# 线性代数入门 Introduction to Linear Algebra MATH103

0

Let $[., .]$ denote an indefinite inner product defined on $\mathrm{R}^{n}$ by a real symmetric invertible matrix $H$ of size $n$, i.e.,,
$$[x, y]=(H x, y), \quad \text { for all } x, y \in \mathrm{R}^{n} .$$
Recall that (… ) stands for the standard inner product in $\mathrm{R}^{n}$ :
$$(x, y)=\sum_{j=1}^{n} x^{(j)} y^{(j)}$$
where $x=\left\langle x^{(1)}, \ldots, x^{(n)}\right\rangle, y=\left\langle y^{(1)}, \ldots, y^{(n)}\right\rangle$ are vectors in $\mathrm{R}^{n}$. The adjoint $A^{[]}$ of a real $n \times n$ matrix $A$ is defined just as in (4.1.2) and, as in (4.1.3) it is easily seen that $A^{[]}=H^{-1} A^{} H$. Since $A^{}$ is now just the transpose of $A, A^{[]}$ is obviously real. The following facts and definitions are all formally identical with predecessors in earlier chapters: A real matrix $A$ is $H$-selfadjoint if $A^{}=A$, i.e., if $H A=A^{} H$. A real matrix $A$ is $H$-unitary if $A^{} A=I$, i.e., if $A^{} H A=H$. A real matrix $A$ is $H$-normal if $A^{} A=A A^{}$, i.e., if $\left(H^{-1} A^{} H\right) A=A\left(H^{-1} A^{*} H\right)$.

## MATH103COURSE NOTES ：

only if there is a $(B, G) \in S_{r}$ which can be transformed to $\left(J, P_{e, J}\right)$ by both $T_{+}$ and $T_{-}$, say, with $\operatorname{det} T_{+}>0$ and $\operatorname{det} T_{-}<0$. Thus,
$$B=T_{+}^{-1} J T_{+}=T_{-}^{-1} J T_{-}, \quad G=T_{+}^{} P_{\varepsilon, J} T_{+}=T_{-}^{} P_{\varepsilon, J} T_{-},$$
and it follows immediately that $U S_{+}=U S_{-}$if and only if there is a real $T$ with negative determinant such that
$$J=T^{-1} J T, \quad P_{\varepsilon, J}=T^{*} P_{\varepsilon, J} T .$$

# 线性代数入门 Introductory Linear Algebra MATH106001

0

Let $F$ be the given hermitian matrix (3.4.14) and $w \in C$. Assume that
$d_{1}:=\operatorname{det} F(1, \ldots, n-1) \neq 0, \quad d_{2}:=F(2, \ldots, n) \neq 0, \quad d_{3}:=\operatorname{det} F(2, \ldots, n-1) \neq 0 .$
Then
$$\operatorname{det} F(w)=\frac{d_{1} d_{2}}{d_{3}}\left(1-\frac{\left|w-w_{0}\right|^{2} d_{3}^{2}}{d_{1} d_{2}}\right)$$

where
$$w_{0}=\frac{1}{w_{1 n}} \sum_{j=2}^{n-1} f_{n j} w_{j n}$$
and
$$[F(1, \ldots, n-1)]^{-1}=\left(w_{j k}\right)_{j, k=1}^{n-1} .$$

## MATH106001COURSE NOTES ：

It follows from these equations that
$$\rho_{n}=\frac{\operatorname{det} F(w)}{d_{2}} \text { and } \mu_{n}=\frac{\operatorname{det} F(w)}{d_{1}} .$$
Hence $\rho_{n}=\rho_{n-1}\left(1-\varphi_{n} \psi_{n}\right)$ and $\mu_{n}=\mu_{n-1}\left(1-\varphi_{n} \psi_{n}\right)$.
If we define (this is the same $w_{0}$ as in formula (3.4.15))
$$w_{0}=-\sum_{j=2}^{n-1} f_{n j} \nu_{j}=\overline{-\sum_{j=2}^{n-1} f_{1 j} \mu_{j}}$$
then $\delta_{n}=w-w_{0}, \Delta_{n}=\overline{w-w_{0}}$ and
$$\left|w-u_{0}\right|^{2}=\varphi_{n} \psi_{n} \rho_{n-1} \mu_{n-1} .$$

# 线性代数入门 Introductory Linear Algebra MATH106001

0

$$T=\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right)$$
we take it as the representation of a transformation with respect to the standard basis $T=\operatorname{Rep} \varepsilon_{2}, \mathcal{\varepsilon}{2}(t)$ and we look for a basis $B=\left\langle\vec{\beta}{1}, \vec{\beta}{2}\right\rangle$ such that $$\operatorname{Rep}{B, B}(t)=\left(\begin{array}{cc} \lambda_{1} & 0 \ 0 & \lambda_{2} \end{array}\right)$$
that is, such that $t\left(\vec{\beta}{1}\right)=\lambda{1} \vec{\beta}{1}$ and $t\left(\vec{\beta}{2}\right)=\lambda_{2} \vec{\beta}{2}$. $$\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) \vec{\beta}{1}=\lambda_{1} \cdot \vec{\beta}{1} \quad\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right) \vec{\beta}{2}=\lambda_{2} \cdot \vec{\beta}_{2}$$

We are looking for scalars $x$ such that this equation
$$\left(\begin{array}{ll} 3 & 2 \ 0 & 1 \end{array}\right)\left(\begin{array}{l} b_{1} \ b_{2} \end{array}\right)=x \cdot\left(\begin{array}{l} b_{1} \ b_{2} \end{array}\right)$$
has solutions $b_{1}$ and $b_{2}$, which are not both zero. Rewrite that as a linear system.
$$(3-x) \cdot b_{1}+\begin{array}{r} 2 \cdot b_{2}=0 \ (1-x) \cdot b_{2}=0 \end{array}$$

$$S=\left(\begin{array}{ll} \pi & 1 \ 0 & 3 \end{array}\right)$$
(here $\pi$ is not a projection map, it is the number $3.14 \ldots$ ) then
$$\left|\left(\begin{array}{cc} \pi-x & 1 \ 0 & 3-x \end{array}\right)\right|=(x-\pi)(x-3)$$
so $S$ has eigenvalues of $\lambda_{1}=\pi$ and $\lambda_{2}=3$. To find associated eigenvectors, first plug in $\lambda_{1}$ for $x$ :
$$\left(\begin{array}{cc} \pi-\pi & 1 \ 0 & 3-\pi \end{array}\right)\left(\begin{array}{l} z_{1} \ z_{2} \end{array}\right)=\left(\begin{array}{l} 0 \ 0 \end{array}\right) \quad \Longrightarrow \quad\left(\begin{array}{l} z_{1} \ z_{2} \end{array}\right)=\left(\begin{array}{l} a \ 0 \end{array}\right)$$
for a scalar $a \neq 0$, and then plug in $\lambda_{2}$ :
$$\left(\begin{array}{cc} \pi-3 & 1 \ 0 & 3-3 \end{array}\right)\left(\begin{array}{l} z_{1} \ z_{2} \end{array}\right)=\left(\begin{array}{l} 0 \ 0 \end{array}\right) \quad \Longrightarrow \quad\left(\begin{array}{l} z_{1} \ z_{2} \end{array}\right)=\left(\begin{array}{c} -b /(\pi-3) \ b \end{array}\right)$$
where $b \neq 0$.