# Linear transformations (algebra)

1. Jan 24, 2010

### Idyllic

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution
Just wondering if i got it right, i dont want to learn anything the wrong way and im hoping to learn lots while doing my assignment so i'd like to get it checked asap.

Last edited by a moderator: May 4, 2017
2. Jan 25, 2010

### blkqi

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!

3. Jan 25, 2010

### Idyllic

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

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: Jan 25, 2010
4. Jan 25, 2010

### blkqi

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!