Induction proof that $a^n=na-\left(n-1\right)I$

Question

I need to know how to finish this proof with induction, i got stuck and i do not know how to finish. Thanks.

$a^n=na-\left(n-1\right)I$ $\space\space\space\space\space\space,\forall n\in \mathbb{N},\:n\ge 2$

$a^{n-1}=\left(a-1\right)a-\left(\left(n-1\right)-1\right)I\:=\:an\:-a\:-\:\left(n-2\right)I\:=\:an-a-In+2I$

$a^n=a^{n-1}\cdot a$

$a^n=\left(a^n-a+In+2I\right)a$

Answer 1

Well, if you assume $n>1$ and that $$ a^{n-1}=(n-1)a-(n-1-1)I $$ you have $$ a^n=(n-1)a^2-(n-2)a $$ which should equal $na-(n-1)I$. Now $$ (n-1)a^2-(n-2)a=na-(n-1)I $$ if and only if $$ (n-1)a^2-2(n-1)a+(n-1)I=0 $$ which is equivalent to $$ a^2-2a+I=0 $$ or, as well, to $(a-I)^2=0$. This is false, in general.

What we have proved is actually the statement

if $a^2=2a-I$, then, for every nonnegative integer $n$, $a^n=na-(n-1)I$.

Answer 2

You need to check the base case with $n=2$. That gives you a formula for $a^2$. Betcha that comes in handy when proving the induction step as well.

As to how to prove the base case? You haven't told us what $a$ is, so it is kinda difficult to say how to do that. Presumably it is straight forward.