# Thursday November 14th

## Equivalence to Canonical Forms

Let $D$ be a division ring and $k$ a field.

Recall that a matrix $A$ is *equivalent* to $B \iff \exists P, Q$ such that $PBQ=A$.
From a previous theorem, if $\rank(A) = r$, then $A$ is equivalent to a matrix with $I_r$ in the top-left block.

**Theorem:**
Let $A$ be a matrix over a PID $R$.
Then $A$ is equivalent to a matrix with $L_r$ in the top-left corner, where $L_r = \mathrm{diag}(d_1, d_2, \cdots, d_r)$ and $d_1 \divides d_2 \divides \cdots \divides d_r$, and the $d_i$ are uniquely determined.

**Theorem:**
Let $A$ be an $n\times n$ matrix over a division ring $D$.
TFAE:

1. $\rank A = n$.

2. $A$ is equivalent to $I_n$.

3. $A$ is invertible.

$1\implies 2$: Use Gaussian elimination.

$2\implies 3$: $A = PI_n Q = PQ$ where $P, Q$ are invertible, so $PQ = A$ is invertible.

$3\implies 1$: If $A$ is invertible, then $A: D^n \to D^n$ is bijective and thus surjective, so $\dim \im A = n$.

> Note: the image is now *row space* because we are taking *left* actions.

$\qed$

## Determinants

**Definition:**
Let $M_1, \cdots, M_n$ be $R\dash$modules, and then $f: \prod M_i \to R$ is $n\dash$linear iff

\begin{align*}
f(
m_1, m_2, \cdots, rm_k + sm_k', \cdots, m_n
) = \\
r f(
	m_1, \cdots, m_k, \cdots m_k) + sf(m_1, \cdots, m_k', \cdots, m_n
)
.\end{align*}

*Example:*
The inner product is a 2-linear form.

**Definition:**
$f$ is **symmetric** iff
$$
f(m_1, \cdots, m_n) = f(m_{\sigma(1)}, \cdots, m_{\sigma(n)}) ~~\forall \sigma \in S_n
.$$

**Definition:**
$f$ is **skew-symmetric** iff
$$
f(m_1, \cdots, m_n) = \mathrm{sgn}(\sigma) f(m_{\sigma(1)}, \cdots, m_{\sigma(n)}) ~~\forall \sigma \in S_n
,$$

where

$$
\mathrm{sgn}(\sigma) =
\begin{cases}
1 & \sigma \text{ is even } \\
-1 & \sigma \text{ is odd }
\end{cases}
.$$

**Definition:**
$f$ is **alternating** iff

$$
m_i = m_j \text{ for some pair } (i, j) \implies f(m_1, \cdots, m_n) = 0
.$$

**Theorem:**
Let $f$ be an $n\dash$linear form.
If $f$ is alternating, then $f$ is skew-symmetric.

*Proof:*
It suffices to show the $n=2$ case.
We have

\begin{align*}
0
&= f(m+1 + m_2, m_1 + m_2) \\
&= f(m_1, m_1) + f(m_1, m_2) + f(m_2, m_1) + f(m_2, m_2) \\
&= f(m_1, m_2) + f(m_2, m_1)\\
\implies f(m_1, m_2) &= - f(m_2, m_1)
.\end{align*}

$\qed$

**Theorem:**
Let $R$ be a unital commutative ring and let $r\in R$ be arbitrary.

Then
$$
\exists! f: \bigoplus_{i=1}^n R^n \to R
,$$

where $f$ is an alternating $R\dash$form such that $f(\vector e_i) = r$ for all $i$, where $\vector e_i = [0, 0, \cdots, 0, 1, 0, \cdots, 0, 0]$.

> $R^n$ is a free module, so $f$ can be identified with a matrix once a basis is chosen.

*Proof*:

*Existence:*
Let $x_i = [a_{i1}, a_{i2}, \cdots, a_{in}]$ and define
$$
f(x_1, \cdots, x_n) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) r \prod_i a_{i \sigma(i)}
.$$

*Exercise:*
Check that $f(\vector e_1, \cdots, \vector e_n) = r$ and $f$ is $n\dash$linear.

Moreover, $f$ is alternating.
Consider $f(x_1, \cdots, x_n)$ where $x_i = x_j$ for some $i\neq j$.

Letting $\phi = (i, j)$, we can write $S_n = A_n \disjoint A_n \rho$.

