Is phi(C(u,v))=C(phi(u,v,)) a linear transformation?

[cad2blender]

Let phi(u,v)=(u-2v,-v) is this a R^2->R^2 a linear transformation?

I know that there must be two rules that must be met in order to be a linear transformation, after doing the first part, it seems that it may be linear. But I do not know how to show whether or not the second rule is satisfied. Any tips?

 

[Hurkyl]

Let phi(u,v)=(u-2v,-v) is this a R^2->R^2 a linear transformation?

I know that there must be two rules ... But I do not know how to show whether or not the second rule is satisfied. Any tips?

What is the second rule? Have you plugged this problem and the definition of phi into the rule to see what the result would look like? What problems have you had proving it? How far did you get?

 

[cad2blender]

the thing is that I don't know what to do to use the second rule. The first rule says that phi(u1,v1)+(u2,v2)=phi(u1,v1)+phi(u2,v2) must be true, so far I think i got this to work, after some time I got the right side equal to the left side. That is how far I got.

EDIT: 2nd rule is phi(C(u,v))=C(phi(u,v,))

 

[Hurkyl]

the thing is that I don't know what to do to use the second rule.

EDIT: 2nd rule is phi(C(u,v))=C(phi(u,v,))

Substituting what you know is always a good thing to try.

You have this equation you want to prove/disprove is an identity. (i.e. it's true for all values of C and (u,v))

Have you tried plugging in some specific values yet? You might get lucky and find a disproof quickly. Always a good thing to try when considering disproving an identity.

You already know two rules that allow you to rewrite parts of this equation. (in particular, the rule for scalar multiplication of vectors, and the definition of phi)

Applying these rules to rewrite the equation you are studying may or may not turn it into something you understand better. But you won't know until you try it.