# 终其一生，我们只不过在寻找自己

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MSBD5004 Mathematical Methods for Data Analysis Homework 1
20635526 LiYuan

# ( a )

By norm definition, we can see:

• Zero vector:
$\forall x \in V, |\boldsymbol{x}|_{\infty}=\max _{1 \leq i \leq n}\left|x_{i}\right|\geq0$ .
And only when $X=0$ then $|x|_{\infty}=0$.
• positive homogeneity:
$\forall x \in V$ and $\alpha \in \mathbb{R}$, $|\boldsymbol{\alpha x}|_{\infty}=\max _{1 \leq i \leq n}\left|\alpha x_{i}\right|=\left|\alpha \right||\boldsymbol{ x}|_{\infty}$.
• triangle inequality:
$\forall x, y \in V$, $|\boldsymbol{x+y}|_{\infty}= \max _{1 \leq i \leq n}\left|x_{i}+y_i\right| \leq \max _{1 \leq i \leq n}\left|x_{i}\right|+\max _{1 \leq j \leq n}\left|y_{j}\right| =|\boldsymbol{x}|_{\infty}+|\boldsymbol{y}|_{\infty}$

So, $|\boldsymbol{x}|_{\infty}=\max _{1 \leq i \leq n}\left|x_{i}\right|$ is indeed a norm on $\mathbb{R}^{n}$.◾

## ( b )

• Proof.

Let $|x|_{p}=\left(\sum_{i}\left|x_{i}\right|^{p}\right)^{\frac{1}{p}} .$ Observe that for all $x$ we have that

and that

Thus,

but then,

Thus

however, $\lim _{p \rightarrow \infty} n^{\frac{1}{p}}=1,$ so

or rather, that $|x|_{\infty}=\lim _{p \rightarrow \infty}|x|_{p}=\max _{i}\left|x_{i}\right|$.

# ( c )

• Proof.
$|\boldsymbol{x}|_{\infty}=\max _{1 \leq i \leq n}\left|x_{i}\right|\leq\sum_{i=1}^n\left|x_{i}\right|=|\boldsymbol{x}|_{1}$
$|\boldsymbol{x}|_{1}=\sum_{i=1}^n\left|x_{i}\right|\leq n*\max _{1 \leq i \leq n}\left|x_{i}\right|=n|\boldsymbol{x}|_{\infty}$

So that $|\boldsymbol{x}|_{\infty} \leq|\boldsymbol{x}|_{1} \leq n|\boldsymbol{x}|_{\infty}, \quad \forall \boldsymbol{x} \in \mathbb{R}^{n}$.◾

# Q2

## ( a )

• Proof:

Running through all $x \neq 0$ is equivalent to running through all $y:=\frac{x}{|x|_2}$ with $|y|_2=1$

## ( b )

By norm definition, we can see:

• Zero vector:
$\forall A \in V^{m \times n}$,Because $|\boldsymbol{x}|_2=1$, so not all $x_j=0$, So $|\boldsymbol{x}|_2 \geq 0$.
And only when all $a_{i,j}=0$,then $|\boldsymbol{A}|_{2}=0$, So $|\boldsymbol{A}|_{2}=0\Leftrightarrow \mathbf{A}=0$
• positive homogeneity:
$\forall A \in V$ and $\alpha \in R$, $\|\alpha\boldsymbol{A}\|_{2}=\sup _{\boldsymbol{x} \in \mathbb{R}^{n}, \boldsymbol{x} \neq \mathbf{0}} \frac{\|\alpha\boldsymbol{A} \boldsymbol{x}\|_{2}}{\|\boldsymbol{x}\|_{2}}=\alpha\sup _{\boldsymbol{x} \in \mathbb{R}^{n}, \boldsymbol{x} \neq \mathbf{0}} \frac{\|\boldsymbol{A} \boldsymbol{x}\|_{2}}{\|\boldsymbol{x}\|_{2}}=\alpha\|\boldsymbol{A}\|_{2}$
• triangle inequality:
$\forall A,B \in V$,So, $|\cdot|_{2}$is indeed a norm on $\mathbb{R}^{m\times n}$.◾

## ( c )

From the definition we have $\forall \boldsymbol{A} \in \mathbb{R}^{m \times n},\forall\boldsymbol{|x|}\neq \mathbf{0}$,

## ( d )

$\forall \boldsymbol{A} \in \mathbb{R}^{m \times n},\forall\boldsymbol{B} \in \mathbb{R}^{m \times n}$,

by ( c ) above. Thus

# Q3

• Proof.

We’re basically after:

One should notice that $\frac{\mathrm{d} \left | x \right | }{\mathrm{d} x} = \operatorname{sign} \left( x \right)$ (Being more rigorous would say it is a Sub Gradient of the non smooth $L_1$ Norm function).
Hence, deriving the sum above yields $\sum_{i = 1}^{m} \operatorname{sign} \left( {a}_{i} - b \right)$.
This equals to zero only when the number of positive items equals the number of negative which happens when $b = \operatorname{median} \left\{ {a}_{1}, {a}_{2}, \cdots, {a}_{m} \right\}$.

# Q4

## ( a )

Because $z_j$ is average of some $x_i$, so $z_j$ located at center of these $x_i$ , so $z_j$ is in the entries. So $z_j$ also are nonnegative.◾

## ( b )

If each vector sums to one, $\mathbf{1}^{T} x_{k}=1$ for all $k$ then the same is true for the average:

$\mathbf{1}^{T} z_{j}=\frac{1}{\left|G_{j}\right|} \sum_{k \in G_{j}} \mathbf{1}^{T} x_{k}=\frac{\left|G_{j}\right|}{\left|G_{j}\right|}=1$

## ( c )

The $i$th entry of group represenative $z_j$is the fraction of the vectors in group $j$that have $i$th entry one. If it is equal to one, all vectors in the group have $i$thentry one. If it is close to one, most vectors in the group haveith entry one. If it zero, no vectors in the group haveith entry one.

# Q5

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