谱半径

设${\lambda }_1,\cdots ,{\lambda }_n$为$\mathbf{A}\in {\mathbb{C}}^{n\times n}$的特征值，则谱半径定义为
ρ ( A ) = max ⁡ { ∣ λ 1 ∣ , ⋯   , ∣ λ n ∣ } \rho\left(\mathbf{A}\right)=\max\left\{\left|\lambda_1\right|,\cdots, \left|\lambda_n\right|\right\} ρ(A)=max{∣λ1​∣,⋯,∣λn​∣}

注意$\Vert \mathbf{Av}\Vert \le \rho \left(\mathbf{A}\right)\Vert \mathbf{v}\Vert$不一定成立
C r = ( 0 1 r r 0 ) \mathbf{C}_r=

Cr​=(0r​r1​0​)
其中$r>1$
则${\mathbf{C}}_r$的特征值为$\pm 1$
${\mathbf{C}}_r{\mathbf{e}}_1=r{\mathbf{e}}_2$
$$\Vert {\mathbf{C}}_r{\mathbf{e}}_1\Vert =r>1=\rho \left({\mathbf{C}}_r\right)\Vert {\mathbf{e}}_1\Vert$$
如果$\mathbf{A}$是Hermitian矩阵的话则$\Vert \mathbf{Av}\Vert \le \rho \left(\mathbf{A}\right)\Vert \mathbf{v}\Vert$

性质

性质1

$$\rho \left(\mathbf{A}\right)\le \Vert \mathbf{A}\Vert$$
其中$\Vert \cdot \Vert$是任意算子范数

证明：
设$\lambda$是$\mathbf{A}$的特征值，$\mathbf{x}$是对应的特征向量
$$\left|\lambda \right|\Vert \mathbf{x}\Vert =\Vert \lambda \mathbf{x}\Vert =\Vert \mathbf{Ax}\Vert \le \Vert \mathbf{A}\Vert \Vert \mathbf{x}\Vert \Rightarrow \rho \left(\mathbf{A}\right)\le \Vert \mathbf{A}\Vert$$

性质2

对于$ϵ>0$，存在某种矩阵范数$\Vert \cdot \Vert$,使得
$$\Vert \mathbf{A}\Vert \le \rho \left(\mathbf{A}\right)+ϵ$$

证明：
由若尔当分解定理
A = S ( J n 1 ( λ 1 ) 0 ⋯ 0 0 J n 2 ( λ 2 ) ⋱ ⋮ ⋮ ⋱ ⋱ 0 0 ⋯ 0 J n k ( λ k ) ) S − 1 \mathbf{A}=\mathbf{S}

\mathbf{S}^{-1} A=S⎝ ⎛​Jn1​​(λ1​)0⋮0​0Jn2​​(λ2​)⋱⋯​⋯⋱⋱0​0⋮0Jnk​​(λk​)​⎠ ⎞​S−1
其中$\mathbf{S}$是可逆矩阵，${\lambda }_1,\cdots ,{\lambda }_k$是$\mathbf{A}$的特征值，$n_1+\cdots +n_k=n$

设
D ( η ) = ( D n 1 ( η ) 0 ⋯ 0 0 D n 2 ( η ) ⋱ ⋮ ⋮ ⋱ ⋱ 0 0 ⋯ 0 D n k ( η ) ) \mathbf{D}\left(\eta\right)=

D(η)=⎝ ⎛​Dn1​​(η)0⋮0​0Dn2​​(η)⋱⋯​⋯⋱⋱0​0⋮0Dnk​​(η)​⎠ ⎞​
其中
D m ( η ) = ( η 0 ⋯ 0 0 η 2 ⋱ ⋮ ⋮ ⋱ ⋱ 0 0 ⋯ 0 η m ) \mathbf{D}_m\left(\eta\right)=

(η0⋯00η2⋱⋮⋮⋱⋱00⋯0ηm)" role="presentation" style="position: relative;">⎛⎝⎜⎜⎜⎜⎜⎜η0⋮00η2⋱⋯⋯⋱⋱00⋮0ηm⎞⎠⎟⎟⎟⎟⎟⎟$$\left(\begin{array}{cccc}\eta & 0& \cdots & 0\\ 0& {\eta }^2& \ddots & ⋮\\ ⋮& \ddots & \ddots & 0\\ 0& \cdots & 0& {\eta }^m\end{array}\right)$$

