Bijective Linear Transformation Proof

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• Nov 17th 2009, 06:59 PM
amm345
Bijective Linear Transformation Proof
Let
T : V W be a linear transformation that is bijective, that is, it is

injective1 and surjective2. For each w W let T1(w) be the unique v V such that
T(v) = w. v exists because T is surjective and it is unique because T is injective. T1 is characterized by the two conditions3 that T T1 = idW and T1 T = idV .

We obtain a map T1 : W V

Show that T1 is again a linear transformation.
• Nov 17th 2009, 08:05 PM
tonio
Quote:

Originally Posted by amm345
Let
T : V W be a linear transformation that is bijective, that is, it is

injective1 and surjective2. For each w W let T1(w) be the unique v V such that

T(v) = w. v exists because T is surjective and it is unique because T is injective. T1 is characterized by the two conditions3 that T T1 = idW and T1 T = idV .

We obtain a map T1 : W V

Show that T1 is again a linear transformation.

Let $w_1\,,\,w_2\in W\Longrightarrow\,\exists\,\,v_1\,,\,v_2\,\in\,V\, \,s.t.\,\,Tv_1=w_1\,,\,Tv_2=w_2\,\,(why?)$ , so:

As $T(v_1+v_2)=Tv_1+Tv_2=w_1+w_2\,,\,then\,\,T^{-1}(w_1+w_2)=v_1+v_2$ ....take it from here, and very simmilar with multiplication by a scalar.

Tonio