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Theorem 2.22

$T:V \to V, \dim V$ is finite.

Let $V$ be a finite-dimensional vector space, and $\beta$ ordered basis of $V$.

Let $I_V: V \to V$ is defined as $I_V(x) = x \forall x \in V, $ i.e. $I_V =$ identity matrix. Let $Q = [I_V]$, then

  1. $Q$ is invertible
  2. $Q [ v ]_{ \beta' } = [ \beta ]$

The matrix $Q= [ I_V ]_{ \beta' }^{ \beta }$ is called the change of coordinate matrix from $\beta'$ -coordinate to $\beta$-coordinate.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Since the identity transformation $I_V $ is clearly invertible, its matrix $[ I_V ]_{ \beta' }^{ \beta } = Q$ is also invertible.

(ii) $[ I_v(v) ]_{ \beta } = [ v ]_{ \beta }$

$[ I_V ]_{ \beta' }^ \beta [ v ]_{ \beta' }^{ \beta } = [ v ]_{ \beta } $

$\therefore Q[v]_\beta' = [ v ]_{ \beta }$

Theorem 2.25

$T: V \to V, \dim V $ is finite and $\beta$ and $\beta' $ is ordered basis of $V$.

$$[ T ]_{ \beta' } = \inv{ Q } [ T ]_{ \beta } Q$$

where $Q $ is the change of coordinate matrix from $\beta'$ to $\beta $.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

$I_V T = T I_V $

$[ I_V T ]_{ \beta' }^{ \beta } = [ T I_V ]_{ \beta' }^{ \beta }$

$[ I_V ]_{ \beta' }^{ \beta } [ T ]_{ \beta' } = [ T ]_{ \beta } [ I_V ]_{ \beta' }^{ \beta }$ $Q [ T ]_{ \beta' } = [ T ]_{ \beta } Q$

$\implies [ T ]_{ \beta' } = \inv{ Q } [ T ]_{ \beta } Q$

Corollary

Let $A \in M_{n \times n} (F) $, and let $\gamma $ be an ordered basis for $F^n $, then $[ L_A ]_{ \gamma } = \inv{ Q } A Q $, where $Q$ is the $n \times n $ matrix whose $j $th column is the $j$th vector of $\gamma $.

Definition: Similar Matrices

Let $A $ and $B $ be two $n \times n $ matrices, then $A $ and $B $ are similar if $\exists $ and invertible matrix $Q$ s.t. $B = \inv{ Q } AQ $.

Remarks

Two matrices $T $ with respect to different ordered basis are similar to each other.

Theorem

$A$ is invertible matrix if and only if it is a change of coordinate matrix.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

$\implies$

If $A$ is invertible, so is $L_A$, and therefore the columns of $A$, which span the range of $L_A$, must be linearly independent. Then $A$ is the matrix that changes $\beta$ coordinates into standard coordinates in $F^n$, where $\beta$ is the ordered basis consisting of the columns of $A$.

$\impliedby $

The inverse is the matrix that changes coordinates back again.