Matrix Algebra (2)

det M = 0 => ad - bc = 0 => ad = bc.

Let p(n) be the proposition that M^(n+1) = (a+d)^n M for postivie integers n.

n = 1:
M^(1+1) = M^2

圖片參考：http://latex.codecogs.com/gif.latex?\left( \begin{array}{cc}a&b\\c&d\end{array} \right)\left( \begin{array}{cc}a&b\\c&d\end{array} \right)=\left( \begin{array}{cc}a&b\\c&d\end{array} \right)=\left( \begin{array}{cc}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{array} \right)=\left( \begin{array}{cc}a^2+ad&ab+bd\\ac+cd&ad+d^2\end{array} \right)


圖片參考：http://latex.codecogs.com/gif.latex?=\left( \begin{array}{cc}a(a+d)&b(a+d)\\c(a+d)&d(a+d)\end{array} \right)=(a+d)M


Therefore p(n) is true for n=1.

Assume n = k is true. i.e.,
圖片參考：http://latex.codecogs.com/gif.latex?M^{n+1}=(a+d)^nM


Then for n = k+1,

圖片參考：http://latex.codecogs.com/gif.latex?M^{n+2}=(a+d)^nM^2=(a+d)^n(a+d)M=(a+d)^{n+1}M


By MI, p(n) is true for all positive integers n.

這條不是已經給解決了嗎?