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By admin 数学代写 2022年2月22日

Proposition 2.14.5 The greatest common divisor of $\left{p_{1}(\lambda), \cdots, p_{\mathrm{m}}(\lambda)\right}$ exists and is characterized as the monic polynomial of smallest degree equal to an expression of the form
$$
\sum_{k=1}^{m} a_{k}(\lambda) p_{k}(\lambda), \text { the } a_{k}(\lambda) \text { being polynomials. }
$$
Proof: First I need show that if $q(\lambda)$ is monic of the above form with smallest degree, then it is the greatest common divisor. If $q(\lambda)$ fails to divide $p_{k}(\lambda)$, then $p_{k}(\lambda)=q(\lambda) l(\lambda)+r(\lambda)$ where the degree of $r(\lambda)$ is smaller than the degree of $q(\lambda)$. Thus,
$$
r(\lambda)=p_{k}(\lambda)-l(\lambda) \sum_{k=1}^{m} a_{k}(\lambda) p_{k}(\lambda)
$$
which violates the condition that $q(\lambda)$ has smallest degree. Thus $q(\lambda) / p_{k}(\lambda)$ for each $k$. If $\hat{q}(\lambda)$ divides each $p_{k}(\lambda)$ then it must divide $q(\lambda)$ because $q(\lambda)$ is given by $2.1$. Hence $q(\lambda)$ is the greatest common divisor.
Next, why does such greatest common divisor exist? Simply pick the monic polynomial which has smallest degree which is of the form $\sum_{k-1}^{m} a_{k}(\lambda) p_{k}(\lambda)$. Then from what was just shown, it is the greatest common divisor.
Proposition 2.14.6 Let $p(\lambda)$ be a polynomial. Then there are polynomials $p_{1}(\lambda)$ such that
$$
p(\lambda)=a \prod_{i=1}^{m} p_{i}(\lambda)^{m_{i}}
$$
where $m_{1} \in \mathbb{N}$ and $\left{p_{1}(\lambda), \cdots, p_{m}(\lambda)\right}$ are monic and relatively prime or else there is no polynomial of degree larger than 0 which divides $p(\lambda)$.
Proof: If there is no polynomial of degree larger than 0 dividing $p(\lambda)$, then we are done. Simply pick $a$ such that $p(\lambda)$ is monic. Otherwise $p(\lambda)=a p_{1}(\lambda) p_{2}(\lambda)$ where $p_{1}(\lambda)$ is monic and each has degree at least 1. These could be the same polynomial. If some nonconstant polynomial divides each $p_{1}(\lambda)$, factor further. Continue doing this. Eventually the process must end with a factorization as described in $2.2$ because the degrees of the factors are decreasing-
$2.15$ The Cauchy Schwarz Inequality
This fundamental inequality takes several forms. I will present the version first given by Cauchy although I am not sure if the proof is the same.
Proposition 2.15.1 Let $z_{j}, w_{j}$ be complex numbers. Then
$$
