathbf{x}=\mathbf{0}$, which involves two unknowns $(\lambda$ and $\mathbf{x})$, into the scalar equation $\lambda^2+4 \lambda-21=0$, which involves only one unknown. The same idea works for $n \times n$ matrices. However, before turning to larger matrices, we summarize the properties of determinants needed to study eigenvalues.

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