Dm​(η)=⎝ ⎛​η0⋮0​0η2⋱⋯​⋯⋱⋱0​0⋮0ηm​⎠ ⎞​

D ( 1 ϵ ) S − 1 A S D ( ϵ ) = ( B n 1 ( λ 1 , ϵ ) 0 ⋯ 0 0 B n 2 ( λ 2 , ϵ ) ⋱ ⋮ ⋮ ⋱ ⋱ 0 0 ⋯ 0 B n k ( λ k , ϵ ) ) \mathbf{D}\left(\frac{1}{\epsilon}\right)\mathbf{S}^{-1}\mathbf{A}\mathbf{S}\mathbf{D}\left(\epsilon\right)=

D(ϵ1​)S−1ASD(ϵ)=⎝ ⎛​Bn1​​(λ1​,ϵ)0⋮0​0Bn2​​(λ2​,ϵ)⋱⋯​⋯⋱⋱0​0⋮0Bnk​​(λk​,ϵ)​⎠ ⎞​
其中
B m ( λ , ϵ ) = D m ( 1 ϵ ) J m ( λ ) D m ( ϵ ) = ( λ ϵ 0 ⋯ 0 0 λ ϵ 0 ⋮ 0 ⋱ ⋱ ⋱ 0 ⋮ ⋱ ⋱ λ ϵ 0 ⋯ 0 0 λ ) \mathbf{B}_{m}\left(\lambda,\epsilon\right)=\mathbf{D}_m\left(\frac{1}{\epsilon}\right)\mathbf{J}_m\left(\lambda\right)\mathbf{D}_m\left(\epsilon\right)=

(λϵ0⋯00λϵ0⋮0⋱⋱⋱0⋮⋱⋱λϵ0⋯00λ)" role="presentation" style="position: relative;">⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜λ00⋮0ϵλ⋱⋱⋯0ϵ⋱⋱0⋯0⋱λ00⋮0ϵλ⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟$$\left(\begin{array}{ccccc}\lambda & ϵ& 0& \cdots & 0\\ 0& \lambda & ϵ& 0& ⋮\\ 0& \ddots & \ddots & \ddots & 0\\ ⋮& \ddots & \ddots & \lambda & ϵ\\ 0& \cdots & 0& 0& \lambda \end{array}\right)$$

Bm​(λ,ϵ)=Dm​(ϵ1​)Jm​(λ)Dm​(ϵ)=⎝ ⎛​λ00⋮0​ϵλ⋱⋱⋯​0ϵ⋱⋱0​⋯0⋱λ0​0⋮0ϵλ​⎠ ⎞​

定义矩阵范数
$$\Vert \mathbf{A}\Vert =\Vert \mathbf{D}\left(\frac{1}{ϵ}\right){\mathbf{S}}^{-1}\mathbf{ASD}\left(ϵ\right){\Vert }_1$$
接下来验证$\Vert \cdot \Vert$是矩阵范数

容易验证非负性和正齐次性

三角不等式：
∥ A + B ∥ = ∥ D ( 1 ϵ ) S − 1 ( A + B ) S D ( ϵ ) ∥ 1 = ∥ D ( 1 ϵ ) S − 1 A S D ( ϵ ) + D ( 1 ϵ ) S − 1 B S D ( ϵ ) ∥ 1 ≤ ∥ D ( 1 ϵ ) S − 1 A S D ( ϵ ) ∥ 1 + ∥ D ( 1 ϵ ) S − 1 B S D ( ϵ ) ∥ 1 = ∥ A ∥ + ∥ B ∥

∥A+B∥​=∥D(ϵ1​)S−1(A+B)SD(ϵ)∥1​=∥D(ϵ1​)S−1ASD(ϵ)+D(ϵ1​)S−1BSD(ϵ)∥1​≤∥D(ϵ1​)S−1ASD(ϵ)∥1​+∥D(ϵ1​)S−1BSD(ϵ)∥1​=∥A∥+∥B∥​

