Linear transformations (algebra)

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Homework Help Overview

The discussion revolves around the properties of linear transformations, specifically examining a function defined on subsets of R². The original poster presents a function F and seeks to prove that it is not a linear transformation, while also exploring related concepts in linear algebra.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the necessity of generalizing proofs and the implications of one-dimensional subspaces. There are attempts to use counterexamples to demonstrate the non-linearity of F. Questions arise regarding the implications of specific vector combinations and their relationships to the kernel of the transformation.

Discussion Status

The discussion is active, with participants providing feedback on proof strategies and exploring various interpretations of linear transformations. Some guidance is offered regarding the use of counterexamples and the nature of direct versus indirect proofs.

Contextual Notes

Participants are navigating the constraints of homework expectations, including the need to avoid learning incorrect methods and the challenge of proving statements about linear transformations. There is mention of specific vector relationships and the properties of orthonormal bases in relation to the line y = 3x.

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Homework Statement



Let V be a subset of R2 and some fixed 1-dimensional subspace of R2.

F:R2->R2 by F(v) = v if v is in V, 0 otherwise

Prove that F is not a linear transformation.

Homework Equations





The Attempt at a Solution


Just wondering if i got it right, i don't want to learn anything the wrong way and I am hoping to learn lots while doing my assignment so i'd like to get it checked asap.

http://www.nm3210.com/upload2/files/7/maths.jpg
 
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That's the idea. You could have been a little more general in the proof, for instance:

Let [tex]a \in V[/tex] and [tex]b \notin V[/tex]... blah blah.

The fact that u is one-dimensional is a necessary but not sufficient condition for u in V. There are many one-dimension subsets of R!
 
blkqi said:
That's the idea. You could have been a little more general in the proof, for instance:

Let [tex]a \in V[/tex] and [tex]b \notin V[/tex]... blah blah.

The fact that u is one-dimensional is a necessary but not sufficient condition for u in V. There are many one-dimension subsets of R!

I think it should be ok, to disprove something you just need to show that it is false by example, so i just chose a convienient u vector.

I have another question:

Let F: R3 -> R5 be a L.T. Let {u1, u2, u3} be a lin. indep. set of vectors in R3.

If u1 + u2 is in ker(F) and F(u1) = v =/ 0. What is F(u2)?

I did F(u1 + u2) = 0
F(u1) + F(u2) = 0
v + F(u2) = 0
F(u2) = -v

And if u1 + u2 + u3 is in ker(F). what is F(u3)?

I got u3 = 0

Is this ok? Thanks for your reply.Also, what would an orthonormal basis look like for the line y = 3x?
Would it be {[1, 0], [0, 3]} or {1/(10^0.5)*[1, 3]}?
 
Last edited:
Counter example is fine if you are going for proof by contradiction. Remember the statement you want to demonstrate is that

"F is not a linear transformation."

In proof by contradiction you can assume that F is a linear transformation, and use a counterexample to draw the contradiction (just like you did). Though it's widely used proof by contradiction is not as strong as a direct proof. In fact it is usually just a contrapositive proof in disguise.

The direct proof would derive "F is not a linear transformation" from the definition of F directly, for arbitrary V. But I'm being picky, what you did is fine! One of my old math instructors hated proof by contradiction and its stuck with me.The other stuff you did looks good. The line y=3x is just one dimensional so its basis should be just one vector. Find something on the line, like [1,3] and normalize it by dividing by its magnitude sqrt(1^2+3^2)=sqrt(10), just like you did! That way your basis is a minimal generating set for the space given by y=3x. Notice that your first basis spans more than just y=3x!
 

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