Several conclusions on diagonalization/conjugation

The universal multiplication of 'ten'

For every $x \in \mathbb{R}$ and $y \in \mathbb{R}$, there exists some function $f : \mathbb{R} \to \mathbb{R}$ such that $y = f(x)$

$f = f_k^{-1} \circ g \circ f_k \text{ where } f_k(x) = x - k, g(x) = 10x$

$\displaystyle{k = \frac{10x - y}{9}}$

$map : (x, y) \mapsto k = \frac{10x - y}{9} \text{ is a well-defined mapping}$ .

Arbitrary Power of a Square Matrix $A$

For any square matrix $A \in \mathbb{C}^{n \times n}$, assume it can be diagonalized, i.e., there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^{-1}$. Then for any integer $k$, the $k$-th power of $A$ can be expressed as $A^k = PD^kP^{-1}$, where $D^k$ represents each diagonal element of $D$ raised independently to the power of $k$.

Iteration of Arbitrary Fractional Linear Function

Assume $ad - bc \neq 0$ for all fractional linear mappings forming a group $G$. Define a mapping $\phi: G \rightarrow GL_2(\mathbb{R})$ as $\phi(\frac{ax + b}{cx + d}) = \left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right)$.

The bijection $\phi$ is an isomorphism between the group $G$ and $GL_2(\mathbb{R})$. Therefore, $f^{(n)} = \phi^{-1}(\phi(f)^n) = \phi^{-1}(\left(\begin{array}{cc} a & b \\ c & d \\ \end{array}\right)^n)$.

Using the bridge function method taught in high school competitions, construct $\displaystyle \varphi = \frac{x - x_2}{x - x_1}$ from the two fixed points $x_1, x_2$ of $f$. At this time, simply calculate the iteration of $g = \varphi \circ f \circ \varphi^{-1}$.

Lie Group $SO(3)$ Describing Rigid Body Rotation Losses One Degree of Freedom

Rotation around the $X$ and $Y$ axes implies rotation around the $Z$ axis: For example, a counterclockwise rotation of $90^\circ$ around the X axis is represented as $r_{90^\circ}(x)$, then $r_{90^\circ}(z) = r_{90^\circ}(x) r_{90^\circ}(y) r_{90^\circ}(x)^{-1}$.

Similar conclusions apply to other angles, so the $n$-times angle of $90^\circ$ around the $Z$ axis just needs to change the power of $r(y)$ in the middle.

$D$ as a Differential Operator

$D - \lambda = e^{\lambda t} \circ D \circ e^{-\lambda t}$

Lie bracket: $[D, e^{-\lambda t}] = -\lambda e^{-\lambda t}$, directly expand the left side $[D, e^{-\lambda t}] = D \circ e^{-\lambda t} - e^{-\lambda t} \circ D = -\lambda e^{-\lambda t}$

$D \circ e^{-\lambda t} = e^{-\lambda t} \circ (D - \lambda)$, also multiplying by $e^{\lambda t}$ in the left proves the result. The corollary $\displaystyle (D - \lambda)^n = e^{\lambda t} \circ \int^n \circ e^{-\lambda t}$

Permuted indicator function

Define $f : S_n \rightarrow \{\pm 1\} \\ \sigma \mapsto \prod_{\{i,j\} \in P_2(n)} \frac{\sigma(j) - \sigma(i)}{j - i}$ then $\sigma, \tau \in S_n$ , then $f(\sigma \tau \sigma^{-1}) = f(\tau)$

and $(i_1...i_k)$ is a $k$-cycle in $S_n$, where $\sigma \in S_n$. Then:

is also a $k$-cycle.

More Examples to Follow...