A question on linear maps

  • Thread starter kostas230
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  • #1
kostas230
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Suppose that V and W are finite dimensional and that U is a subspace of V. If dimU≥dimV-dimW prove that there exists a linear map T from V to W such that kerT=U.

My answer is this:

Consider the following linear map:
T|u>=|u> if |u> belongs to V-U and T|u>=|0> if |u> belongs to U​

Therefore kerT=U
Is this correct?
 

Answers and Replies

  • #2
tiny-tim
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hi kostas230! :smile:
Consider the following linear map:
T|u>=|u> if |u> belongs to V-U and T|u>=|0> if |u> belongs to U​

Therefore kerT=U
Is this correct?

nooo …

i] that's a map from V to V, not to W
ii] what do you mean by V-U ? :confused: (eg what is R3 - R2 ?)
 
  • #3
kostas230
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i] that's a map from V to V, not to W

Silly me, I overlooked it xD

ii] what do you mean by V-U ? :confused: (eg what is R3 - R2 ?)

I mean the elements of V that do not belong in U.
 
  • #4
tiny-tim
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I mean the elements of V that do not belong in U.

then R3 - R2 would include all elements with z ≠ 0 :confused:
 
  • #6
tiny-tim
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Well, the correct answer would be: ℝ^3-ℝ^2={(0,0,z):z is real

that's a line

add a line to R2 and you don't get R3 :redface:
 
  • #7
HallsofIvy
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R2 direct sum a line gives R3 but that is not at all what you said!


R3- R2 is all (x, y, z) such that [itex]z\ne 0[/itex] which is not a subspace.
I think you need to review your definitions!
 

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