# Tuesday November 12th

## Equivalence and Similarity

Recall from last time:

If $V, W$ are free left $R\dash$modules of ranks $m,n$ respectively with bases $\beta_v, \beta_w$ respectively, then
$$
\hom_R(V, W) \cong M_{m, n}(R)
.$$

**Definition:**
Two matrices $A, B \in M_{m \times n}(R)$ are **equivalent** iff
\begin{align*}
\exists P \in \GL(m, R),~ \exists Q \in \GL(n, R)
\quad \text{ such that } \quad
A = PBQ
.\end{align*}

**Definition:**
Two matrices $A, B \in M_m(R)$ are **similar** iff
\begin{align*}
\exists P \in \GL(m, R)
\quad \text{ such that } \quad
A = P\inv B P
.\end{align*}

**Theorem:**
Let $T: V\to W$ be an $R\dash$module homomorphism.

Then $T$ has an $m\times n$ matrix relative to other bases for $V, W$ $\iff$
$$
B = P [T]_{\beta_v, \beta_w} Q
.$$

*Proof*:
$\implies$:

Let $\beta_v', \beta_w'$ be other bases.
Then we want $B = [T]_{\beta_v', \beta_w'}$, so just let

\begin{align*}
P = [\id]_{\beta_v', \beta_v} \quad Q = [\id]_{\beta_w, \beta_w'}
.\end{align*}

$\qed$

$\impliedby$:

Suppose $B = P [T]_{\beta_v, \beta_w} Q$ for some $P, Q$.

Let $g: V\to V$ be the transformation associated to $P$, and $h: W \to W$ associated to $Q\inv$.

Then
\begin{align*}
P &= [\id]_{g(\beta_v), \beta_v}  \\
\implies Q\inv &= [\id]_{h(\beta_w), \beta_w} \\
\implies Q &= [\id]_{\beta_w, h(\beta_w)} \\
\implies B &= [T]_{g(\beta_v), h(\beta_w)}
.\end{align*}

$\qed$

**Corollary:**
Let $V$ be a free $R\dash$module and $\beta_v$ a basis of size $n$.

Then $T: V\to V$ has an $n\times n$ matrix relative to $\beta_v$ relative to another basis $\iff$
$$
B = P [T]_{\beta_v, \beta_v} P\inv
.$$

> Note how this specializes to the case of linear transformations, particularly when $B$ is diagonalizable.

## Review of Linear Algebra:

Let $D$ be a division ring.
Recall the notions of rank and nullity, and the statement of the rank-nullity theorem.

Note that we can always factor a linear transformation $\phi: E\to F$ as the following short exact sequence:

$$
0 \to \ker \phi \to E \mapsvia{\phi} \im \phi \to 0,
$$

and since every module over a division ring is free, this sequence splits and $E \cong \ker\phi \oplus \im \phi$.
Taking dimensions yields the rank-nullity theorem.

Let $A\in M_{m, n}(D)$ and define

- $R(A) \in D^n$ is the span of the rows of $A$, and

- $C(A) \in D^m$ is the span of the columns of $A$.

Recall that finding a basis of the **row space** involves doing Gaussian Elimination and taking the rows which have nonzero pivots.

For a basis of the **column space**, you take the corresponding columns in the *original* matrix.

> Note that in this case, $\dim R(A) = \dim C(A)$, and in fact these are always equal.

**Theorem (Rank and Equivalence):**
Let $\phi: V\to W$ be a linear transformation and $A$ be the matrix of $\phi$ relative to $\beta_v, \beta_v'$.

Then $\dim \im \pi = \dim C(A) = \dim R(A)$.

*Proof*:
Construct the matrix $A = [\phi]_{\beta_v, \beta_w}$.

Then $\phi: V \to W$ descends to a map $A: D^m \to D^n$.
Writing the matrix $A$ out and letting $v\in D^m$ a row vector act on $A$ from the *left* yields a column vector $Av \in D^n$.

But then $\im \phi$ corresponds to $R(A)$, and so
$$
\dim \im \phi = \dim R(A) = \dim C(A)
.$$

$\qed$

## Canonical Forms

Let $1 \leq r \leq \min(m, n)$, and define $E_r$ to be the $m\times n$ matrix with the $r\times r$ identity matrix in the top-left block.

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

1. $A$ is equivalent to $E_r \iff \rank A = r$
     - That is, $\exists P,Q$ such that $E_r = PAQ$

2. $A$ is equivalent to $B$ iff $\rank A = \rank B$.

3. $E_r$ for $r = 0, 1, \cdots, \min(m,n)$ is a complete set of representatives for the relation of matrix equivalence on $M_{m, n}(D)$.

Let $X = M_{m, n}(D)$ and $G = \GL_m(D) \cross \GL_n(D)$, then
$$
G \actson X \text{ by } (P, Q) \actson A \definedas PAQ\inv
.$$
Then the orbits under this action are exactly $\theset{E_r \mid 0 \leq r \leq \min(m, n)}$.

*Proof*:
Note that 2 and 3 follow from 1, so we'll show 1.

$\implies$:

Let $A$ be an $m\times n$ matrix for some linear transformation $\phi: D^m \to D^n$ relative to some basis.
Assume $\rank A = \dim \im \phi = r$.
We can find a basis such that $\phi(u_i) = v_i$ for $1 \leq i \leq r$, and $\phi(u_i) = 0$ otherwise.
Relative to this basis, $[\phi] = E_r$.
But then $A$ is equivalent to $E_r$.

$\impliedby$:

If $A = PE_r Q$ with $P, Q$ invertible, then $\dim \im A = \dim \im E_r$, and thus $\rank A = \rank E_r = r$.


How do we do this?
Recall the row operations:

- Interchange rows

- Multiply a row by a unit

- Add one row to another

But each corresponds to left-multiplication by an elementary matrix, each of which is invertible.
If you proceed this way until the matrix is in RREF, you produce $P \prod P_i A$.
You can now multiply on the *right* by elementary matrices to do column operations and move all pivots to the top-left block, which yields $E_r$.

$\qed$

**Theorem:**
Let $A \in M_{m, n}(R)$ where $R$ is a PID.

Then $A$ is equivalent to a matrix with $L_r$ in the top-left block, where $L_r$ is a diagonal matrix with $L_{ii} = d_i$ such that $d_1 \divides d_2 \divides \cdots \divides d_r$.
Each $(d_i)$ is uniquely determined by $A$.