次乘性：

因为$\mathbf{D}\left(\frac{1}{ϵ}\right)\mathbf{D}\left(ϵ\right)=\mathbf{I}$
∥ A B ∥ = ∥ D ( 1 ϵ ) S − 1 A B S D ( ϵ ) ∥ 1 = ∥ D ( 1 ϵ ) S − 1 A S D ( ϵ ) D ( 1 ϵ ) S − 1 B S D ( ϵ ) ∥ 1 ≤ ∥ D ( 1 ϵ ) S − 1 A S D ( ϵ ) ∥ 1 ∥ D ( 1 ϵ ) S − 1 B S D ( ϵ ) ∥ 1 = ∥ A ∥ ∥ B ∥

∥AB∥​=∥D(ϵ1​)S−1ABSD(ϵ)∥1​=∥D(ϵ1​)S−1ASD(ϵ)D(ϵ1​)S−1BSD(ϵ)∥1​≤∥D(ϵ1​)S−1ASD(ϵ)∥1​∥D(ϵ1​)S−1BSD(ϵ)∥1​=∥A∥∥B∥​
所以$\Vert \cdot \Vert$是矩阵范数

于是
∥ A ∥ = max ⁡ i ∈ { 1 , 2 , ⋯   , n } ( ∣ λ i ∣ + ϵ ) = ρ ( A ) + ϵ \|\mathbf{A}\|=\max_{i\in\left\{1,2,\cdots,n\right\}}\left(\left|\lambda_i\right|+\epsilon\right)=\rho\left(\mathbf{A}\right)+\epsilon ∥A∥=i∈{1,2,⋯,n}max​(∣λi​∣+ϵ)=ρ(A)+ϵ

性质3

$\rho \left(\mathbf{A}\right)<1\iff \underset{k\to \infty }{\lim }{\mathbf{A}}^k=0$

证明：
设$\mathbf{A}$的特征值为$\lambda$,对应的特征向量为$\mathbf{v}$
假设$\underset{k\to \infty }{\lim }{\mathbf{A}}^k=0$
0 = ( lim ⁡ k → ∞ A k ) v = lim ⁡ k → ∞ ( A k v ) = lim ⁡ k → ∞ ( λ k v ) = v lim ⁡ k → ∞ λ k

0​=(k→∞lim​Ak)v=k→∞lim​(Akv)=k→∞lim​(λkv)=vk→∞lim​λk​
因为$\mathbf{v}\ne 0$,
$$\underset{k\to \infty }{\lim }{\lambda }^k=0\Rightarrow \left|\lambda \right|<1\Rightarrow \rho \left(\mathbf{A}\right)<1$$

假设$\rho \left(\mathbf{A}\right)<1$
由若尔当分解定理
A = S ( J n 1 ( λ 1 ) 0 ⋯ 0 0 J n 2 ( λ 2 ) ⋱ ⋮ ⋮ ⋱ ⋱ 0 0 ⋯ 0 J n k ( λ k ) ) S − 1 \mathbf{A}=\mathbf{S}

\mathbf{S}^{-1} A=S⎝ ⎛​Jn1​​(λ1​)0⋮0​0Jn2​​(λ2​)⋱⋯​⋯⋱⋱0​0⋮0Jnk​​(λk​)​⎠ ⎞​S−1
其中$\mathbf{S}$是可逆矩阵，${\lambda }_1,\cdots ,{\lambda }_k$是$\mathbf{A}$的特征值，$n_1+\cdots +n_k=n$
由若尔当块的性质,对于充分大的$k$,有
J n i k ( λ i ) = [ λ i k ( k 1 ) λ i k − 1 ( k 2 ) λ i k − 2 ⋯ ( k n i − 1 ) λ i k − n i + 1 0 λ i k ( k 1 ) λ i k − 1 ⋯ ( k n i − 2 ) λ i k − n i + 2 ⋮ ⋮ ⋱ ⋱ ⋮ 0 0 … λ i k ( k 1 ) λ i k − 1 0 0 ⋯ 0 λ i k ] \mathbf{J}_{n_{i}}^{k}\left(\lambda_{i}\right)=\left[

\begin{array}{ccccc} \lambda_{i}^{k} & \left(\begin{array}{c} k \\ 1 \end{array}" role="presentation" style="position: relative;">\begin{array}{ccccc} \lambda_{i}^{k} & \left(\begin{array}{c} k \\ 1 \end{array}$$\text{begin{array}{ccccc} lambda\_i^k \& left(begin{array}{c} k 1 end{array}}$$

