Matrix Power

hbghlyj

$\frac{a_{100}}{a_0}=\frac{b_{100}}{b_0}=\lambda$,
$$\cases{a_{n+1}=5a_n+b_n\\b_{n+1}=-a_n+3b_n}$$
求$\lambda$.

hbghlyj

$A=\pmatrix{5&1\\-1&3}$
$A$的特征值是$4$
$A^{100}$的特征值是$λ=4^{100}$.

hbghlyj

$\ker(A-4I)=\operatorname{span}\left\{\pmatrix{1\\-1}\right\}$
由$(A-4I)v=\pmatrix{1\\-1}$解得$v=\pmatrix{1\\0}$
所以$A$的Jordan标准型
\[A=\left(
\begin{array}{cc}
1 & 1 \\
-1 & 0 \\
\end{array}
\right)\left(
\begin{array}{cc}
4 & 1 \\
0 & 4 \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & 1 \\
-1 & 0 \\
\end{array}
\right)^{-1}\]
而$J^2=0⇒(4I+J)^n=(4I)^n+n(4I)^{n-1}J=4^nI+4^{n-1}nJ$
所以
\[A^n=\left(
\begin{array}{cc}
1 & 1 \\
-1 & 0 \\
\end{array}
\right)\left(
\begin{array}{cc}
4^n & 4^{n-1}n \\
0 & 4^n \\
\end{array}
\right)\left(
\begin{array}{cc}
1 & 1 \\
-1 & 0 \\
\end{array}
\right)^{-1}=\left(
\begin{array}{cc}
4^{n-1} n+4^n & 4^{n-1} n \\
-4^{n-1} n & 4^n-4^{n-1} n \\
\end{array}
\right)\]