91影视

Consider the transformation as follows:

$T:U鈫�V$

Suppose that assume the domain (U) of T be a finite dimension when the image (V) of a linear transformation T is infinite dimensional.

Let dimension of U be spanned by the set $\left\{u_1,u_2,...,u_n\right\}$of dimension n then only every element of u then there exists an element v such that .$T\left(u\right)=v$

Since,$u鈭�U$then :

$$\begin{gathered} u=a_1{u}_1+a_2{u}_2+...+a_n{u}_n \\ T\left(u\right)=T\left(a_1{u}_1+a_2{u}_2+...+a_n{u}_n\right) \\ v=a_{1}T\left(u_1\right)+a_{2}T\left(u_2\right)+...+a_{n}T\left(u_n\right) \end{gathered}$$

Here,$T\left(u_1\right),鈥�T\left(u_2\right),T\left(u_n\right)鈭�V$

Thus, image element of T is a linear combination of finite element of V.

This is a contradiction as the image $\left(V\right)$of linear transformation T is infinite dimensional.

Hence, the assumption is wrong.

Therefore, the given statement 鈥渋f the image of a linear transformation T is infinite dimensional then the domain of T must be infinite dimensional鈥� is true.