# 线性代数网课代修|交换代数代写Commutative Algebra代考|MATH483

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• 数值分析
• 高等线性代数
• 矩阵论
• 优化理论
• 线性规划
• 逼近论

## 线性代数作业代写linear algebra代考|First Applications of Groebner Bases

If we combine Groebner bases with the division algorithm, we get the following ideal membership algorithm: given an ideal $I=\left\langle f_{1}, \ldots, f_{s}\right\rangle$, we can decide whether a given polynomial $f$ lies in $I$ as follows. First, using an algorithm similar to Theorem 2 of $\S 7$, find a Groebner basis $G=\left{g_{1}, \ldots, g_{t}\right}$ for $I$. Then Corollary 2 of $\S 6$ implies that
$f \in I$ if and only if $\bar{f}^{G}=0$.
Example 1. Let $I=\left\langle f_{1}, f_{2}\right\rangle=\left\langle x z-y^{2}, x^{3}-z^{2}\right\rangle \in \mathbb{C}[x, y, z]$, and use the grlex order. Let $f=-4 x^{2} y^{2} z^{2}+y^{6}+3 z^{5}$. We want to know if $f \in I$.

The generating set given is not a Groebner basis of $I$ because $\operatorname{LT}(I)$ also contains polynomials such as $\operatorname{LT}\left(S\left(f_{1}, f_{2}\right)\right)=\operatorname{LT}\left(-x^{2} y^{2}+z^{3}\right)=-x^{2} y^{2}$ that are not in the ideal $\left\langle\operatorname{Lr}\left(f_{1}\right), \operatorname{LT}\left(f_{2}\right)\right\rangle=\left\langle x z, x^{3}\right\rangle$. Hence, we begin by computing a Groebner basis for $I$. Using a computer algebra system, we find a Groebner basis
$$G=\left(f_{1}, f_{2}, f_{3}, f_{4}, f_{5}\right)=\left(x z-y^{2}, x^{3}-z^{2}, x^{2} y^{2}-z^{3}, x y^{4}-z^{4}, y^{6}-z^{5}\right) .$$
Note that this is a reduced Groebner basis.
We may now test polynomials for membership in $I$. For example, dividing $f$ above by $G$, we find
$$f=\left(-4 x y^{2} z-4 y^{4}\right) \cdot f_{1}+0 \cdot f_{2}+0 \cdot f_{3}+0 \cdot f_{4}+(-3) \cdot f_{5}+0 .$$
Since the remainder is zero, we have $f \in I$.
For another example, consider $f=x y-5 z^{2}+x$. Even without completely computing the remainder on division by $G$, we can see from the form of the elements in $G$ that $f \notin I$. The reason is that $\operatorname{LT}(f)=x y$ is clearly not in the ideal $\langle\operatorname{LT}(G)\rangle=\left\langle x z, x^{3}, x^{2} y^{2}, x y^{4}, y^{6}\right\rangle$. Hence, $\bar{f}^{G} \neq 0$, so that $f \notin I$.

This last observation illustrates the way the properties of an ideal are revealed by the form of the elements of a Groebner basis.

## 线性代数作业代写linear algebra代考|The Problem of Solving Polynomial Equations

Next, we will investigate how the Groebner basis technique can be applied to solve systems of polynomial equations in several variables. Let us begin by looking at some specific examplès.
Example 2. Consider the equations
\begin{aligned} x^{2}+y^{2}+z^{2} &=1 \ x^{2}+z^{2} &=y \ x &=z \end{aligned}
in $\mathbb{C}^{3}$. These equations determine $I=\left\langle x^{2}+y^{2}+z^{2}-1, x^{2}+z^{2}-y, x-z\right\rangle \subset$ $\mathbb{C}[x, y, z]$, and we want to find all points in $\mathbf{V}(I)$. Proposition 9 of $\S 5$ implies that we can compute $\mathbf{V}(I)$ using any basis of $I$. So let us see what happens when we use a Groebner basis.

Though we have no compelling reason as of yet to do so, we will compute a Groebner basis on $I$ with respect to the lex order. The basis is
\begin{aligned} &g_{1}=x-z, \ &g_{2}=-y+2 z^{2}, \ &g_{3}=z^{4}+(1 / 2) z^{2}-1 / 4 . \end{aligned}
If we examine these polynomials closely, we find something remarkable. First, the polynomial $g_{3}$ depends on $z$ alone, and its roots can be found by first using the quadratic formula to solve for $z^{2}$, then, taking square roots,
$$z=\pm \frac{1}{2} \sqrt{\pm \sqrt{5}-1}$$
This gives us four values of $z$. Next, when these values of $z$ are substituted into the equations $g_{2}=0$ and $g_{1}=0$, those two equations can be solved uniquely for $y$ and $x$, respectively. Thus, there are four solutions altogether of $g_{1}=g_{2}=g_{3}=0$, two real and two complex. Since $\mathbf{V}(I)=\mathbf{V}\left(g_{1}, g_{2}, g_{3}\right)$ by Proposition 9 of $\S 5$, we have found all solutions of the original equations (1).

## 线性代数作业代写linear algebra代考|First Applications of Groebner Bases

$f \in I$ 当且仅当 $\bar{f}^{G}=0$.

$$G=\left(f_{1}, f_{2}, f_{3}, f_{4}, f_{5}\right)=\left(x z-y^{2}, x^{3}-z^{2}, x^{2} y^{2}-z^{3}, x y^{4}-z^{4}, y^{6}-z^{5}\right) .$$

$$f=\left(-4 x y^{2} z-4 y^{4}\right) \cdot f_{1}+0 \cdot f_{2}+0 \cdot f_{3}+0 \cdot f_{4}+(-3) \cdot f_{5}+0 .$$

$\langle\mathrm{LT}(G)\rangle=\left\langle x z, x^{3}, x^{2} y^{2}, x y^{4}, y^{6}\right\rangle$. 因此， $\bar{f}^{G} \neq 0$ ，以便 $f \notin I$.

## 线性代数作业代写linear algebra代考|The Problem of Solving Polynomial Equations

$$x^{2}+y^{2}+z^{2}=1 x^{2}+z^{2} \quad=y x=z$$

$$g_{1}=x-z, \quad g_{2}=-y+2 z^{2}, g_{3}=z^{4}+(1 / 2) z^{2}-1 / 4 .$$

$$z=\pm \frac{1}{2} \sqrt{\pm \sqrt{5}-1}$$

# 计量经济学代写

## 在这种情况下，如何学好线性代数？如何保证线性代数能获得高分呢？

1.1 mark on book

【重点的误解】划重点不是书上粗体，更不是每个定义，线代概念这么多，很多朋友强迫症似的把每个定义整整齐齐用荧光笔标出来，然后整本书都是重点，那期末怎么复习呀。我认为需要标出的重点为

A. 不懂，或是生涩，或是不熟悉的部分。这点很重要，有的定义浅显，但证明方法很奇怪。我会将晦涩的定义，证明方法标出。在看书时，所有例题将答案遮住，自己做，卡住了就说明不熟悉这个例题的方法，也标出。

B. 老师课上总结或强调的部分。这个没啥好讲的，跟着老师走就对了

C. 你自己做题过程中，发现模糊的知识点

1.2 take note

1.3 understand the relation between definitions