# Tuesday November 19th

## Determinants

Let $A\in M_n(R)$, where $R$ is a commutative unital ring.

Given $A = (a_{ij})$, recall that
$$
\det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod a_{i, \sigma(i)}
.$$

This satisfies a number of properties:

- $\det(AB) = \det A \det B$

- $A$ invertible $\implies$ $\det A$ is a unit in $R$

- $A \sim B \implies \det(A) = \det(B)$

- $\det A^t = \det A$

- $A$ is triangular $\implies \det A = \prod a_{ii}$.

### Calculating Determinants

1. **Gaussian Elimination**

	a. $B$ is obtained from $A$ by interchanging rows: $\det B = -\det A$

	b. $B$ is obtained from $A$ by multiplying $\det B = r \det A$

	c. $B$ is obtained from $A$ by adding a scalar multiple of one row to another: $\det B = \det A$.

2. **Cofactors**
	Let $A_{ij}$ be the $(n-1)\times (n-1)$ minor obtained by deleting row $i$ and column $j$, and $C_{ij} = (-1)^{i+j} \det A_{ij}$.

	Then **(theorem)** $\det A = \sum_{j=1}^n a_{ij} C_{ij}$ by expanding along either a row or column.

**Theorem**:
$$
A \mathrm{Adj}(A) = \det (A) I_n
,$$
where $\mathrm{Adj} = (C_{ij})^t$.

If $A\inv$ is a unit, then $A\inv = \mathrm{Adj}(A) / \det(A)$.


### Decomposition of a Linear Transformation:

Let $\phi: V \to V$ be a linear transformation of vector spaces.
and $R = \hom_k(V, V)$.
Then $R$ is a ring.

Let $f(x) = \sum a_j x^j \in k[x]$ be an arbitrary polynomial.
Then for $\phi \in R$, it makes sense to evaluate $f(\phi)$ where $\phi^n$ denotes an $n\dash$fold composition,
and $f(\phi): V \to V$.

**Lemma:**

- There exists a unique monic polynomial $q_\phi(x) \in k[x]$ such that $q_\phi(\phi) = 0$ and $f(\phi) = 0 \implies q_\phi \divides f$.
	$q_\phi$ is referred to as the **minimal polynomial** of $\phi$.

- The exact same conclusion holds with $\phi$ replaced by a matrix $A$, yielding $q_A$.

- If $A$ is the matrix of $\phi$ relative to a fixed basis, then $q_\phi =  q_A$.

*Proof of a and b:*
Fix $\phi$, and define

\begin{align*}
\Gamma: k[x] &\to \hom_k(V, V) \\
f &\mapsto f(\phi)
.\end{align*}

Since $\dim_k V\dual = \dim_k V < \infty$ and $\dim_k k[x] = \infty$, we must have $\ker \Gamma \neq 0$.

Since $k[x]$ is a PID, we have $\ker \Gamma = (q)$ for some $q\in k[x]$.
Then if $f(\phi) = 0$, we have $f(x) \in \ker \Gamma \implies q \divides f$.
We can then rescale $q$ to be monic, which makes it unique.

> Note: for (b), just replace $\phi$ with $A$ everywhere.

$\qed$

*Proof of c:*
Suppose $A = [\phi]_\mathcal{B}$ for some fixed basis $\mathcal B$.

Then $\hom_k(V, V) \cong M_n(k)$, so we have the following commutative diagram:

```{=latex}
\begin{center}
\begin{tikzcd}
{k[x]} \arrow[rr, "\Gamma_\phi"] \arrow[rrdd, "\Gamma_A"] &  & {\hom_k(V, V)} \arrow[dd, "\cong", two heads, hook] \\
                                                          &  &                                                     \\
                                                          &  & M_n(k)
\end{tikzcd}
\end{center}
```
$\qed$

### Finitely Generated Modules over a PID

Let $M$ be a finitely generated module over $R$ a PID.
Then

\begin{align*}
M &\cong F \oplus \bigoplus_{i=1}^n R/(r_i)
\quad r_1 \divides r_2 \divides \cdots r_n \\
M &\cong F \oplus \bigoplus_{i=1}^n R/(p_i^{s_i})
\quad p_i \text{ not necessarily distinct primes. }
.\end{align*}

Letting $R = k[x]$ and $\phi: V\to V$ with $\dim_k V < \infty$,
$V$ becomes a $k[x]\dash$module by defining
$$
f(x) \actson \vector v \definedas f(\phi)(\vector v)
$$

Note that $W$ is a $k[x]\dash$submodule iff $\phi: W \to W$.

