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