Cayley-Hamilton

Theorem 16

Every square matrix $A$ satisfies its own characteristic polynomial if there are no repeated eigenvalues.

$\Delta(A) = 0$

$\Delta(\lambda) = |\lambda I - A| = \lambda^n + \sum_{i=0}^{n-1} c_i \lambda^i$

In the case where $A$ is diagonalizable (i.e $A = P\Lambda P^{-1}$),

$\Delta(A) = P\left[ \Lambda^n + \sum_{i=0}^{n-1}c_i \Lambda^i \right]P^{-1}.$

$\Lambda^n + \sum_{i=0}^{n-1}c_i \Lambda^i$ is itself a diagonal matrix where the jth entry on the diagonal is

$\lambda_j^n + \sum_{i=0}^{n-1}c_i\lambda_j = 0$

since $\lambda_j$ is a root of the characteristic polynomial. Thus $\Delta(A) = P \cdot 0 \cdot P^{-1} = 0$, and

$-A^n = \sum_{i=0}^{n-1}c_iA^i. \qquad (20)$

This also gives us a new way to find $e^{At}$ because by its Taylor series expansion,

$e^{At} = \sum_{k=0}^{\infty} \frac{1}{k!}A^k.$

By Equation 20, all $A^k = A^{n}A^{k-n}$ for $k>n$ can be expressed in terms of the lower powers $A^i$ for $i\in[0, n)$.

Theorem 17

$e^{At} = \sum_{i=0}^{n-1}\alpha_i(t)A^i$

for some $\alpha_i$ which are solutions to the equations

$e^{\lambda_jt} = \sum_{i=0}^{n-1}\alpha_i(t)\lambda_j^i.$