# Cayley-Hamilton

{% hint style="info" %}

#### Theorem 16

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

$$\Delta(A) = 0$$
{% endhint %}

$$\Delta(\lambda) = |\lambda I - A| = \lambda^n + \sum\_{i=0}^{n-1} c\_i \lambda^i$$

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)$$.

{% hint style="info" %}

#### Theorem 17

$$e^{At} = \sum\_{i=0}^{n-1}\alpha\_i(t)A^i$$

for some $$\alpha\_i$$ which are solutions to the equations

$$e^{\lambda\_jt} = \sum\_{i=0}^{n-1}\alpha\_i(t)\lambda\_j^i.$$
{% endhint %}


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