\left|\sum_{j=1}^{\mathrm{P}} z_{j} \overline{w_{j}}\right| \leq\left(\sum_{j=1}^{p}\left|z_{j}\right|^{2}\right)^{1 / 2}\left(\sum_{j=1}^{\mathrm{P}}\left|w_{j}\right|^{2}\right)^{1 / 2}
$$
Proof: First note that $\sum_{j-1}^{p} z_{j} \overline{z_{j}}=\sum_{j-1}^{p}\left|z_{j}\right|^{2} \geq 0 .$ Next, if $a+i b$ is a complex number, consider $\theta=1$ if both $a, b$ are zero and $\theta=\frac{a-1 b}{\sqrt{a^{2}+b^{2}}}$ if the complex number is not zero. Thus, in either case, there exists a complex number $\theta$ such that $|\theta|=1$ and $\theta(a+i b)=|a+i b| \equiv \sqrt{a^{2}+b^{2}}$. Now let $|\theta|=1$ and
$$
\theta \sum_{j=1}^{p} z_{j} \overline{w_{j}}=\left|\sum_{j=1}^{p} z_{j} \overline{w_{j}}\right|
$$

命题 2.14.5 $\left\{p_{1}(\lambda), \cdots, p_{\mathrm{m}}(\lambda)\right\}$ 的最大公约数存在并且被表征为最小次数多项式等于以下形式的表达式 $$ \sum_{k=1}^{m} a_{k}(\lambda) p_{k}(\lambda), \text { } a_{k}(\lambda) \text { 是多项式。 } $$ 证明：首先我需要证明如果$q(\lambda)$是上述形式的最小度数的二元，那么它是最大公约数。如果$q(\lambda)$不能除$p_{k}(\lambda)$，则$p_{k}(\lambda)=q(\lambda) l(\lambda)+r(\lambda)$其中 $r(\lambda)$ 的度数小于 $q(\lambda)$ 的度数。因此， $$ r(\lambda)=p_{k}(\lambda)-l(\lambda) \sum_{k=1}^{m} a_{k}(\lambda) p_{k}(\lambda) $$ 这违反了 $q(\lambda)$ 具有最小度数的条件。因此，对于每个 $k$，$q(\lambda) / p_{k}(\lambda)$。如果 $\hat{q}(\lambda)$ 除以每个 $p_{k}(\lambda)$，那么它必须除以 $q(\lambda)$，因为 $q(\lambda)$ 由 $2.1$ 给出。因此 $q(\lambda)$ 是最大公约数。其次，为什么存在这样的最大公约数？只需选择次数最小的一元多项式，其形式为 $\sum_{k-1}^{m} a_{k}(\lambda) p_{k}(\lambda)$。那么从刚刚显示的内容来看，它是最大公约数。命题 2.14.6 令 $p(\lambda)$ 是一个多项式。然后有多项式 $p_{1}(\lambda)$ 使得 $$ p(\lambda)=a \prod_{i=1}^{m} p_{i}(\lambda)^{m_{i}} $$ 其中 $m_{1} \in \mathbb{N}$ 和 $\left\{p_{1}(\lambda), \cdots, p_{m}(\lambda)\right\}$ 是一元且互质的否则没有大于 0 的多项式除以 $p(\lambda)$。证明：如果没有大于 0 的多项式除以 $p(\lambda)$，那么我们就完成了。只需选择 $a$ 使得 $p(\lambda)$ 是 monic。否则 $p(\lambda)=a p_{1}(\lambda) p_{2}(\lambda)$ 其中 $p_{1}(\lambda)$ 是 monic 且每个度数至少为 1。这些可能是同一个多项式。如果某个非常数多项式除以每个 $p_{1}(\lambda)$，则进一步分解。继续这样做。最终，该过程必须以 $2.2$ 中所述的因式分解结束，因为因子的度数正在减少- $2.15$ 柯西施瓦茨不等式这种根本的不平等有多种形式。我将介绍 Cauchy 首先给出的版本，尽管我不确定证明是否相同。命题 2.15.1 令 $z_{j}, w_{j}$ 为复数。然后 $$ \left|\sum_{j=1}^{\mathrm{P}} z_{j} \overline{w_{j}}\right| \leq\left(\sum_{j=1}^{p}\left|z_{j}\right|^{2}\right)^{1 / 2}\left(\sum_{j=1}^ {\mathrm{P}}\left|w_{j}\right|^{2}\right)^{1 / 2} $$ 证明：首先注意$\sum_{j-1}^{p} z_{j} \overline{z_{j}}=\sum_{j-1}^{p}\left|z_{j}\right |^{2} \geq 0 .$ 接下来，如果 $a+ib$ 是一个复数，考虑 $\theta=1$ 如果 $a, b$ 都为零并且 $\theta=\frac{a-1 b}{\sqrt{a^{2}+b^{2}}}$ 如果复数不为零。因此，在任何一种情况下，都存在一个复数 $\theta$ 使得 $|\theta|=1$ 并且 $\theta(a+i b)=|a+i b| \equiv \sqrt{a^{2}+b^{2}}$。现在让 $|\theta|=1$ 和 $$ \theta \sum_{j=1}^{p} z_{j} \overline{w_{j}}=\left|\sum_{j=1}^{p} z_{j} \overline{w_{j }}\对| $$