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# 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|MATH221 Preliminary Results from Algebra and Analysis

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## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|Preliminary Results from Algebra and Analysis

For future reference we collect here several fundamental concepts and results from algebra and analysis.
A function $P_n(x)$ defined by
$$P_n(x)=a_0+a_1 x+\cdots+a_n x^n=\sum_{i=0}^n a_i x^i, \quad a_n \neq 0$$
where $a_i \in \mathbb{R}, 0 \leq i \leq n$, is called a polynomial of degree $n$ in $x$. If $P_n\left(x_1\right)=0$, then the number $x=x_1$ is called a zero of $P_n(x)$. The following fundamental theorem of algebra of complex numbers is valid.

Theorem 13.1. Every polynomial of degree $n \geq 1$ has at least one zero.

Thus, $P_n(x)$ has exactly $n$ zeros; however, some of these may be the same, i.e., $P_n(x)$ can be written as
$$P_n(x)=a_n\left(x-x_1\right)^{r_1}\left(x-x_2\right)^{r_2} \cdots\left(x-x_k\right)^{r_k}, \quad r_i \geq 1, \quad 1 \leq i \leq k,$$
where $\sum_{i=1}^k r_i=n$. If $r_i=1$, then $x_i$ is called a simple zero, and if $r_i>$ 1 , then multiple zero of multiplicity $r_i$. Thus, if $x_i$ is a multiple zero of multiplicity $r_i$, then $P^{(j)}\left(x_i\right)=0,0 \leq j \leq r_i-1$ and $P^{\left(r_i\right)}\left(x_i\right) \neq 0$.

A rectangular table of $m \times n$ elements arranged in $m$ rows and $n$ columns
$$\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \ a_{21} & a_{22} & \cdots & a_{2 n} \ \cdots & & & \ a_{m 1} & a_{m 2} & \cdots & a_{m n} \end{array}\right]$$
is called an $m \times n$ matrix and in short represented as $A=\left(a_{i j}\right)$. We shall mainly deal with square matrices $(m=n)$, row matrices or row vectors $(m=1)$, and column matrices or column vectors $(n=1)$.

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|Preliminary Results from Algebra and Analysis (Contd.)

The number $\lambda$, real or complex, is called an eigenvalue of the matrix $A$ if there exists a nonzero real or complex vector $v$ such that $A v=\lambda v$. The vector $v$ is called an eigenvector corresponding to the eigenvalue $\lambda$. From Theorem 13.2, $\lambda$ is an eigenvalue of $A$ if and only if it is a solution of the characteristic equation $p(\lambda)=\operatorname{det}(A-\lambda I)=0$. Since the matrix $A$ is $n \times n, p(\lambda)$ is a polynomial of degree exactly $n$, and it is called the characteristic polynomial of $A$. Hence, from Theorem $13.1$ it follows that $A$ has exactly $n$ eigenvalues counting their multiplicities.

In the case when the eigenvalues $\lambda_1, \ldots, \lambda_n$ of $A$ are distinct it is easy to find the corresponding eigenvectors $v^1, \ldots, v^n$. For this, first we note that for the fixed eigenvalue $\lambda_j$ of $A$ at least one of the cofactors of $\left(a_{i i}-\lambda_j\right)$ in the matrix $\left(A-\lambda_j I\right)$ is nonzero. If not, then from (13.5) it follows that $p^{\prime}(\lambda)=-\left[\right.$ cofactor of $\left.\left(a_{11}-\lambda\right)\right]-\cdots-\left[\operatorname{cofactor}\right.$ of $\left.\left(a_{n n}-\lambda\right)\right]$, and hence $p^{\prime}\left(\lambda_j\right)=0$; i.e., $\lambda_j$ was a multiple root, which is a contradiction to our assumption that $\lambda_j$ is simple. Now let the cofactor of $\left(a_{k k}-\lambda_j\right)$ be different from zero, then one of the possible nonzero solution of the system $\left(A-\lambda_j I\right) v^j=0$ is $v_i^j=$ cofactor of $a_{k i}$ in $\left(A-\lambda_j I\right), 1 \leq i \leq n, i \neq k$, $v_k^j=$ cofactor of $\left(a_{k k}-\lambda_j\right)$ in $\left(A-\lambda_j I\right)$. Since for this choice of $v^j$, it follows from (13.2) that every equation, except the $k$ th one, of the system $\left(A-\lambda_j I\right) v^j=0$ is satisfied, and for the $k$ th equation from (13.1), we have $\sum_{i \neq \hbar}^n a_{k i}\left[\right.$ cofactor of $\left.a_{k i}\right]+\left(a_{k k}-\lambda_j\right)\left[\operatorname{cofactor}\right.$ of $\left.\left(a_{k k}-\lambda_j\right)\right]=\operatorname{det}\left(A-\lambda_j I\right)$,
which is also zero. In conclusion this $v^j$ is the eigenvector corresponding to the eigenvalue $\lambda_j$.

# 多变量微积分和常微分方程代考

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus \& Ordinary Differential Equations代写|Preliminary Results from Algebra and Analysis

$$P_n(x)=a_0+a_1 x+\cdots+a_n x^n=\sum_{i=0}^n a_i x^i, \quad a_n \neq 0$$

$$P_n(x)=a_n\left(x-x_1\right)^{r_1}\left(x-x_2\right)^{r_2} \cdots\left(x-x_k\right)^{\tau_k}, \quad r_i \geq 1, \quad 1 \leq i \leq k,$$

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus \& Ordinary Differential Equations代写|Preliminary Results from Algebra and Analysis (Contd.)

$\left.\left(a_{11}-\lambda\right)\right]-\cdots-\left[\right.$ cofactor的 $\left(a_{n n}-\lambda\right)$ ， 因此 $p^{\prime}\left(\lambda_j\right)=0 ; \mathrm{E}^2$ 。 $\lambda_j$ 是 个多重根，这与我们的假设相矛盾 $\lambda_j$ 很简单。现在 让 $\left(a_{k k}-\lambda_j\right)$ 不为零，则系统可能的非零解之一 $\left(A-\lambda_j I\right) v^j=0$ 是 $v_i^j=$ 的辅因子 $a_{k i}$ 在 $\left(A-\lambda_j I\right), 1 \leq i \leq n, i \neq k ， v_k^j=$ 的辅因子 $\left(a_{k k}-\lambda_j\right)$ 在 $\left(A-\lambda_j I\right)$. 因为对于这个选择 $v^j$ ，从 (13.2) 可以得出，每个方程，除了 $k$ 一系统的 $\left(A-\lambda_j I\right) v^j=0$ 是满意的，并且对于 $k(13.1)$ 中的方程，我们有 $\sum_{i \neq h}^n a_{k i}\left[\right.$ 的辅因子 $\left.a_{k i}\right]+\left(a_{k k}-\lambda_j\right)[$ cofactor的 $\left.\left(a_{k k}-\lambda_j\right)\right]=\operatorname{det}\left(A-\lambda_j I\right)$

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## MATLAB代写

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