Let $v\in V$, and $\generators{v} = \theset{\phi^i(v) \suchthat i = 0,1,2,\cdots}$ is the **cyclic submodule generated by $v$**, and we write $\generators{v} = k[x].v$.

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

1. There exist cyclic $k[x]\dash$submodules $V_i$ such that $V = \bigoplus_{i=1}^t V_i$, where for each $i$ there exists a $q_i: V_i \to V_i$ such that $q_1 \divides q_2 \divides \cdots \divides q_t$.

2. There exist cyclic $k[x]\dash$submodules $V_j$ such that $V = \bigoplus_{j=1}^\nu$ and $p_j^{m_j}$ is the minimal polynomial of $\phi: V_j \to V_j$.

*Proof:*
Apply the classification theorem to write $V = \bigoplus R/(r_i)$ as an invariant factor decomposition.

Then $R/(q_i) \cong V_i$, some vector space, and since there is a direct sum decomposition, the invariant factors are minimal polynomials for $\phi_i: V_i \to V_i$, and thus $k[x]/(q_i)$.

$\qed$

### Canonical Forms for Matrices

We'll look at

- Rational Canonical Form

- Jordan Canonical Form

**Theorem**:
Let $\phi: V\to V$ be linear, then $V$ is a cyclic $k[x]\dash$module and $\phi: V\to V$ has minimal polynomial $q(x) = \sum_j a_j x^j$ iff $\dim V = n$ and $V$ has an ordered basis of the form

$$
[\phi]_{\mathcal{B}} = \left[
\begin{array}{ccccc}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_0 & -a_1 & -a_2 & \cdots & -a_{n-1}
\end{array}
\right]
$$

with ones on the super-diagonal.


*Proof:*

$\impliedby$:

Let $V = k[x].v = \generators{v, \phi(v), \cdots, \phi^{n-1}(v)}$ where $\deg q(x) = n$.
The claim is that this is a linearly independent spanning set.

Linear independence: suppose $\sum_{j=0}^{n-1} k_j \phi^j(v) = 0$ with some $k_j \neq 0$.
Then $f(x) = \sum k_j x^j$ is a polynomial where $f(\phi) = 0$, but this contradicts the minimality of $q(x)$.

But then we have $n$ linearly independent vectors in $V$ which is dimension $n$, so this is a spanning set.

$\implies$:

We can just check where basis elements are sent.
Set $\mathcal{B} = \theset{v, \phi(v), \cdots, \phi^{n-1}(v)}$. Then

\begin{align*}
v &\mapsto \phi(v) \\
\phi(v) &\mapsto \phi^2(v) \\
&\vdots \\
\phi{n-1}(v) &\mapsto \phi^n(v)  = -\sum a_i \phi^i(v) \\
.\end{align*}

$\impliedby$
Fix a basis $B = \theset{v_1, \cdots, v_n}$ and $A = [\phi]_B$, then

\begin{align*}
v_1 &\mapsto v_2 = \phi(v_1) \\
v_1 &\mapsto v_3 = \phi^2(v_1) \\
v_{n-2} &\mapsto v_{n-1} = \phi^2(v_1)
.\end{align*}

and
$$
\phi^n(v) = -a_k v_1 \neq -a_1 \phi(v_1), \cdots -a_{n-1} \phi^{n-1}(v_1)
.$$

Thus $V = k[x].v_1$, since $\dim V = n$ with $\theset{v_1, \phi(v_1), \cdots, \phi^{n-1}(v_1)}$ as a basis.

$\qed$