How can I use Cayley-Hamilton's Theorem to find B5?

[annoymage]

Homework Statement

let x² + x + 1 be the characteristic polynomial of matrix B

find B⁵ using Cayley-Hamilton's Theorem

Homework Equations

The Attempt at a Solution

from what i have learn, cayley hamilton theorem is something like this

B² + B + I =0

B(B+I)=I

so, B^-1 = (B+1)

how can i apply this to make B⁵? help me please owho

 

[annoymage]

hmm, is my work here valid?

B^-1B = (B+1)B = I

IB⁵ = (B+1)B⁵

 

[irycio]

As far as I'm concerned, you mustn't use B^(-1) because you don't know whether B i reversible. Quite likely it is, but I don't know if the fact that the characteristic polynomial has no solutions is enough.

I have never done such exercise before, but let me try:

B^2+B+I=0

B^2=-I-B /*B
B^3=-B-B^2=-B+I+B=I
B^4=B^3*B=B
B^5=B^2=-I-B

but again, completely not sure :/

 

[annoymage]

yeaaaa, i don't know B is invertible or not... Silly me.. ahaha

i guess that is the answer

 

[Mark44]

Homework Statement

let x² + x + 1 be the characteristic polynomial of matrix B

find B⁵ using Cayley-Hamilton's Theorem

Homework Equations

The Attempt at a Solution

from what i have learn, cayley hamilton theorem is something like this

B² + B + I =0

B(B+I)=I

The second equation doesn't follow from the first. The first is equivalent to B² + B = -I, so B(B + I) = -I

so, B^-1 = (B+1)

how can i apply this to make B⁵? help me please owho

 

[hgfalling]

You have this from Cayley-Hamilton:

B^2 + B + I = 0
B^2 = -(B + I)

Then:

B^3 = B^2(B) = ..., substitute -(B+I) for B^2, etc, and so on through B^5.