线性代数 linear algebra essay代写

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线性代数 linear algebra 的历史

Linear algebra was initiated in principle by the Persian mathematician Al-Khwârizmî who was inspired by Indian mathematical texts and who completed the work of the Greek school, which continued to develop for centuries1. It was taken up by René Descartes who posed geometric problems, such as the determination of the intersection of two lines, in terms of a linear equation, thus establishing a bridge between two previously separate branches of mathematics: algebra and geometry. Although he did not define the basic notion of linear algebra, that of vector space, he already used it successfully, and this natural use of the linear aspects of the manipulated equations will remain used in an ad hoc manner, based essentially on the underlying geometric ideas. After this discovery, progress in linear algebra was limited to one-off studies such as the definition and analysis of the first properties of determinants by Jean d’Alembert.

线性代数 linear algebra 课后作业代写

In a vector space, the span of any subset is a subspace.

Proof. Call the subset $S$. If $S$ is empty then by definition its span is the trivial subspace. If $S$ is not empty then by Lemma $2.9$ we need only check that the span $[S]$ is closed under linear combinations. For a pair of vectors from that span, $\vec{v}=c_{1} \vec{s}{1}+\cdots+c{n} \vec{s}{n}$ and $\vec{w}=c{n+1} \vec{s}{n+1}+\cdots+c{m} \vec{s}{m}$, a linear combination \begin{aligned} p \cdot\left(c{1} \vec{s}{1}+\cdots+c{n} \vec{s}{n}\right)+r & \cdot\left(c{n+1} \vec{s}{n+1}+\cdots+c{m} \vec{s}{m}\right) \ =& p c{1} \vec{s}{1}+\cdots+p c{n} \vec{s}{n}+r c{n+1} \vec{s}{n+1}+\cdots+r c{m} \vec{s}{m} \end{aligned} ( $p, r$ scalars) is a linear combination of elements of $S$ and so is in $[S]$ (possibly some of the $\vec{s}{i}$ ‘s forming $\vec{v}$ equal some of the $\vec{s}_{j}$ ‘s from $\vec{w}$, but it does not matter).

The converse of the lemma holds: any subspace is the span of some set, because a subspace is obviously the span of the set of its members. Thus a subset of a vector space is a subspace if and only if it is a span. This fits the intuition that a good way to think of a vector space is as a collection in which linear combinations are sensible.

Taken together, Lemma $2.9$ and Lemma 2.15 show that the span of a subset $S$ of a vector space is the smallest subspace containing all the members of $S$.

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