Question

A=[1, -5, 9; 2, 0, -1; 1, 2 , -4]
A^20=? using Cayley–Hamilton theorem!

Answer 1

did you compute the characteristic polynomial? maybe we can use that, and some gimmick to find \(A^{20}\)

Answer 2

looks like it is
\[P(x)=-x^3-3 x^2+x+3\]

then \(-A^3-3A^2+A+3=0\) and so
\[A^3=-3A^2+A+3\]

then multiply both sides by \(A\) and get
\[A^4=-3A^3+A^2+3A+3\] now replace \(A^3\) in this expression by \(-3A^2+A+3\) and get
\[A^4=-3(-3A^2+A+3)+A^2+3A+3=10A^2-6\]

Answer 3

well that took a while, but maybe now there is some other trick to find \(A^{20}\) without continuing along this line of work, but i am pretty sure this is one way to do it

Answer 4

the point being that by repeating this process you will be able to compute \(A^{20}\) as some quadratic polynomial in \(A\)

Answer 5

now maybe multiply both sides by \(A^2\) and get
\[A^6=10A^4-6A^2=10(10A^2-6)-6A^2=96A^2-60\] hmm this is going to be ugly, maybe my algebra is wrong

Answer 6

I have this problem solved. From characteristic polynomial \[A ^{3}=-3A ^{2}+A+3E\]
and then here is written \[A ^{n}=a _{n}A ^{2}+bnA+c _{n}E\]
and after computing by A I have \[A ^{n+1}=a _{n}A ^{3}+b _{n}A ^{2}+c _{n}A\]\[A ^{n+1}=(b _{n}-3a _{n})A ^{2}+(b _{n}+c _{n})A+3a _{n}E\]
\[a _{n+1}=b _{n}-3a _{n} \rightarrow a _{n+1}=a _{n-1}+3a _{n-2}-3a _{n} \]
\[b _{n+1}=a _{n}+c _{n} \rightarrow b _{n+1}=a _{n}+3a_{n-1}\]
\[c _{n+1}=3a _{n}\]

now, using \[A ^{n}=a _{n}A ^{2}+b _{n}A+c _{n}E\] I have to compute coefficients for n=0, 1, ..., 20
I don't know how did they get those coefficients above

Answer 7

*multiplying by A

Answer 8

which coefficients?

Answer 9

we have them for \(n=3\) because \(A^3=-3A^2+A+3E\) so
\[a_3=-3,b_3=1,c_3=3\]

Answer 10

an+1=bn−3an→an+1=an−1+3an−2−3an
bn+1=an+cn→bn+1=an+3an−1
cn+1=3an
these recurrent formulas, I don't know how did they get them

Answer 11

oh ok lets go step by step
is this line
\[A ^{3}=-3A ^{2}+A+3E\] clear?

Answer 12

yes :D

Answer 13

did you see above how i write \(A^4\) as a quadratic in \(A\)?

Answer 14

I did

Answer 15

is it clear then that \(A^n\) can be written as a quadratic in \(A\)

Answer 16

yes

Answer 17

then this line is clear right?
\[A ^{n+1}=a _{n}A ^{3}+b _{n}A ^{2}+c _{n}A\]

Answer 18

oh i meant
\[A^n=a_nA^2+b_nA+c_nE\] that says \(A^n\) is a quadratic in \(A\)

Answer 19

multiply by \(A\) and get
\[A ^{n+1}=a _{n}A ^{3}+b _{n}A ^{2}+c _{n}A\]

Answer 20

that part is ok or no?

Answer 21

yes, that's ok

Answer 22

ok now we have
\[A^{n+1}=a_{n+1}A^2+b_{n+1}A+c_{n+1}E=a_nA^3+b_nA^2+c_nA\]

Answer 23

i.e. two expressions for \(A^{n+1}\)
is that ok?

Answer 24

yes

Answer 25

so now to relate the coefficients, you can replace \(A^3\) on the right hand side by
\[-3A^2+A+3E\] analagous to how i wrote \(A^4\) in term of \(A^2\)

Answer 26

i am going to make a guess that it is this step that was not obvious, since the just wrote the result of that computation, rather than the actual computation
i could be wrong

Answer 27

Thank you, I got it :) Thank you very much :)

Answer 28

yw glad to help