2022年10月10日

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
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Via the inner product, one can relate with a linear map another linear map (called the adjoint). On a vector space over the reals, the adjoint of multiplication with a matrix $A$ corresponds to multiplication with the transpose $A^T$ of the matrix $A$. Over the complex numbers, it also involves taking a complex conjugate. We now provide you with the definition.

Let $\left(V,\langle\cdot, \cdot\rangle_V\right)$ and $\left(W,\langle\cdot, \cdot\rangle_W\right)$ be inner product spaces, and let $T: V \rightarrow W$ be linear. We call a map $T^{\star}: W \rightarrow V$ the adjoint of $T$ if
$$\langle T(\mathbf{v}), \mathbf{w}\rangle_W=\left\langle\mathbf{v}, T^{\star}(\mathbf{w})\right\rangle_V \text { for all } \mathbf{v} \in V, \mathbf{w} \in W .$$
Notice that the adjoint is unique. Indeed if $S$ is another adjoint for $T$, we get that $\left\langle\mathbf{v}, T^{\star}(\mathbf{w})\right\rangle_V=\langle\mathbf{v}, S(\mathbf{w})\rangle_V$ for all $\mathbf{v}, \mathbf{w}$. Choosing $\mathbf{v}=T^{\star}(\mathbf{w})-S(\mathbf{w})$ yields
$\left\langle T^{\star}(\mathbf{w})-S(\mathbf{w}), T^{\star}(\mathbf{w})-S(\mathbf{w})\right\rangle_V=0$,
and thus $T^{\star}(\mathbf{w})-S(\mathbf{w})=0$. As this holds for all $\mathbf{w}$, we must have that $T^{\star}=S$.
Lemma 5.3.1 If $T: V \rightarrow W$ is an isometry, then $T^{\star} T=i d_V$.
Proof. Since $T$ is an isometry, we have that
$$\langle T(\mathbf{v}), T(\hat{\mathbf{v}})\rangle_W=\langle\mathbf{v}, \hat{\mathbf{v}}\rangle_V \text { for all } \mathbf{v}, \hat{\mathbf{v}} \in V .$$
But then, we get that
$$\left\langle T^{\star} T(\mathbf{v}), \hat{\mathbf{v}}\right\rangle_W=\langle\mathbf{v}, \hat{\mathbf{v}}\rangle_V \text { for all } \mathbf{v}, \hat{\mathbf{v}} \in V,$$
or equivalently,
$$\left\langle T^{\star} T(\mathbf{v})-\mathbf{v}, \hat{\mathbf{v}}\right\rangle_W=\langle\mathbf{v}, \hat{\mathbf{v}}\rangle_V \text { for all } \mathbf{v}, \hat{\mathbf{v}} \in V .$$
Letting $\dot{\mathbf{v}}=T^{\star} T(\mathbf{v})-\mathbf{v}$, this yields $T^{\star} T(\mathbf{v})-\mathbf{v}=0$ for all $\mathbf{v} \in V$. Thus $T^{\star} T=i d_V$.

## 数学代写|高等线性代数代写Advanced Linear Algebra代考|Unitary matrices, QR, and Schur triangularization

Unitary transformations are ones where a pair of vectors is mapped to a new pair of vectors without changing their lengths or the angle between them. Thus, one can think of a unitary transformation as viewing the vector space from a different viewpoint. Using unitary transformations (represented by unitary matrices) can be used to put general transformations in a simpler form. These simpler forms give rise to $\mathrm{QR}$ and Schur triangular decompositions.
Let $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. We call a matrix $A \in \mathbb{F}^{n \times m}$ an isometry if $A^* A=I_m$.
Notice that necessarily we need to have that $n \geq m$. The equation $A^* A$ can also be interpreted as that the columns of $A$ are orthonormal. When $A \in \mathbb{F}^{n \times n}$ is square, then automatically $A^* A=I_n$ implies that $A A^*=I_n$. Such a matrix is called unitary. Thus a square isometry is a unitary. From the Gram-Schmidt process we can deduce the following.

