Exercise on expression with matrices using Cayley-Hamilton theorem

Question

For the following matrix $$A= \begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 1 \end{bmatrix} $$ I need to use the Cayley-Hamilton theorem to calculate $(A+I_3)^{10}(A-I_3)^2+A$

First I try writing the characteristic polynomial $f(t)=\det(A-tI_3)=-t^3+t^2+2t-1$. Then, by using Caylely-Hamilton, I have that $f(A)=O_3$, that is, $$-A^3+A^2+2A-I_3=O_3$$ $$A^3-A^2-2A+I_3=O_3$$ I decompose the previous expression and I get that $$(A+I_3)(A-I_3)^2=A$$ Now I replace the last expression in the initial expression $(A+I_3)^{10}(A-I_3)^2+A$ and get $$(A+I_3)^9(A+I_3)(A-I_3)^2+A=$$ $$=(A+I_3)^9A+A$$

Now what should the next step be?

Answer 1

Hint: The result is a matrix with integer coeffcients, namely $$ \begin{pmatrix} 2070 & 3722 & 4631 \cr 3722 & 6701 & 8353 \cr 4631 & 8353 & 10423 \end{pmatrix} $$ So you need to evaluate at a certain point.

Answer 2

I am not sure how fruitful your approach will be. One method that will work is as follows: with polynomial long-division, we find that $$ (t + 1)^{10}(t - 1)^2 + t = q(t)(t^3 - t^2 - 2t + 1) + 4631 t^2 + 3721 t - 2560. $$ Thus, we have $$ (A + I)^{10}(A - I)^2 + A = 4631 A^2 + 3721A - 2560 I. $$

Answer 3

You can calculate the last expression as $$(A+I)^9A+A = (((A+I)^3)^2+I)A = \begin{bmatrix} 2070 & 3722 & 4631 \cr 3722 & 6701 & 8353 \cr 4631 & 8353 & 10423 \end{bmatrix}$$ with only $4$ matrix multiplications and $2$ additions with identity $I$ which shouldn't be hard.

Answer 4

\begin{align} \det(\lambda I-A)&=\left| \begin{array}{ccc} \lambda & -1 & 0 \\ -1 & \lambda & -1 \\ 0 & -1 & \lambda-1 \end{array} \right| \\ & = \lambda(\lambda(\lambda-1)-1)+((-1)(\lambda-1)) \\ & = \lambda^2(\lambda-1)-(\lambda-1) \\ & =(\lambda-1)^2(\lambda+1). \end{align} Therefore $(A-I)^2(A+I)=0$. You are asked to compute $(A+I)^{10}(A-I)^2+A$, which must be the same as $A$ because $(A+I)^{10}(A-I)^2$ has $(A-I)^2(A+I)$ as a factor.