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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|Invertibility of linear mappings as a generic property

Let $U$ be a finite-dimensional vector space with norm $|\cdot|$. It has been seen that the space $L(U)$ may be equipped with an induced norm which may also be denoted by $|\cdot|$ since there is no risk of confusion. The availability of a norm of $L(U)$ allows one to perform analysis on $L(U)$ so that a deeper understanding of $L(U)$ may be achieved.

As an illustration, in this subsection, we will characterize invertibility of linear mappings by using the norm.

Theorem 2.24 Let $U$ be a finite-dimensional normed space. An element $T \in$ $L(U)$ is invertible if and only if there is a constant $c>0$ such that $$|T(u)| \geq c|u|, \quad u \in U .$$ Proof Assume (2.6.9) is valid. Then it is clear that $N(T)={0}$. Hence $T$ is invertible. Conversely, assume that $T$ is invertible and $T^{-1} \in L(U)$ is its inverse. Then $1=|I|=\left|T^{-1} \circ T\right| \leq\left|T^{-1}\right||T|$ implies that the norm of an invertible mapping can never be zero. Thus, for any $u \in U$, we have $|u|=\left|\left(T^{-1} \circ T\right)(u)\right| \leq\left|T^{-1}\right||T(u)|$ or $|T(u)| \geq\left(\left|T^{-1}\right|\right)^{-1}|u|, u \in U$, which establishes (2.6.9).

We now show that invertibility is a generic property for elements in $L(U)$. Theorem 2.25 Let $U$ be a finite-dimensional normed space and $T \in L(U)$. (1) For any $\varepsilon>0$ there exists an invertible element $S \in L(U)$ such that $|S-T|<\varepsilon$. This property says that the subset of invertible mappings in $L(U)$ is dense in $L(U)$ with respect to the norm of $L(U)$.

## 线性代数作业代写linear algebra代考|Exponential of a linear mapping

Let $T \in L(U)$. For a positive integer $m$, we consider $T_{m} \in L(U)$ given by $$T_{m}=\sum_{k=0}^{m} \frac{1}{k !} T^{k},$$ with the understanding $T^{0}=I$. Therefore, for $l<m$, we have the estimate $$\left|T_{l}-T_{m}\right| \leq \sum_{k=l+1}^{m} \frac{|T|^{k}}{k !} .$$ In particular, $\left|T_{l}-T_{m}\right| \rightarrow 0$ as $l, m \rightarrow \infty$. Hence we see that the limit $$\lim {m \rightarrow \infty} \sum{k=0}^{m} \frac{1}{k !} T^{k}$$ is a well-defined element in $L(U)$ and is naturally denoted as $$\mathrm{e}^{T}=\sum_{k=0}^{\infty} \frac{1}{k !} T^{k}$$ and is called the exponential of $T \in L(U)$. Thus $\mathrm{e}^{0}=I .$ As in calculus, if $S, T \in L(U)$ are commutative, we can verify the formula $$\mathrm{e}^{S} \mathrm{e}^{T} \equiv \mathrm{e}^{S} \circ \mathrm{e}^{T}=\mathrm{e}^{S+T}$$ A special consequence of this simple property is that the exponential of any mapping $T \in L(U)$ is invertible. In fact, the relation (2.6.15) indicates that $$\left(\mathrm{e}^{T}\right)^{-1}=\mathrm{e}^{-T} .$$ More generally, with the notation $\Phi(t)=\mathrm{e}^{t T}(t \in \mathbb{R})$, we have (1) $\Phi(s) \Phi(t)=\Phi(s+t), s, t \in \mathbb{R}$, (2) $\Phi(0)=I$, and we say that $\Phi: \mathbb{R} \rightarrow L(U)$ defines a one-parameter group.

## 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