Theorem 5.4.1 (QR factorization) Let $A \in \mathbb{F}^{n \times m}$ with $n \geq m$. Then there exists an isometry $Q \in \mathbb{F}^{n \times m}$ and an upper triangular matrix $R \in \mathbb{F}^{m \times m}$ with nonnegative entries on the diagonal, so that
$$A=Q R .$$
If $A$ has rank equal to $m$, then the diagonal entries of $R$ are positive, and $R$ is invertible. If $n=m$, then $Q$ is unitary.

Proof. First we consider the case when $\operatorname{rank} A=m$. Let $\mathbf{v}1, \ldots, \mathbf{v}_m$ denote the columns of $A$, and let $\mathbf{z}_1, \ldots, \mathbf{z}_m$ denote the resulting vectors when we apply the Gram-Schmidt process to $\mathbf{v}_1, \ldots, \mathbf{v}_m$ as in Theorem 5.2.3. Let now $Q$ be the matrix with columns $\frac{\mathbf{z}_1}{\left|\mathbf{z}_1\right|}, \ldots, \frac{\mathbf{z}_m}{\left|\mathbf{z}_m\right|}$. Then $Q^* Q=I_m$ as the columns of $Q$ are orthonormal. Moreover, we have that $$\mathbf{v}_k=\left|\mathbf{z}_k\right| \frac{\mathbf{z}_k}{\left|\mathbf{z}_k\right|}+\sum{j=1}^{k-1} r_{k j} \mathbf{z}j,$$ for some $r{k j} \in \mathbb{F}, k>j$. Putting $r_{k k}=\left|\mathbf{z}k\right|$, and $r{k j}=0, k<j$, and letting $R=\left(r_{k j}\right)_{k, j=1}^m$, we get the desired upper triangular matrix $R$ yielding $A=Q R$.

# 高等线性代数代考

$$\langle T(\mathbf{v}), \mathbf{w}\rangle_W=\left\langle\mathbf{v}, T^{\star}(\mathbf{w})\right\rangle_V \text { for all } \mathbf{v} \in V, \mathbf{w} \in W .$$
，我们称地图$T^{\star}: W \rightarrow V$为$T$的伴随符。注意，伴随符是唯一的。实际上，如果$S$是$T$的另一个伴随词，我们就会得到所有$\mathbf{v}, \mathbf{w}$的$\left\langle\mathbf{v}, T^{\star}(\mathbf{w})\right\rangle_V=\langle\mathbf{v}, S(\mathbf{w})\rangle_V$。选择$\mathbf{v}=T^{\star}(\mathbf{w})-S(\mathbf{w})$会产生
$\left\langle T^{\star}(\mathbf{w})-S(\mathbf{w}), T^{\star}(\mathbf{w})-S(\mathbf{w})\right\rangle_V=0$，
，因此是$T^{\star}(\mathbf{w})-S(\mathbf{w})=0$。因为这对所有$\mathbf{w}$都成立，我们必须有那个$T^{\star}=S$ .

$$\langle T(\mathbf{v}), T(\hat{\mathbf{v}})\rangle_W=\langle\mathbf{v}, \hat{\mathbf{v}}\rangle_V \text { for all } \mathbf{v}, \hat{\mathbf{v}} \in V .$$

$$\left\langle T^{\star} T(\mathbf{v}), \hat{\mathbf{v}}\right\rangle_W=\langle\mathbf{v}, \hat{\mathbf{v}}\rangle_V \text { for all } \mathbf{v}, \hat{\mathbf{v}} \in V,$$

$$\left\langle T^{\star} T(\mathbf{v})-\mathbf{v}, \hat{\mathbf{v}}\right\rangle_W=\langle\mathbf{v}, \hat{\mathbf{v}}\rangle_V \text { for all } \mathbf{v}, \hat{\mathbf{v}} \in V .$$

$$A=Q R .$$

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## MATLAB代写

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