Find $A^{1000}$ by using Cayley-Hamilton Theorem

Solution 1:

Your formula tells you, after you multiply through by $A^{997}$, that $$A^{1000}=2A^{999}.$$ Similarly, $$2A^{999}=4A^{998}.$$

This process can be repeated to find $A^{1000}$ in terms of $A^2$, which you can then compute.

There is another way of approaching this.

You could divide $x^{1000}$ by the characteristic polynomial:

$x^{1000} = (-x^3+2x^2)Q+R$ where $R$ is a polynomial of degree less than 3 with unknown coefficients.

write down $R=ax^2+bx+c$ and evaluate $R$ at the roots of the characteristic polynomial.

Meaning, write down $\lambda^{1000}=a\lambda ^2+b\lambda+c$

and

$\xi^{1000} = a\xi ^2+b\xi+c$

and

$\rho^{1000} = a\rho ^2+b\rho+c$

where $\lambda$ and $\xi$ and $\rho$ are roots of the characteristic polynomial. as you can see, $Q$ wont matter because it is multiplied by zero.

Do this to find the coeffiecents of the remainder, $R$.

after you have done that, insert $x=A$ to get $A^{1000}=aA^2+bA+c$ with the coeffiecents $a,b,c$ that you found.

Edit: The problem here, is that you have a double root, so you need to use the derivative as well.

Full answer:

divide $x^{1000}$ by $(-x^3+2x^2)$ to get:

$x^{1000} = (-x^3+2x^2)Q+ax^2+bx+c$ where $Q$ is some polynomial unknown to us.

the roots of the char. polynomial are $0,2$. put $x=0$ to get:

$0^{1000}=0=0*Q+c=c$ so $c=0$.

now derive $x^{1000} = (-x^3+2x^2)Q+ax^2+bx$ to get:

$1000x^{999}=(-3x^2+4x)Q+Q'(-x^3+2x^2)+2ax+b$ and insert $x=0$ again t oget:

$1000*0^{999} = 0 =b$ meaning $b=0$.

Now back to our original formula with $b=c=0$:

$x^{1000} = (-x^3+2x^2)Q+ax^2$

Insert $x=2$ to get:

$2^{1000} = 4a$ meaning $a=2^{998}$.

Now our original formula looks like $x^{1000} = (-x^3+2x^2)Q+2^{998}x^2$

Inserts $x=A$ to get:

$A^{1000} = 2^{998}A^2$