Linear transformations (algebra)

AI Thread Summary
The discussion focuses on proving that the function F defined from R2 to R2 is not a linear transformation, particularly by examining its behavior on the subset V. Participants suggest using examples and counterexamples to illustrate the failure of linearity, emphasizing that a direct proof is preferable to proof by contradiction. There is also a related question about the linear transformation F from R3 to R5 and the implications of vectors being in the kernel of F. The conversation includes clarification on constructing an orthonormal basis for the line y = 3x, highlighting the importance of normalization. Overall, the thread emphasizes understanding the definitions and properties of linear transformations in algebra.
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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 a \in V and b \notin V... 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 a \in V and b \notin V... 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]}?
 
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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!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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