Linear transformation (minor clarification)

[negation]

Homework Statement

The Attempt at a Solution

I don't think I'm interpreting the question correctly. Maybe someone can point me in the right direction?

There are 2 conditions: if y =/=0 then f(x,y) = x^2/y and if y=0 then f(x,y) = 0

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

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

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

f(u) + f(v) = 2

f(u+v) = 2

testing for the second condition: if y=0 then f(x,y) = 0

let u = (1,0) v = (1,0)

f(u) = 0 since if y = 0, f(x,y) = 0

f(v) = 0

f(u) + f(v) = 0

f(u+v) = f(2,0) = 0

 

[jbunniii]

Homework Statement

View attachment 68313

The Attempt at a Solution

I don't think I'm interpreting the question correctly. Maybe someone can point me in the right direction?

I assume that the goal is to check whether $f$ is linear. So your approach is fine:

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

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

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

f(u) + f(v) = 2

OK so far.

f(u+v) = 2

No, $u+v = (1,1) + (1,1) = (2,2)$. What do you get when you evaluate $f$ at this point?

[edit] Oops, I miscalculated. Yes, you get 2. So this shows that $f(u+v) = f(u) + f(v)$ for that particular choice of $u$ and $v$. But to conclude linearity, you need to show that it is true for all choices of $u$ and $v$.

Hint: try another choice of $u$ and $v$.

 

[negation]

I assume that the goal is to check whether $f$ is linear. So your approach is fine:

OK so far.

No, $u+v = (1,1) + (1,1) = (2,2)$. What do you get when you evaluate $f$ at this point?

[edit] Oops, I miscalculated. Yes, you get 2. So this shows that $f(u+v) = f(u) + f(v)$ for that particular choice of $u$ and $v$. But to conclude linearity, you need to show that it is true for all choices of $u$ and $v$.

Hint: try another choice of $u$ and $v$.

I presume we are referring to the first condition in the question where x^2/y ?

I tried a few values but they all preserve addition.

 

[jbunniii]

I presume we are referring to the first condition in the question where x^2/y ?

Yes, the $y=0$ case is unlikely to provide a counterexample for additivity.

I tried a few values but they all preserve addition.

You should be able to find a simple counterexample using points of the form $(a,1)$ and $(b,1)$

 

[HallsofIvy]

That "$x^2$" makes me suspicious! So I would try u= (2, 1) and v= (3, 1).

What are f(u) and f(v)? What is f(u)+ f(v)?

u+ v= (5, 2). What is f(u+ v)?

 

[negation]

That "$x^2$" makes me suspicious! So I would try u= (2, 1) and v= (3, 1).

What are f(u) and f(v)? What is f(u)+ f(v)?

u+ v= (5, 2). What is f(u+ v)?

Halls!

Any reason as to why you chose those values for u and v? Are they just arbitrary?

f(u) = f(2,1) = 4 f(v) = f(3,1) = 9
f(u) + f(v) = 13

u + v = (5,1)

f(u+v) = 25

 

[Fredrik]

Any reason as to why you chose those values for u and v? Are they just arbitrary?

Almost any choice of u and v will work, but the calculation of $\frac{(u_1)^2}{u_2}+\frac{(v_1)^2}{v_2}$ is easier when $u_2=v_2$, and is especially easy when $u_2=v_2=1$.

 

[jbunniii]

Halls!

Any reason as to why you chose those values for u and v? Are they just arbitrary?

f(u) = f(2,1) = 4 f(v) = f(3,1) = 9
f(u) + f(v) = 13

u + v = (5,1)

You added $(2,1)$ to $(3,1)$ and got $(5,1)$?