[Math] Cayley-Hamilton theorem-eigenvalues

Question

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 2×2 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

Answer 1

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