# Tuesday November 26th

## Minimal and Characteristic Polynomials

**Theorem**

a. ? (Todo)

b. **(Cayley Hamilton)** If $p$ is the minimal polynomial of a linear transformation $\phi$, then $p(\phi) = 0$

c. For any $f(x) \in k[x]$ that is irreducible, $f(x) \divides p_\phi(x) \iff f(x) \divides q_\phi(x)$.

*Proof of (a):*
?

$\qed$

*Proof of (b):*

If $q_\phi(x) \divides p_\phi(x)$ and $q_\phi(\phi) = 0$, then $p_\phi(\phi) = 0$ as well.

$\qed$

*Proof of (c):*
We have $f(x) \divides q_\phi(x) \implies f(x) \divides p_\phi(x)$ and $f(x) \divides p_\phi(x) \implies f(x) \divides q_i(x)$ for some $i$, and so $f(x) \divides q_\phi(x)$.

$\qed$

## Eigenvalues and Eigenvectors

**Definition:**
Let $\phi: V\to V$ be a linear transformation. Then

1. An **eigenvector** is a vector $\vector v = \vector 0$ such that $\phi(\vector v) = \lambda \vector v$ for some $\lambda \in k$.

2. If such a $\vector v$ exists, then $\lambda$ is called an **eigenvalue** of $\phi$.

**Theorem:**
The eigenvalues of $\phi$ are the roots of $p_\phi(x)$ in $k$.

*Proof:*
Let $[\phi]_B = A$, then

\begin{align*}
&p_A(\lambda) = p_\phi(\lambda) = \det(\lambda I - A) = 0 \\
&\iff \exists \vector v\neq \vector 0 \text{ such that } (\lambda I - A)\vector v = \vector 0 \\
&\iff \lambda I\vector v = A \vector v \\
&\iff A\vector v = \lambda \vector v \\
&\iff \lambda \text{ is an eigenvalue and } \vector v \text{ is an eigenvector}
.\end{align*}

$\qed$