\right) \lambda_{i}^{k-1} & \left(

k2" role="presentation" style="position: relative;">k2$$\begin{array}{c}k\\ 2\end{array}$$

\right) \lambda_{i}^{k-2} & \cdots & \left(

kni−1" role="presentation" style="position: relative;">kni−1$$\begin{array}{c}k\\ n_i-1\end{array}$$

\right) \lambda_{i}^{k-n_{i}+1} \\ 0 & \lambda_{i}^{k} & \left(

k1" role="presentation" style="position: relative;">k1$$\begin{array}{c}k\\ 1\end{array}$$

\right) \lambda_{i}^{k-1} & \cdots & \left(

kni−2" role="presentation" style="position: relative;">kni−2$$\begin{array}{c}k\\ n_i-2\end{array}$$

\right) \lambda_{i}^{k-n_{i}+2} \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \ldots & \lambda_{i}^{k} & \left(

k1" role="presentation" style="position: relative;">k1$$\begin{array}{c}k\\ 1\end{array}$$

\right) \lambda_{i}^{k-1} \\ 0 & 0 & \cdots & 0 & \lambda_{i}^{k} \end{array}\right] Jni​k​(λi​)=⎣ ⎡​λik​0⋮00​(k1​)λik−1​λik​⋮00​(k2​)λik−2​(k1​)λik−1​⋱…⋯​⋯⋯⋱λik​0​(kni​−1​)λik−ni​+1​(kni​−2​)λik−ni​+2​⋮(k1​)λik−1​λik​​⎦ ⎤​
所以
$$\underset{k\to \infty }{\lim }{\mathbf{J}}_{n_i}^k=0\Rightarrow \underset{k\to \infty }{\lim }{\mathbf{J}}^k=0\Rightarrow \underset{k\to \infty }{\lim }{\mathbf{A}}^k=\underset{k\to \infty }{\lim }\mathbf{S}{\mathbf{J}}^k{\mathbf{S}}^{-1}=\mathbf{S}\left(\underset{k\to \infty }{\lim }{\mathbf{J}}^k\right){\mathbf{S}}^{-1}=0$$

Gelfand定理

$$\rho \left(\mathbf{A}\right)=\underset{k\to \infty }{\lim }\Vert {\mathbf{A}}^k{\Vert }^{\frac{1}{k}}$$

证明：$k\ge 0$
$$\rho {\left(\mathbf{A}\right)}^k=\rho \left({\mathbf{A}}^k\right)=\Vert {\mathbf{A}}^k\Vert \Rightarrow \rho \left(\mathbf{A}\right)\le \Vert {\mathbf{A}}^k{\Vert }^{\frac{1}{k}}\Rightarrow \rho \left(\mathbf{A}\right)\le \underset{k\to \infty }{\lim }\Vert {\mathbf{A}}^k{\Vert }^{\frac{1}{k}}$$

存在范数$\Vert \cdot {\Vert }_M$,使得$\Vert \mathbf{A}{\Vert }_M\le \rho \left(\mathbf{A}\right)+ϵ$
($\Vert \cdot {\Vert }_M$的$M$主要是为了和上面的范数区分)
由范数的等价性
$$\exists C>0,s.t.\Vert \cdot \Vert \le C\Vert \cdot {\Vert }_M$$
所以
$$\begin{gathered} \Vert {\mathbf{A}}^k\Vert \le C\Vert {\mathbf{A}}^k{\Vert }_M\le C\Vert \mathbf{A}{\Vert }_M^k\le C{\left(\rho \left(\mathbf{A}\right)+ϵ\right)}^k \\ \Vert {\mathbf{A}}^k{\Vert }^{\frac{1}{k}}\le C^{\frac{1}{k}}\left(\rho \left(\mathbf{A}\right)+ϵ\right) \\ \underset{k\to \infty }{\lim }\Vert {\mathbf{A}}^k{\Vert }^{\frac{1}{k}}\le \rho \left(\mathbf{A}\right)+ϵ \end{gathered}$$

参考：
https://www.math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect6.pdf
https://en.wikipedia.org/wiki/Spectral_radius