How to find $A^{100}$ by using the Cayley Hamilton theorem?

Question

How to find $A^{100}$ as a linear function of A by use of the Cayley Hamilton theorem?

The characteristic equation leads to $A^2-4A+3I=0$. I can't find any way to solve this.

Answer 1

Answer $$\color{blue}{~~ A^{100} =\frac12(3^{100}-1)(A -I)~~+~~I}$$

Blatant way of doing.

$$A^2 =4A-3I \implies A^{100 } = A^{2*50} =(4A-3I)^{50}$$

that is by Binomial formula I have, $$A^{100 } =(4A-3I)^{50} =\sum_{j=0}^{50} {50\choose j }3^{50-j}4^j A^j$$ From here you get the result by induction. But I guess it is Harder and studios.

General setting In a more general setting.

Problem I want to compute $A^n$ for all $n$ given that $A^2 =4A-3I $

Let $X_n =A^n$. Then $$X_{n+2} =A^{n+2} =A^{n}A^2 = A^n(A-3I) = 4X_{n+1} -3X_n$$

Hence we have the recurssive relation $$\color{blue}{X_{n+2} =4X_{n+1} -3X_n\Longleftrightarrow X_{n+2}-X_{n+1} =3(X_{n+1} -X_n) }$$

By Hence $(X_{n+1} -X_n)_n$ is a geometric sequence with coefficient 3 so that,

$$\color{blue}{ X_{n+1}-X_{n} =3^n(X_{1} -X_0) =3^n(A -I)}$$

Thus for very $n$ $$\color{red}{\forall~n,~~~ X_{n+1}-X_{n} =3^n(A -I) \implies X_n -X_0 = \sum_{j=0}^{n-1}(X_{j+1}-X_j) =(A -I)\sum_{j=0}^{n-1}3^j }$$

But, $$\sum_{j=0}^{n-1}3^j =\frac{3^n-1}{3-1}= \frac12(3^n-1)$$

Finally since $X_n =A^n $ we have, $$\color{red}{\forall~n,~~ A^n =\frac12(3^n-1)(A -I)~~+~~I}$$

Particularly for n$ n =100 $ $$\color{blue}{~~ A^{100} =\frac12(3^{100}-1)(A -I)~~+~~I}$$