# 计算机代写|机器学习代写machine learning代考|COMP3670

## 计算机代写|机器学习代写machine learning代考|Normed Vector Spaces

In order to define how close two vectors or two matrices are, and in order to define the convergence of sequences of vectors or matrices, we can use the notion of a norm. Recall that $\mathbb{R}_{+}={x \in \mathbb{R} \mid x \geq 0}$. Also recall that if $z=a+i b \in \mathbb{C}$ is a complex number, with $a, b \in \mathbb{R}$, then $\bar{z}=a-i b$ and $|z|=\sqrt{z \bar{z}}=\sqrt{a^2+b^2}(|z|$ is the modulus of $z$ ).

Definition 9.1. Let $E$ be a vector space over a field $K$, where $K$ is either the field $\mathbb{R}$ of reals, or the field $\mathbb{C}$ of complex numbers. A norm on $E$ is a function ||$: E \rightarrow \mathbb{R}_{+}$, assigning a nonnegative real number $|u|$ to any vector $u \in E$, and satisfying the following conditions for all $x, y \in E$ and $\lambda \in K$ :
(N1) $|x| \geq 0$, and $|x|=0$ iff $x=0$.
(positivity)
(N2) $|\lambda x|=|\lambda||x|$.
(homogeneity (or scaling))
(N3) $|x+y| \leq|x|+|y|$.
(triangle inequality)
A vector space $E$ together with a norm || is called a normed vector space.
By (N2), setting $\lambda=-1$, we obtain
$$|-x|=|(-1) x|=|-1||x|=|x| ;$$
that is, $|-x|=|x|$. From (N3), we have
$$|x|=|x-y+y| \leq|x-y|+|y|,$$
which implies that
$$|x|-|y| \leq|x-y| .$$
By exchanging $x$ and $y$ and using the fact that by (N2),
$$|y-x|=|-(x-y)|=|x-y|,$$

we also have
$$|y|-|x| \leq|x-y| .$$
Therefore,
$$||x|-|y|| \leq|x-y|, \quad \text { for all } x, y \in E .$$
Observe that setting $\lambda=0$ in (N2), we deduce that $|0|=0$ without assuming (N1). Then by setting $y=0$ in $(*)$, we obtain
$$||x|| \leq|x|, \quad \text { for all } x \in E \text {. }$$

## 计算机代写|机器学习代写machine learning代考|Matrix Norms

For simplicity of exposition, we will consider the vector spaces $M_n(\mathbb{R})$ and $M_n(\mathbb{C})$ of square $n \times n$ matrices. Most results also hold for the spaces $\mathrm{M}{m, n}(\mathbb{R})$ and $\mathrm{M}{m, n}(\mathbb{C})$ of rectangular $m \times n$ matrices. Since $n \times n$ matrices can be multiplied, the idea behind matrix norms is that they should behave “well” with respect to matrix multiplication.

Definition 9.3. A matrix norm || on the space of square $n \times n$ matrices in $\mathrm{M}n(K)$, with $K=\mathbb{R}$ or $K=\mathbb{C}$, is a norm on the vector space $\mathrm{M}_n(K)$, with the additional property called submultiplicativity that $$|A B| \leq|A||B|$$ fơ all $A, B \in \mathrm{M}_n(K)$. A normón mátricés satisfying thè aboové property is often calleed a submultiplicative matrix norm. Since $I^2=I$, from $|I|=\left|I^2\right| \leq|I|^2$, we get $|I| \geq 1$, for every matrix norm. Before giving examples of matrix norms, we need to review some basic definitions about matrices. Given any matrix $A=\left(a{i j}\right) \in \mathrm{M}{m, n}(\mathbb{C})$, the conjugate $\bar{A}$ of $A$ is the matrix such that $$\bar{A}{i j}=\overline{a_{i j}}, \quad 1 \leq i \leq m, 1 \leq j \leq n .$$
The transpose of $A$ is the $n \times m$ matrix $A^{\top}$ such that
$$A_{i j}^{\top}=a_{j i}, \quad 1 \leq i \leq m, 1 \leq j \leq n .$$
The adjoint of $A$ is the $n \times m$ matrix $A^$ such that $$A^=\overline{\left(A^{\top}\right)}=(\bar{A})^{\top} .$$
When $A$ is a real matrix, $A^=A^{\top}$. A matrix $A \in \mathrm{M}_n(\mathbb{C})$ is Hermitian if $$A^=A .$$
If $A$ is a real matrix $\left(A \subset \mathrm{M}_n(\mathbb{R})\right)$, we say that $A$ is symmetric if
$$A^{\top}=A \text {. }$$
A matrix $A \in \mathrm{M}_n(\mathbb{C})$ is normal if
$$A A^-A^ A$$
and if $A$ is a real matrix, it is normal if
$$A A^{\top}=A^{\top} A .$$
A matrix $U \in \mathrm{M}_n(\mathbb{C})$ is unitary if
$$U U^=U^ U=I$$

# 机器学习代考

## 计算机代写|机器学习代写machine learning代考|Normed Vector Spaces

(积极性)
(N2) $|\lambda x|=|\lambda||x|$.
(同质性 (或缩放) )
(N3) $|x+y| \leq|x|+|y|$.
(三角不等式)

$$|-x|=|(-1) x|=|-1||x|=|x| ;$$

$$|x|=|x-y+y| \leq|x-y|+|y|,$$

$$|x|-|y| \leq|x-y| .$$

$$|y-x|=|-(x-y)|=|x-y| \text {, }$$

$$|y|-|x| \leq|x-y| .$$

$$|| x|-| y|| \leq|x-y|, \quad \text { for all } x, y \in E .$$

$$|x| \leq|x|, \quad \text { for all } x \in E$$

## 计算机代写|机器学习代写machine learning代考|Matrix Norms

$$A_{i j}^{\top}=a_{j i}, \quad 1 \leq i \leq m, 1 \leq j \leq n .$$

$$A^{=} \overline{\left(A^{\top}\right)}=(\bar{A})^{\top} .$$

$$A^{=} A \text {. }$$

$$A^{\top}=A .$$

$$A A^{-} A^A$$

$$A A^{\top}=A^{\top} A .$$

$$U U^{=} U^U=I$$

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