Q:

Cayley-Hamilton theorem-eigenvalues

I am studying and doing exercises about the Cayley-Hamilton Theorem but I am having difficulties to understand this question:
I have the matrix $A$ as
\begin{bmatrix}
 2&1\\
1&2\\
\end{bmatrix}
I found the characteristic equation which is : $\lambda^2-4\lambda+3=0$ and eigenvalues of $\lambda=1$ and  and $\lambda=3$
Then to apply Cayley-Hamilton I replaced the $\lambda$ in the characteristic equeation by A and obtained:
$A^2-4A+3=0$ and then I calculated matrix $A^2$ and verify that the Cayley-Hamilton is verified since, 
$$
A^2-4A+3I=\begin{bmatrix}
5&4\\
4&5\\
\end{bmatrix}
-
4\begin{bmatrix}
2&1\\
1&2\\
\end{bmatrix}
+
3\begin{bmatrix}
1&0\\
0&1\\
\end{bmatrix}
=
\begin{bmatrix}
0&0\\
0&0\\
\end{bmatrix}$$ 
Considering the matrices $A,A^2,0$ the 2x2 zero matrix  and $I$ the identity matrix and all linear combinations of them, so that they form a vector space. 
How can I use the Cayley-Hamilton Theorem to give basis for this vector space?
I do not understand this part, should I found the eigenvectors for matrices $A,A^2,0$ and $I$?
Can anyone help on this I do not understand how Cayley-Hamilton can give a basis for this vector.
Thank you

A:

Hint: The matrices $I$ and $A$ are linearly independent.  However, the matrices $I,A,A^2$ are not linearly independent.

