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Example of a linear transformation L which is injective but not surj, or vice versa

  • Thread starter damabo
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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 finitedimensional 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.
 

Answers and Replies

  • #2
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(continued)
2. example of an injective, but not surjective function:
how about I, if I is the integration function, such that I: a_0+a_1X+a_2X^2 ->a_0X + a_1(X^2)/2+ a_2(X^3)/3 + ...
not surjective: is there a p in ℝ[X] such that there is no p_0 in ℝ[X] so that I(p_0)=p (or equivalently D(p)=p_0)?
In other words, is there a p whose derivative is not a real polynomial?
I don't think this is possible.
 
  • #3
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Let [itex]V=\mathbb{R}^2[/itex]. Every vector in such space can be described as [itex]\mathbf{v}=(r\cos{\phi},r\sin{\phi})[/itex], where [itex]r\in[0;\infty)\: \wedge \:\phi\in[0;2\pi)[/itex]. Let:
[itex]L:V\rightarrow V;\quad L((r\cos{\phi},r\sin{\phi}))=(r\cos{\frac{\phi}{2}},r\sin{\frac{\phi}{2}})[/itex]
Due to function [itex]f(x)=\frac{x}{2}[/itex] being injective, [itex]L[/itex] is also injective. Furthermore, because none of [itex]L(\mathbf{v})[/itex] have [itex]\phi > \pi[/itex], [itex]L[/itex] is not surjective.
 
  • #4
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none of the L(v) have [itex]\phi[/itex]/2 > [itex]\pi[/itex] ?
 
  • #5
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Ah, yes, sorry, I've written it too quickly. More precisely: none of vectors from [itex]L[/itex] image has angle greater than [itex]\pi[/itex]. Basically the image is a half-plane.
 
  • #6
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ah ok. Get it.
 

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