Matrix and Cayley Hamilton

Let $A= \begin{pmatrix}
2 &-1  &1 \\
 0&-2  &-1 \\
 0& 3 &2
\end{pmatrix}$.

a) Find the characteristic polynomial of $A$ and deduce that $A$ is invertible.
b) Express $A^{-1}$ as a linear combination of $\mathbb{I}_3, \; A, \; A^2$.
c) Prove that: $A^{2006} - 2A^{2005} = A^2 -2A$.

Solution



a) The characteristic polynomial of $A$ will be found by evaluating the determinant:

$$\begin{vmatrix}
2-\ell &-1  &-1 \\
 0& 2 -\ell&-1 \\
 0& 3 &2 -\ell
\end{vmatrix}=0$$

Doing the process we find that the characteristic polynomial of $A$ is $$\chi_A(x)= -x^3 + 2x^2 +x-2$$ Since the constant term of the polynomial is $a_0=-2$ we deduce that $\det A = -2 \neq 0$ and the matrix is invertible.

b) We are using Cayley Hamilton's theorem. Therefore:
$$\begin{aligned} -A^3 +2A^2 +A-2\mathbb{I}_3 =0 & \Leftrightarrow -A^3 +2A^2 +A -2 A A^{-1} =0 \\
& \Leftrightarrow \frac{1}{2}\left(-A^2 +2A +\mathbb{I}_3 \right) =A^{-1} \end{aligned}$$

c) From Cayley Hamilton and the characteristical polynomial we have that:

$$A^3 - 2A^2 =A -2\mathbb{I}_3$$

Now we easily deduce that:

$$ \begin{matrix}
A^4-2A^2 =A^2 -2A\\
A^5 - 2A^4 =A^3 -2A^2 = A-2\mathbb{I}_3\\
A^6 - 2A^5 =A^2 -2A\\
\cdots \cdots \cdots
\end{matrix}$$

Inductively we prove that $A^{2n} - 2A^{2n-1} =A^2 - 2A$ hence the result follows.