I need to find the characteristic equation of the matrix $ A = \begin{bmatrix} 2&1&1\\ 0&1&0\\ 1&1&2\\ \end{bmatrix} $ and find the matrix represented by $ A^{12}-5A^{11}+7A^{10}-3A^{9}-A^8+5A^7-7A^6+3A^5+A^4-5A^3+8A^2-4A+3I $

My attempt:

I started by finding the character equation which was $ \lambda^3-5\lambda^2+7\lambda-3=0 $ where $\lambda$ is the eigenvalue, and found $\lambda = 5,\frac{9+\sqrt{69}}{2},\frac{9-\sqrt{69}}{2} $

$\lambda = 5$ satisfies the equation and I would take $\lambda = 5$ let $\lambda = A$ and $ A^3-5 A^2+7A -3=0 $.
I don't know how to proceed further. All help would be appreciated.

Edit: What I can do, is multiply the matrix $A$ by itself and consequently the resultant matrix would be $A^2$ and so on. However, I'm not sure if this will give an accurate answer.

  • 1
    $\begingroup$ I suspect the characteristic equation is $\lambda^3 - 5\lambda^2 + 7\lambda - 3 = 0$. You might want to double check your work on that part. $\endgroup$ – JimmyK4542 Dec 9 '18 at 21:43
  • $\begingroup$ Yes, you're right about the characteristic equation. I took a different problem's matrix for this one. My bad. $\endgroup$ – tNotr Dec 9 '18 at 21:48

The characteristic polynomial of $A$ is $p(\lambda) = \det(\lambda I - A) = \lambda^3-5\lambda^2+7\lambda-3$.

So by the Cayley-Hamilton theorem $p(A) = A^3-5A^2+7A-3I = 0$ (*).

Multiplying (*) by $A^9$ yields $A^{12}-5A^{11}+7A^{10}-3A^9 = 0$ (1).

Multiplying (*) by $-A^5$ yields $-A^8+5A^7-7A^6+3A^5 = 0$ (2).

Multiplying (*) by $A$ yields $A^4-5A^3+7A^2-3A = 0$ (3).

If we add (1), (2), and (3), we get $$A^{12}-5A^{11}+7A^{10}-3A^9-A^8+5A^7-7A^6+3A^5+A^4-5A^3+7A^2-3A = 0$$

Can you take it from here?

  • $\begingroup$ I find ${}+8\lambda$ for the characteristic polynomial. $\endgroup$ – Bernard Dec 9 '18 at 21:53
  • $\begingroup$ yes absolutely! However, I am not allowed to multiply the matrix A by any matrix greater than $A^3$ in the exam for some reason. Is this an attempt to baffle the student into making a mistake? I will never know but nonetheless I will try your solution though regardless of the risks. $\endgroup$ – tNotr Dec 9 '18 at 21:55
  • $\begingroup$ You don't have to actually compute $A^3$ or any other higher power of $A$. Just add $A^2-A+3I$ to both sides of the last equation I wrote. Then, the left side is the expression you are looking for, and the right side is something that should be easy enough to compute. $\endgroup$ – JimmyK4542 Dec 9 '18 at 22:25

Check you computation for the characteristic polynomial $\chi_A$.


Divide the polynomial $p(x)=x^{12}-5x^{11}+\dots+3$ by the characteristic polynomial: $$p(x)=q(x)\chi_A(x)+r(x)\qquad (\deg r \le 2),$$ to get $$p(A)=q(A)\chi_A(A)+r(A)=r(A).$$:


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