0
$\begingroup$

I really understand the definition of linear transformation, but I'm not sure about the definition of matrix transformation. Could it be that a matrix transformation is defined as a linear transformation that is defined by: $$ T(x)= Ax $$ where $A$ is a matrix over some field?

$\endgroup$
2
  • 2
    $\begingroup$ i think so :) that's indeed a common definition $\endgroup$
    – gt6989b
    Apr 15, 2016 at 4:20
  • $\begingroup$ That's the most common definition I have heard. $\endgroup$ Apr 15, 2016 at 4:22

1 Answer 1

0
$\begingroup$

In fact, any linear transformation $\mathscr{A}: V \to V$, where $V$ is an arbitrary vector space over a field $F$ has its "matrix transformation" representation. In detail, suppose $\dim(V) = n$, then we can choose a fixed basis $\{v_1, \ldots, v_n\}$ of $V$. Since the transformation is linear and $\{v_1, \ldots, v_n\}$ is a basis of $V$, it can be shown that for each $i \in \{1, \ldots, n\}$, there exist $a_{i1}, \ldots, a_{in} \in F$, such that $$\mathscr{A}v_i = a_{i1}v_1 + \cdots + a_{in}v_n.$$ Define matrix $A$ as: $$A = \begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix}$$ Then for any $v \in V$, suppose its coordinate with respect to $\{v_1, \ldots, v_n\}$ is $x = (x_{1}, \ldots, x_{n})^T$, then it is easily seen that $$\mathscr{A}v = x_1\mathscr{A}v_1 + \cdots + x_n\mathscr{A}v_n = (v_1, \ldots, v_n)Ax. \tag{1}$$

From $(1)$ it can be seen that the space of all linear transformations on $V$ and the space of all $n \times n$ matrices are isometric. In this sense "linear transformation" and "matrix transformation" can be treated equally.

Regarding linear transformation and matrix further, it maybe also worth noting that the matrix representations for one linear transformation $\mathscr{A}$ under different bases are similar.

$\endgroup$
2
  • $\begingroup$ $\mathscr{C} \mathscr{K}$ That's a neat trick for cursive in Latex $\endgroup$ Mar 22, 2020 at 14:25
  • $\begingroup$ "arbitrary [finite dimensional] vector space" $\endgroup$
    – JBL
    Jul 30, 2022 at 23:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .