# Simple linear transformation

1. Apr 4, 2014

### negation

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

f(x,y) -> |x+y|

3. The attempt at a solution

The answer is that the above transformation is not linear but my working shows otherwise.

Here's my go:

let u = (1,1) and v = (1,1)

f(u) = f(1,1) = 2

f(v) = f(1,1) = 2

f(u) + f(v) = 4

f(u+v) = f(2,2) = 4

f(u+v) = f(u) + f(v)

So a linear transformation exists.

2. Apr 4, 2014

### LCKurtz

The answer to what? You haven't stated a question.

So if it works for those two u and v values, it works for all of them eh?

3. Apr 4, 2014

### negation

Woops, I submitted the thread too fast.

The question asks if the function is a linear transformation.

Not necessarily. How should I demonstrate? counter example?

Edit:

Things are different should I allow u = (1,0) and v = (-1,0)

then,

f(u) = 1 and f(v) = 1, then, f(u) + f(v) = 2

but f((u+v)) = f((1-1,0+0)) = f(0,0) = 0

4. Apr 4, 2014

### LCKurtz

That's the right idea.

5. Apr 4, 2014

### negation

The value for u and v was randomly pluck. It is obvious that in an examination, trial and error is the last thing I want to assume.

Would you mind shedding some light how I should, without trial and error, arrive at the value for u and v such that they satisfy the condition?

6. Apr 4, 2014

### LCKurtz

You should start by writing the general thing you are trying to prove or disprove. In your example it was whether $f(x,y)=|x+y|$ is a linear transformation from $R^2\rightarrow R^1$. So you are checking whether or not if $u=(a,b)$ and $v = (c,d)$ does $f(u+v)=f(u)+f(v)$. Since $u+v = (a+c,b+d)$ this would mean $|(a+b)+(c+d)| = |a+b|+|c+d|$.

You need to do that before you start trying to solve the problem, and this is a step that you are routinely skipping. Once you have it clear what you are proving or disproving you can usually tell whether it is an identity or not just by looking at it. You will be able to see the algebra to make it work or find numbers that show it doesn't work. Once you really understand the concept of "linear" they will usually be easy to tell.