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?
 A: I wouldn't say that you've got a proof yet, but you're not far off.
The problem is that you have shown that the eigenvector $v$ with eigenvalue $\lambda$ is in the kernel of $c_2 A^2 + c_1 A + c_0 I$ (i.e. $(c_2 A^2 + c_1 A + c_0 I) \vec{v} = 0$), but a priori the entire matrix $c_2 A^2 + c_1 A + c_0 I$ need not be zero if it maps a single vector nonzero to zero (I suppose unless $A$ were a $1 \times 1$).
On the other hand for this specific $2\times2$ matrix, if you knew that $c_2 A^2 + c_1 A + c_0 I$ mapped two linearly independent vectors to zero then it must be zero (if you like, for dimension reasons). So, what if we actually calculate the characteristic polynomial of $A$, and find that it has two distinct roots (it does)? Then the eigenvectors corresponding to these two roots have to be linearly independent (the roots are $4$ and $-2$, and they must be eigenvalues). Then we can use your argument for each eigenvalue/vector, and then this would complete the proof.
A: 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
$$
p(A)
= A^2 - 2A - 8I
= \begin{pmatrix}10&6\\6&10\end{pmatrix} - 2 \begin{pmatrix}1&3\\3&1\end{pmatrix} - 8 \begin{pmatrix}1&0\\0&1\end{pmatrix}
= \begin{pmatrix}0&0\\0&0\end{pmatrix}.
$$
