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