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## Homework Statement

Give an example of a linear vector space V and a linear transformation L: V-> V that is 1.injective, but not surjective (or 2. vice versa)

## Homework Equations

-If L:V-> V is a linear transformation of a

**finite**dimensional vector space, then

L is surjective, L is injective and L is bijective are equivalent

- ker(L)={0} is equivalent with L is injective (in which case ker(L) is 0-dimensional)

-dimV=dim(ker(L))+dim(Im(L)), if (ℝ,V,+),(ℝ,W,+) finite dimensional vector spaces and L maps V on W linearly.

## The Attempt at a Solution

I guess that the word finite is the most relevant, so that the equation of the dimensions does not apply for infinitedimensional vector spaces.

an example of an infinitedimensional vectorspace would be (ℝ,ℝ[X],+) if ℝ[X] is a polynomial of infinite degree, with a basis β={1,X,X^2,...,X^n,...}.

so suppose L:ℝ[X]->ℝ[X] is a transformation.

1. example of surjective, but not injective function:

Perhaps D if D is the differential operator such that a_0+a_1X+a_2X^2+... -> a_1 + 2a_2X + ...

not injective: clearly 1 != 2, yet D(1)=D(2)

surjective: choose p in ℝ[X]. p=a_1+2a_2X+3a_2X^2... for certain a_i in ℝ. is there a p_0 in ℝ[X] such that D(p_0)=p? indeed, choose a_0 randomly, and choose a_1,a_2,a_3,... in ℝ. define p_0= a_0+a_1X+a_2X^2+.... then clearly p_0 [itex]\in[/itex] ℝ[X], and D(p_0)=p.