数学代写|线性代数代写linear algebra代考|MATHS1011

2023年1月2日

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数学代写|线性代数代写linear algebra代考|LINEAR INDEPENDENCE

We present now a topic which is critical in Linear Algebra. It will allow us to define the notions of basis and dimension in an upcoming section. First we present some definitions.

Definition 3.6 Let $v_1, \ldots, v_n$ be elements of a vector space $V$ and $a_1, \ldots, a_n$ be scalars. We call the expression $a_1 v_1+\cdots+a_n v_n$ a linear combination of the vectors $v_1, \ldots, v_n$. The scalars $a_1, \ldots, a_n$ are called the coefficients of the linear combination.

Example $3.22$ In $P_2$, the vector $1+2 x+x^2$ is a linear combination of $1+x, 1-$ $x^2, x+x^2$, since
$$1+2 x+x^2=(-1)(1+x)+(2)\left(1-x^2\right)+(3)\left(x+x^2\right) .$$
Definition 3.7 Let $v_1, \ldots, v_n$ be elements of a vector space $V$. We say these vectors are linearly dependent in $V$, if there exists scalars $a_1, \ldots, a_n$ not all zero such that $a_1 v_1+\cdots+a_n v_n=0$. In other words there is a non-trivial linear combination of $v_1, \ldots, v_n$ which equals 0 . If no such non-trivial linear combination exists, then we say that $v_1, \ldots, v_n$ are linearly independent in $V$. In other words $v_1, \ldots, v_n$ are linearly independent if whenever it should be the case that $a_1 v_1+\cdots+a_n v_n=0$, then it must be that $a_1=0, \ldots, a_n=0$.

The last restatement of linear independence gives us a method for checking linear independence: We assume that $a_1 v_1+\cdots+a_n v_n=0$ and show that this implies that $a_1=0, \ldots, a_n=0$. Some simple results immediately follow from this definition (which we leave as exercises):

1. Any collection of vectors which includes the zero vector is linearly dependent.
2. Any single vector $v \neq 0$ on its own is linearly independent.
3. Two vectors $u, v$ are linearly dependent iff one is a scalar multiple of the other (i.e. there exists a scalar $a$ such that $u-a v$ or $v-a u$ ).

数学代写|线性代数代写linear algebra代考|SPAN

We present in this section a special subspace which plays an important role in the theory of vector spaces as well as introduce the second property necessary for a basis.
Definition $3.8$ Given vectors $v_1, \ldots, v_n$ in a vector space $V$, the span of $v_1, \ldots, v_n$, written $\operatorname{span}\left(v_1, \ldots, v_n\right)$, is the set of all linear combinations of the vectors $v_1, \ldots, v_n$. In other words
$$\operatorname{span}\left(v_1, \ldots, v_n\right)=\left{a_1 v_1+\cdots+a_n v_n \mid a_1, \ldots, a_n \in \mathbb{R}\right} .$$
It is also called the subspace generated by $v_1, \ldots, v_n$ and is sometimes indicated by the notation $\left\langle v_1, \ldots, v_n\right\rangle$. The vectors $v_1, \ldots, v_n$ are called the generators.

We remark that one can define span for infinite sets of vectors as well, but this text does not require such treatment. It wouldn’t be fair to introduce such a nonintuitive object without giving some examples. Later in the section, we will give a method for uncovering a nice description of the span of a collection of vectors. For this reason, our examples at this point will be simple.
Example 3.30 Let $V=\mathbb{R}^3$. The span of $\hat{\imath}$ and $\hat{\jmath}$,
$$\operatorname{span}(\hat{\imath}, \hat{\jmath})={a \hat{\imath}+b \hat{\jmath} \mid a, b \in \mathbb{R}}={[a, b, 0] \mid a, b \in \mathbb{R}} .$$
Hence, this span describes all vectors in $\mathbb{R}^3$ which lie in the xy-plane, or we might just say that this span is the xy-plane. Similarly, the span of $\hat{\imath}, \hat{\jmath}$ and $\hat{k}$ will be all of $\mathbb{R}^3$.

Definition 3.9 Let $A \in M_{m n}$ with rows $r_1, \ldots, r_m \in F^n$ and columns $c_1, \ldots, c_n \in$ $\mathbb{R}^m$. Then

1. $\operatorname{span}\left(r_1, \ldots, r_m\right)$ is called the row space of $A$.
2. $\operatorname{span}\left(c_1, \ldots, c_n\right)$ is called the column space of $A$.
Now we prove an essential fact that the span of a collection of vectors is a subspace of $V$ (and more).

线性代数代考

数学代写|线性代数代写linear algebra代考|LINEAR INDEPENDENCE

$a_1 v_1+\cdots+a_n v_n$ 向量的线性组合 $v_1, \ldots, v_n$. 标量 $a_1, \ldots, a_n$ 称为线性组合的系数。

$1+2 x+x^2=(-1)(1+x)+(2)\left(1-x^2\right)+(3)$

1. 任何包含零向量的向量集合都是线性相关的。
2. 任意单个向量 $v \neq 0$ 本身是线性无关的。
3. 两个向量 $u, v$ 是线性相关的当且仅当一个是另一 个的标量倍数 (即存在一个标量 $a$ 这样 $u-a v$ 要 么 $v-a u)$.

数学代写|线性代数代写linear algebra代考|SPAN

loperatorname ${\operatorname{span}} \backslash l \operatorname{lt}\left(v_{-} 1, \backslash\right.$ dots, $\left.v_{-} n \backslash r i g h t\right)=\backslash l$ eft $\left{a_{-} 1 v_{-} 1+\backslash c c\right.$

$$\operatorname{span}(\hat{\imath}, \hat{\jmath})=a \hat{\imath}+b \hat{\jmath}|a, b \in \mathbb{R}=[a, b, 0]| a, b \in \mathbb{R} \text {. }$$

1. $\operatorname{span}\left(r_1, \ldots, r_m\right)$ 称为行空间 $A$.
2. $\operatorname{span}\left(c_1, \ldots, c_n\right)$ 称为列空间 $A$.
现在我们证明一个基本事实，即向量集合的跨度 是 $V$ (和更多)。

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MATLAB代写

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