## 数学代写|高等线性代数代写Advanced Linear Algebra代考|Normal and Hermitian matrices

In this section we study transformations that interact particularly nicely with respect to the inner product. A main feature of these normal and Hermitian transformations is that its eigenvectors can be used to form an orthonormal basis for the underlying space.
A matrix $A \in \mathbb{F}^{n \times n}$ is called normal if $A^* A=A A^$. Lemma 5.5.1 (a) If $U$ is unitary, then $A$ is normal if and only if $U^ A U$ is normal.
(b) If $T$ is upper triangular and normal, then $T$ is diagonal.
Proof. (a). Let us compute $U^* A U\left(U^* A U\right)^=U^ A U U^* A^* U=U^* A A^* U$, and $\left.\left(U^* A U\right)^* U^* A U\right)=U^* A^* U U^* A U=U^* A^* A U$. The two are equal if and only if $A A^=A^ A$ (where we used that $U$ is invertible). This proves the first part.
(b). Suppose that $T=\left(t_{i j}\right){i, j=1}^n$ is upper triangular. Thus $t{i j}=0$ for $i>j$. Since $T$ is normal we have that $T^* T=T T^$. Comparing the $(1,1)$ entry on both sides of this equation we get $$\left|t_{11}\right|^2=\left|t_{11}\right|^2+\left|t_{12}\right|^2+\cdots+\left|t_{1 n}\right|^2 .$$ This gives that $t_{12}=t_{13}=\cdots=t_{1 n}=0$. Next, comparing the $(2,2)$ entry on both sides of $T^ T=T T^*$ we get
$$\left|t_{22}\right|^2=\left|t_{22}\right|^2+\left|t_{23}\right|^2+\cdots+\left|t_{2 n}\right|^2 .$$
This gives that $t_{23}=t_{24}=\cdots=t_{2 n}=0$. Continuing this way, we find that $t_{i j}=0$ for all $i<j$. Thus $T$ is diagonal.

## 数学代写|高等线性代数代写Advanced Linear Algebra代考|The quotient space

Let $V$ be a vector space over $\mathbb{F}$ and $W \subseteq V$ a subspace. We define the relation $\sim$ via
$$\mathbf{v}_1 \sim \mathbf{v}_2 \Leftrightarrow \mathbf{v}_1-\mathbf{v}_2 \in W .$$
Then $\sim$ is an equivalence relation:
(i) Reflexivity: $\mathbf{v} \sim \mathbf{v}$ for all $\mathbf{v} \in V$, since $\mathbf{v}-\mathbf{v}=\mathbf{0} \in W$.
(ii) Symmetry: Suppose $\mathbf{v}_1 \sim \mathbf{v}_2$. Then $\mathbf{v}_1-\mathbf{v}_2 \in W$. Thus $-\left(\mathbf{v}_1-\mathbf{v}_2\right)=\mathbf{v}_2-\mathbf{v}_1 \in W$, which yields $\mathbf{v}_2 \sim \mathbf{v}_1$.
(iii) Transitivity: Suppose $\mathbf{v}_1 \sim \mathbf{v}_2$ and $\mathbf{v}_2 \sim \mathbf{v}_3$. Then $\mathbf{v}_1-\mathbf{v}_2 \in W$ and $\mathbf{v}_2-\mathbf{v}_3 \in W$. Thus $\mathbf{v}_1-\mathbf{v}_3=\left(\mathbf{v}_1-\mathbf{v}_2\right)+\left(\mathbf{v}_2-\mathbf{v}_3\right) \in W$. This yields $\mathbf{v}_1 \sim \mathbf{v}_3$.
As $\sim$ is an equivalence relation, it has equivalence classes, which we will denote as $\mathbf{v}+W$ :
$$\mathbf{v}+W:={\hat{\mathbf{v}}: \mathbf{v} \sim \hat{\mathbf{v}}}={\hat{\mathbf{v}}: \mathbf{v}-\hat{\mathbf{v}} \in W}=$$
${\hat{\mathbf{v}}:$ there exists $\mathbf{w} \in W$ with $\hat{\mathbf{v}}=\mathbf{v}+\mathbf{w}}$.
Any member of an equivalence class is called a representative of the equivalence class.

