Quick question about Linear Transformations from a space to itselfby Fractal20 Tags: linear, space, transformations 

#1
Aug612, 02:49 PM

P: 74

Hi, I have to take a placement exam in linear algebra this fall so I have been studying some past exams. This is a real basic question. If we have a linear transformation T:W > W does this imply nothing about the injectivity or surjectivity of the transformation? I assume that it does not, but I get confused because if it is not surjective, then it seems like the image of T is not W but some subspace of W.
To phrase it in a different way, does T: W > W only say that T maps vectors in W to other vectors in W and nothing about what the image in W is? Thanks! 



#2
Aug612, 03:09 PM

P: 606

Yes, you can say almost nothing about T without more information, but you can always be sure that if [itex]\,\dim W<\infty\,[/itex] , then T is 11 iff T is onto. 



#3
Aug712, 06:53 PM

Sci Advisor
HW Helper
P: 9,421

Maybe one can say something if the scalar field is the complexes, namely there is at least one subspace of dimension one that maps into itself, which is actually another version of micromass's comment.




#4
Aug712, 09:51 PM

Sci Advisor
P: 1,168

Quick question about Linear Transformations from a space to itselfT_{i}: ℝ^{4}→ℝ^{4} ; i=1,2,..,5: T_{1}(x1,x2,x3,x4)=(0,0,0,0) . What is the dimension of the image T_{2}(x1,x2,x3,x4)=(x1,x1,x1,x1). Is it 11 ? T_{3}(x1,x2,x3,x4)=(x1,x2,0,0). Onto? T_{4}(x1,x2,x3,x4)=(x1,x2,x3,0) . Onto? 



#5
Aug812, 12:06 PM

P: 74

Thanks, that makes it obvious. I guess I assumed as much but felt unsure in the sense of specificity of the image. ie for T^{1} it is really mapped to 0. So if W = the vector space spanned by x1, ... , x5 then in some ways I feel like saying T^{1} W > W is misleading since it is really T: W > 0. But I understand that it is about generality. 


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