If $\sigma$ is even, then the summand is
$$
(+1)r a_{1\sigma(1)} \cdots a_{n\sigma(n)}
.$$

Since $x_i = x_j$, we'll have $\prod_k a_{ik} = \prod a_{jk}$.
Then consider applying $\sigma \rho$.
We have

\begin{align*}
-r \prod a_{i\sigma(i)}
&= -r a_{1\sigma(1)} \cdots \mathbf{a}_{j \sigma(j)} \cdots \mathbf{a}_{i \sigma(i)} \cdots a_{n, \sigma(n)} \\
&=
-r \prod a_{i\sigma(i)}
= -r a_{1\sigma(1)} \cdots \mathbf{a}_{i \sigma(i)} \cdots \mathbf{a}_{j \sigma(j)} \cdots a_{n, \sigma(n)}
,\end{align*}

which permutes the $i,j$ terms.
So these two terms cancel, the remaining terms are untouched.

*Uniqueness*:
Let $x_i = \sum_j a_{ij} \vector e_j$.
Then

\begin{align*}
f(x_1, \cdots, x_n)
&= f(\sum_{j_1} a^1_j \vector e_j, \cdots, \sum_{j_n} a^n_j \vector e_j) \\
&= \sum_{j_1} \cdots \sum_{j_n} f(\vector e_{j_1}, \cdots, \vector e_{j_n} ) a_{1, j_1} \cdots a_{n, j_n} \\
&= \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) f(\vector e_1, \cdots, \vector e_n) a_{1, \sigma(1)} \cdots a_{n, \sigma(n)} \\
&= \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) r a_{1, \sigma(1)} \cdots a_{n, \sigma(n)}
.\end{align*}

$\qed$

**Definition:**
Let $R$ be a commutative unital ring and define $\mathrm{det}: M_n(R) \to R$ is the unique $n\dash$alternating form with $\det(I) = 1$, and is called the *determinant*.

**Theorem:**
Let $A, B \in M_{n}(R)$. Then

a. $\abs{AB} = \abs A \abs B$

b. $A$ is invertible $\iff \abs{A} \in R\units$

c. $A \sim B \implies \abs A = \abs B$.

d. $\abs{A^t} = \abs A$.

e. If $A$ is triangular, then $\abs A$ is the product of the diagonal entries.

*Proof of a:*
Let $B$ be fixed.

Let $\Delta_B: M_n(R) \to R$ be defined as $C \mapsto \abs{CB}$.
Then this is an alternating form, so by the theorem, $\Delta_B = r \mathrm{det}$.
But then $\Delta_B(C) = r\abs{C}$, so $r\abs{C} = \abs{CB}$.
So pick $C = I$, then $r = \abs{B}$.

$\qed$

*Proof of b:*
Suppose $A$ is invertible.

Then $AA\inv = I$, so $\abs{AA\inv} = \abs{A}\abs{A\inv} = 1$, which shows that $\abs A$ is a unit.

$\qed$

*Proof of c:*
Let $A = PBP\inv$.
Then
$$
\abs A = \abs{PBP\inv} = \abs P \abs B \abs{P\inv} = \abs{P} \abs{P\inv} \abs B = \abs B
.$$

$\qed$

*Proof of d:*
Let $A = (a_{ij})$, so $B = (b_{ij}) = (a_{ji})$.
Then

\begin{align*}
\abs{A^t}
&= \sum_{\sigma} \mathrm{sgn}(\sigma) \prod_k b_{k \sigma(k)} \\
&= \sum_\sigma \mathrm{sgn}(\sigma) \prod_k a_{\sigma(k) k} \\
&= \sum_{\sigma\inv} \mathrm{sgn}(\sigma) \prod_k a_{k \sigma\inv(k)} \\
&= \sum_\sigma \mathrm{sgn}(\sigma) \prod_k a_{k \sigma(k)} \\
&= \abs {A}
.\end{align*}

$\qed$

*Proof of e:*
Let $A$ be upper-triangular.
Then
$$
\abs A = \sum_\sigma \mathrm{sgn}(\sigma) \prod_k a_{k \sigma(k)} = a_{11} a_{22} \cdots a_{nn}
.$$

$\qed$

Next time:

- Calculate determinants
	- Gaussian elimination
	- Cofactors
- Formulas for $A\inv$
- Cramer's rule