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

We now give another method for obtaining matrix norms using subordinate norms. First we need a proposition that shows that in a finite-dimensional space, the linear map induced by a matrix is bounded, and thus continuous.

Proposition 9.8. For every norm || on $\mathbb{C}^n$ (or $\mathbb{R}^n$ ), for every matrix $A \in \mathrm{M}n(\mathbb{C}$ ) (or $A \in \mathrm{M}_n(\mathbb{R})$ ), there is a real constant $C_A \geq 0$, such that $$|A u| \leq C_A|u|,$$ for every vector $u \in \mathbb{C}^n$ (or $u \in \mathbb{R}^n$ if $A$ is real). Proof. For every basis $\left(e_1, \ldots, e_n\right)$ of $\mathbb{C}^n$ (or $\left.\mathbb{R}^n\right)$, for every vector $u=u_1 e_1+\cdots+u_n e_n$, we have \begin{aligned} |A u| &=\left|u_1 A\left(e_1\right)+\cdots+u_n A\left(e_n\right)\right| \ & \leq\left|u_1\right|\left|A\left(e_1\right)\right|+\cdots+\left|u_n\right|\left|A\left(e_n\right)\right| \ & \leq C_1\left(\left|u_1\right|+\cdots+\left|u_n\right|\right)=C_1|u|_1, \end{aligned} where $C_1=\max {1 \leq i \leq n}\left|A\left(e_i\right)\right|$. By Theorem 9.5, the norms || and ||$_1$ are equivalent, so there is some constant $C_2>0$ so that $|u|_1 \leq C_2|u|$ for all $u$, which implies that
$$|A u| \leq C_A|u|,$$
where $C_A=C_1 C_2$.
Proposition $9.8$ says that every linear map on a finite-dimensional space is bounded. This implies that every linear map on a finite-dimensional space is continuous. Actually, it is not hard to show that a linear map on a normed vector space $E$ is bounded iff it is continuous, regardless of the dimension of $E$.
Proposition $9.8$ implies that for every matrix $A \in \mathrm{M}n(\mathbb{C})$ (or $A \in \mathrm{M}_n(\mathbb{R})$ ), $$\sup {\substack{x \in \mathbb{C}^n \ x \neq 0}} \frac{|A x|}{|x|} \leq C_A .$$

计算机代写|机器学习代写machine learning代考|Inequalities Involving Subordinate Norms

In this section we discuss two technical inequalities which will be needed for certain proofs in the last three sections of this chapter. First we prove a proposition which will be needed when we deal with the condition number of a matrix.
Proposition 9.11. Let || be any matrix norm, and let $B \in \mathrm{M}_n(\mathbb{C})$ such that $|B|<1$. (1) If || is a subordinate matrix norm, then the matrix $I+B$ is invertible and $$\left|(I+B)^{-1}\right| \leq \frac{1}{1-|B|} .$$ (2) If a matrix of the form $I+B$ is singular, then $|B|>1$ for every matrix norm (not necessarily subordinate).
Proof. (1) Observe that $(I+B) u=0$ implies $B u=-u$, so
$$|u|=|B u| \text {. }$$
Recall that
$$|B u| \leq|B||u|$$
for every subordinate norm. Since $|B|<1$, if $u \neq 0$, then
$$|B u|<|u| \text {, }$$
which contradicts $|u|=|B u|$. Therefore, we must have $u=0$, which proves that $I+B$ is injective, and thus bijective, i.e., invertible. Then we have
$$(I+B)^{-1}+B(I+B)^{-1}=(I+B)(I+B)^{-1}=I,$$

so we get
$$(I+B)^{-1}=I-B(I+B)^{-1}$$
which yields
and finally,
$$\left|(I+B)^{-1}\right| \leq 1+|B|\left|(I+B)^{-1}\right|$$
$$\left|(I+B)^{-1}\right| \leq \frac{1}{1-|B|} .$$

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计算机代写|机器学习代写machine learning代考|Inequalities Involving Subordinate Norms

$$|u|=|B u| \text {. }$$

$$|B u| \leq|B||u|$$

$$|B u|<|u|,$$

$$(I+B)^{-1}+B(I+B)^{-1}=(I+B)(I+B)^{-1}=I,$$

$$(I+B)^{-1}=I-B(I+B)^{-1}$$

，最后，
$$\left|(I+B)^{-1}\right| \leq 1+|B|\left|(I+B)^{-1}\right|$$
$$\left|(I+B)^{-1}\right| \leq \frac{1}{1-|B|} .$$

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