Coordinate Representation

There are more than one way to represent a basis.

Def: Coordinate transformation

Given a fdvs V with a (ordered) basis $B = {b_{1}, b_{2}, . ., b_{n}}$ , the coordinate transformation on V is the map $C_{B} : V \to R^{n}$ that maps $w \in V$ to a vector composed of its unique linear combinations with respect to $B$ .

$C_{B} (w) = [c_{1}, c_{2}, \dots, c_{n}]$ where $w = \sum_{i}^{n} c_{i} b_{i}$

Properties:

$C_{B}$ is a linear map and is an isomorphism
$C_{B} (b_{i}) = {\hat{e}}_{i}$
Basis are ordered

Coordinate Representation and Linear Transformations

\begin{matrix} V \overset{T}{\to} W \\ ↓ C_{B} ↓ C_{D} \\ R^{n} \overset{T_{A}}{\to} R^{m} \end{matrix}

Main Identities:

$T_{A} \circ C_{B} = C_{D} \circ (T)$

pugging $v \in V$ in we get:

LHS: $(T_{A} \circ C_{B}) (v) = T_{A} (C_{B} (v))$
RHS: $(C_{D} \circ T) (v) = C_{D} (T (v))$

other notation: $C_{B} (v) = [v]_{B}$ and $T_{A} (x) = A x$ so $A [v]_{B} = [T (v)]_{D}$

Matrix Representation

Suppose $V, W$ are fdvs with bases $B$ and $D$ respectively. If $T : V \to W$ is linear then the matrix representation of $T$ in the two bases is the matrix A, $T_{A} : R^{dim (V)} \to R^{dim(W)}$ where we write $A = [T]_{B}^{D}$ .

We also define $M_{B}^{D} : L (V, W) \to M_{dim(W) \times d i m (V)} (R)$ where $L (V, W) = {T : V \to W : T is linear}$ .
(or put otherwise $M_{B}^{D} : T \to [T]_{B}^{D}$ )

To compute $[T]_{D}^{B} = [[T (d_{i})]_{B}]_{i = 1}^{dim (V)}$ where $b_{i}$ $i = 1, 2, \dots, dim (V)$ is a basis of $V$

Thm: Transformation Composition

If $U, V, W$ are fdvs with bases $B, D, and F$ respectively and $S : V \to W$ and $T : U \to V$ are linear maps. Then

$[S \circ T]_{B}^{F} = [S]_{D}^{F} [T]_{B}^{D}$

pf:
$C_{F} \circ S \circ T =$
$= (C_{F} \circ S) \circ (T) = (T_{[S]_{D}^{F}} \circ C_{D}) \circ T = T_{[S]_{D}^{F}} \circ (C_{D} \circ T) =$
$= T_{[S]_{D}^{F}} \circ T_{[T]_{B}^{D}} \circ C_{B} = T_{[S \circ T]_{B}^{F}} \circ C_{B}$

Corollary
If S: V $\to$ W is an isomorphism of fdvs and B, D are bases for V and W respectively, then:

$[S^{- 1}]_{D}^{B} = ([S]_{B}^{D})^{- 1}$

pf: S is an isomorphism so there exists a $S^{- 1} S = I_{V}$
$[S^{- 1} \circ S]_{B}^{B} = [S^{- 1}]_{D}^{B} [S]_{B}^{D}$
$[I_{V}]_{B} = [S^{- 1}]_{D}^{B} [S]_{B}^{D}$
$I_{n} = [S^{- 1}]_{D}^{B} [S]_{B}^{D}$

Change of Basis

Def: Change of Basis
If V is fdvs with bases B and D, the change of basis matrix is $P_{D}^{B} = [{id}_{v}]_{B}^{D}$ and $(P_{D}^{B})^{- 1} = P_{B}^{D}$

Identity
$T_{P_{B}^{D}} \circ T_{[S]_{B}} = T_{[S]_{D}} \circ T_{P_{B}^{D}}$

pf:
$T_{P_{B}^{D}} \circ T_{[S]_{B}} \circ C_{B} = T_{P_{B}^{D}} \circ C_{B} \circ S = C_{D} \circ S = T_{[S]_{D}} \circ C_{D} =$
$= T_{[S]_{D}} \circ T_{P_{B}^{D}} \circ C_{B}$

Thm: Matrix Representation of Similar Linear Transformations

If A,B $\in M_{n} (R)$ and $T_{A} : R^{n} \to R^{n}$ s.t x $\mapsto$ Ax and A is similar to B, then there is a basis D of $R^{n}$ such that $[T]_{D} = B$

Let $ϵ$ be the standard basis. $[T_{A}]_{ϵ} = A$

Want a basis so $B = P A P^{- 1} = [T]_{D}$
$[T]_{D} = P_{ϵ}^{D} [T]_{ϵ} P_{D}^{ϵ} = P_{ϵ}^{D} A P_{D}^{ϵ}$

So let $P^{- 1} = P_{D}^{ϵ}$ so that the basis of D is the columns of $P^{- 1}$