Exercise 3.E.4

Answers

In this solution, we are going to create an S which maps from L ( V 1 , W ) × × L ( V m , W ) L ( V 1 × × V m , W )

Define T : V 1 × × V m W as

T ( ( v 1 , . . . , v m ) ) = T 1 v 1 + + T m v m

where T j L ( V j , W ) for j { 1 , . . . , m }

Now define S : L ( V 1 , W ) × × L ( V m , W ) L ( V 1 × × V m , W ) where ( T 1 , . . . , T m ) is mapped to T

To prove S is injective, let l null S Then, v 1 , . . . , v m

T 1 v 1 + + T m v m = 0

Therefore

T 1 = = T m = 0

Therefore null S = { 0 } To prove surjectivity, for any K : V 1 × × V m W , let

T j v j = T ( ( v 1 , 0 , . . . , 0 ) )

For j { 1 , . . . , m } Then T = T 1 + + T m , proving surjectivity. □

User profile picture
2025-05-26 20:16
Comments