<s>[INST] Prove Cayley-Hamilton theorem I have a exercise in my linear algebra textbook:
Let $c_2\lambda^2+c_1\lambda +c_0=0$ be the characteristic equation for the matrix $$A=\begin{pmatrix}1&3\\3&1\end{pmatrix}$$
Prove that $c_2A^2+c_1A +c_0I=0$
This is Cayley-Hamilton theorem.
My solution:
If a root $\lambda$ exists, then it is the eigen value for the eigen vector $\vec{v}$.
If we multiply $c_2A^2+c_1A +c_0I$ with $\vec{v}$ then we get:
$$(c_2\lambda^2+c_1\lambda +c_0)\vec{v}$$
and since the eigen vector is not the zero vector and $c_2\lambda^2+c_1\lambda +c_0=0$ is true, $c_2A^2+c_1A +c_0I=0$ is also true.
Is this enough to prove the theorem and solve the problem?
 [/INST] You have a specific matrix given:
$$
A=\begin{pmatrix}1&3\\3&1\end{pmatrix}.
$$
Its characteristic equation is given by
$$
p(\lambda):=\det(\lambda I - A) = \begin{vmatrix}\lambda-1&-3\\-3&\lambda-1\end{vmatrix}
= (\lambda-1)^2-(-3)^2 = \lambda^2-2\lambda-8.
$$
Now,
$$
A^2=\begin{pmatrix}1&3\\3&1\end{pmatrix}\begin{pmatrix}1&3\\3&1\end{pmatrix}
= \begin{pmatrix}10&6\\6&10\end{pmatrix}
$$
so
$$
p(A)
= A^2 - 2A - 8I
= \begin{pmatrix}10&6\\6&10\end{pmatrix} - 2 \begin{pmatrix}1&3\\3&1\end{pmatrix} - 8 \begin{pmatrix}1&0\\0&1\end{pmatrix}
= \begin{pmatrix}0&0\\0&0\end{pmatrix}.
$$
</s>