Example 6.2.1 Let $V=\mathbb{R}^2$ and $W=\operatorname{Span}\left{\mathbf{e}_1\right}$. Then the equivalence class of $\mathbf{v}=\left(\begin{array}{l}v_1 \ v_2\end{array}\right)$ is the horizontal line through $\mathbf{v}$. In this example it is simple to see how one would add two equivalence classes. Indeed, to add the horizontal line through $\left(\begin{array}{l}0 \ c\end{array}\right)$ to the horizontal line through $\left(\begin{array}{l}0 \ d\end{array}\right)$, would result in the horizontal line through $\left(\begin{array}{c}0 \ c+d\end{array}\right)$. Or, what is equivalent, to add the horizontal line through $\left(\begin{array}{l}5 \ c\end{array}\right)$ to the horizontal line through $\left(\begin{array}{c}10 \ d\end{array}\right)$, would result in the horizontal line through $\left(\begin{array}{c}15 \ c+d\end{array}\right)$. Similarly, one can define scalar multiplication for these equivalence classes. We give the general definition below.
The set of equivalence classes is denoted by $V / W$ :
$$V / W:={\mathbf{v}+W: \mathbf{v} \in V} .$$
We define addition and scalar multiplication on $V / W$ via
$$\begin{gathered} \left(\mathbf{v}_1+W\right)+\left(\mathbf{v}_2+W\right):=\left(\mathbf{v}_1+\mathbf{v}_2\right)+W, \ c(\mathbf{v}+W):=(c \mathbf{v})+W . \end{gathered}$$
These two operations are defined via representatives (namely, $\mathbf{v}_1, \mathbf{v}_2$, and $\mathbf{v}$ ) of the equivalence classes, so we need to make sure that if we had chosen different representatives for the same equivalence classes, the outcome would be the same. We do this in the following lemma.

# 高等线性代数代考

## 数学代写|高等线性代数代写高级线性代数代考|法线和厄米矩阵

(b)如果$T$是上三角形且正规的，那么$T$是对角线的。

(b)。假设$T=\left(t_{i j}\right){i, j=1}^n$是上三角形。因此，$t{i j}=0$为$i>j$。因为$T$是正常的，我们有$T^* T=T T^$。比较这个等式两边的$(1,1)$条目，我们得到$$\left|t_{11}\right|^2=\left|t_{11}\right|^2+\left|t_{12}\right|^2+\cdots+\left|t_{1 n}\right|^2 .$$，这就得到$t_{12}=t_{13}=\cdots=t_{1 n}=0$。接下来，比较$T^ T=T T^*$两边的$(2,2)$条目，我们得到
$$\left|t_{22}\right|^2=\left|t_{22}\right|^2+\left|t_{23}\right|^2+\cdots+\left|t_{2 n}\right|^2 .$$

$$\mathbf{v}_1 \sim \mathbf{v}_2 \Leftrightarrow \mathbf{v}_1-\mathbf{v}_2 \in W .$$

(i)反身性: $\mathbf{v} \sim \mathbf{v}$ 为所有人 $\mathbf{v} \in V$，因为 $\mathbf{v}-\mathbf{v}=\mathbf{0} \in W$.
(ii)对称性:假设 $\mathbf{v}_1 \sim \mathbf{v}_2$。然后 $\mathbf{v}_1-\mathbf{v}_2 \in W$。因此 $-\left(\mathbf{v}_1-\mathbf{v}_2\right)=\mathbf{v}_2-\mathbf{v}_1 \in W$，得到 $\mathbf{v}_2 \sim \mathbf{v}_1$.
(iii)传递性:假设 $\mathbf{v}_1 \sim \mathbf{v}_2$ 和 $\mathbf{v}_2 \sim \mathbf{v}_3$。然后 $\mathbf{v}_1-\mathbf{v}_2 \in W$ 和 $\mathbf{v}_2-\mathbf{v}_3 \in W$。因此 $\mathbf{v}_1-\mathbf{v}_3=\left(\mathbf{v}_1-\mathbf{v}_2\right)+\left(\mathbf{v}_2-\mathbf{v}_3\right) \in W$。这产生了 $\mathbf{v}_1 \sim \mathbf{v}_3$.
As $\sim$ 是等价关系，它有等价类，我们将其表示为 $\mathbf{v}+W$ :
$$\mathbf{v}+W:={\hat{\mathbf{v}}: \mathbf{v} \sim \hat{\mathbf{v}}}={\hat{\mathbf{v}}: \mathbf{v}-\hat{\mathbf{v}} \in W}=$$
${\hat{\mathbf{v}}:$ 存在的 $\mathbf{w} \in W$ 用 $\hat{\mathbf{v}}=\mathbf{v}+\mathbf{w}}$

$$V / W:={\mathbf{v}+W: \mathbf{v} \in V} .$$

$$\begin{gathered} \left(\mathbf{v}_1+W\right)+\left(\mathbf{v}_2+W\right):=\left(\mathbf{v}_1+\mathbf{v}_2\right)+W, \ c(\mathbf{v}+W):=(c \mathbf{v})+W . \end{gathered}$$

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