# 线性代数代写 Linear algebra代考2023

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## 线性代数代写Linear algebra

### 线形图Linear map代写

• Subspaces, span, and basis子空间、跨度和基础
• System of linear equations线性方程组

## 线性代数的历史

The procedure for solving simultaneous linear equations (using counting bars) is known today as Gaussian elimination and appears in Chapter 8 of the Mathematical Texts of Ancient China: Nine Chapters on the Art of Mathematics: Rectangular Matrices. Its use is illustrated in eighteen problems with two to five equations.
In Europe, systems of linear equations originated with the introduction of coordinates into geometry by René Descartes in 1637. Indeed, in this new geometry, known today as Cartesian geometry, lines and planes are represented by linear equations and the calculation of their intersections is equivalent to solving a system of linear equations.
The first systematic approach to solving linear systems was the use of determinants, first considered by Leibniz in 1693, and in 1750 Gabriel Cramer used it to give explicit solutions to linear systems, now known as Cramer’s rules. Later, Gauss further described the elimination method, which was originally classified as an advancement in geodesy.

## 线性代数相关课后作业代写

Let $V$ be the vector space of polynomials of degree at most five with real coefficients. Define a linear map
$$T: V \rightarrow \mathbb{R}^3, \quad T(p)=(p(1), p(2), p(3)) .$$
That is, the coordinates of the vector $T(p)$ are the values of $p$ at 1,2 , and 3 .
a) Find a basis of the null space of $T$.
b) Find a basis of the range of $T$.

(a) The null space of $T$ consists of all polynomials $p$ in $V$ such that $T(p)=(0,0,0)$. This is equivalent to $p(1)=p(2)=p(3)=0$. Thus, the null space of $T$ is the set of all polynomials of degree at most $2$ that have $1,2,$ and $3$ as roots. A basis for this null space is given by ${ (x-1)(x-2), (x-1)(x-3), (x-2)(x-3) }$.

To see why this is a basis, note that any polynomial in the null space can be written as $a(x-1)(x-2) + b(x-1)(x-3) + c(x-2)(x-3)$ for some constants $a,b,c\in\mathbb{R}$. Conversely, any such polynomial is in the null space since it has $1,2,$ and $3$ as roots.

(b) The range of $T$ is a subspace of $\mathbb{R}^3$. To find a basis for the range, we need to find linearly independent vectors in the range that span the entire range. The vectors in the range are of the form $(p(1), p(2), p(3))$ for some polynomial $p$ in $V$.

Consider the polynomials $p_1(x)=1, p_2(x)=x, p_3(x)=x^2$. The corresponding vectors in the range of $T$ are $(1,1,1), (1,2,4),$ and $(1,3,9)$, respectively. We claim that these three vectors form a basis for the range of $T$.

To see why this is true, note that any vector $(a,b,c)$ in the range of $T$ can be written as $(a,b,c) = ap_1(1,2,3) + bp_2(1,2,3) + cp_3(1,2,3)$. Conversely, any such linear combination is in the range of $T$ since $T$ is linear. To show that $p_1, p_2,$ and $p_3$ are linearly independent, consider the equation $ap_1(x) + bp_2(x) + cp_3(x) = 0$ for all $x\in\mathbb{R}$. This implies that $a+b+c=0$, $a+2b+4c=0$, and $a+3b+9c=0$. Solving this system of equations gives $a=b=c=0$, which shows that $p_1, p_2,$ and $p_3$ are linearly independent. Therefore, $(1,1,1), (1,2,4),$ and $(1,3,9)$ form a basis for the range of $T$.

# 线性代数作业代写linear algebra代考

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## 代写线性代数作业代写linear algebra

### 线性映射Linear map代写

• 线性子空间Linear subspace
• 线性生成空间Linear span
• 矩阵 Matrix (mathematics)

## 线性代数的历史

The procedure (using counting rods) for solving simultaneous linear equations now called Gaussian elimination appears in the ancient Chinese mathematical text Chapter Eight: Rectangular Arrays of The Nine Chapters on the Mathematical Art. Its use is illustrated in eighteen problems, with two to five equations..

## 线性代数课后作业代写

This function is one-to-one because if
$$\operatorname{Rep}{B}\left(u{1} \vec{\beta}{1}+\cdots+u{n} \vec{\beta}{n}\right)=\operatorname{Rep}{B}\left(v_{1} \vec{\beta}{1}+\cdots+v{n} \vec{\beta}{n}\right)$$ then $$\left(\begin{array}{c} u{1} \ \vdots \ u_{n} \end{array}\right)=\left(\begin{array}{c} v_{1} \ \vdots \ v_{n} \end{array}\right)$$
and so $u_{1}=v_{1}, \ldots, u_{n}=v_{n}$, and therefore the original arguments $u_{1} \vec{\beta}{1}+\cdots+$ $u{n} \vec{\beta}{n}$ and $v{1} \vec{\beta}{1}+\cdots+v{n} \vec{\beta}{n}$ are equal. This function is onto; any $n$-tall vector $$\vec{w}=\left(\begin{array}{c} w{1} \ \vdots \ w_{n} \end{array}\right)$$