Question

Let $A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right],$ and $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ then prove that $A^n=nA-(n-1)I,n\ge 1.$

Answer 1

$A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

if $n=1$

$A=1\cdot A-(1-1)\cdot I$

or $A=A$

if $n=2$

$A^2=A\times A=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]\times \left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 2& 1\end{array}\right]2\cdot A=\left[\begin{array}{cc}2& 0\\ 2& 2\end{array}\right](n-1)I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\therefore A^2=2\cdot A-(2-1)I$

Let's assume it is true for $n=m$

$A^m=m\cdot A-(m-1)\cdot I=\left[\begin{array}{cc}1& 0\\ m& 1\end{array}\right]A^m\cdot A=\left[\begin{array}{cc}1& 0\\ m& 1\end{array}\right]\cdot \left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ m+1& 1\end{array}\right](m+1)\cdot A-(m+1-1)I=\left[\begin{array}{cc}m+1& 0\\ m+1& m+1\end{array}\right]-\left[\begin{array}{cc}m& 0\\ 0& m\end{array}\right]=\left[\begin{array}{cc}1& 0\\ m+1& 1\end{array}\right]$

$\therefore$ By induction it is proved that

$A^n=nA-(n-1)A$ for $n\ge 1$