If A has distinct eigenvalues $\lambda_1, \cdots, \lambda_n$ , then,

$\displaystyle f(\lambda_i) = g(\lambda_i), \;\;\; i=1, \cdots, n$

The solution will give rise to $\beta_0, \beta_1, \cdots, \beta_{n-1}$ , then

$\displaystyle f(A) = \beta_0 + \beta_1 A + \cdots + \beta_{n-1} A^{n-1}$

If there are multiple roots (multiplicity = 2), then

$\displaystyle f(\lambda_i) = g(\lambda_i)$	(2)
$\displaystyle \frac{\partial}{\partial \lambda_i} f(\lambda_i) = \frac{\partial}{\partial \lambda_i} g(\lambda_i)$	(3)

Example 1:

If $\displaystyle A = \begin{bmatrix}0 & 0 & -2 \\ 0 & 1 & 0 \\ 1 & 0 & 3 \end{bmatrix}$

then compute the state transition matrix using Caley Hamilton Theorem.

(with multiplicity 2) $\displaystyle ,\; \lambda_2= 2$

Let $\displaystyle f(\lambda) = e^{\lambda t} \;\;$ and $\displaystyle \;\; g(\lambda) = \beta_0 + \beta_1 \lambda + \beta_2 \lambda_2^2$

Then using (2) and (3), we can write