Is this transformation a linear operator?

Question

T(x,y) = (x^2, y)
Using the fact that T(u + v) = T(u) + T(v) must hold for it to be a linear operator, I reasoned that this transformation is not a linear operator. However, I would like confirmation on this simply because my prof is horrible and I do not know if I applied the property correctly.

Answer 1

Squaring a quantity is obviously not a "linear" operation. In other words, it can be shown that (a+b)² ≠ a² + b²

If we consider u and v to be Cartesian vectors u = , v = , then by addition of vectors:

(u + v) = < x1+x2, y1+y2>.

Thus T becomes,

T(u + v) = <(x1+x2)², (y1+y2)>.

But,

T(u) + T(v) = + = <(x1² + x2²), (y1 + y2)>.

As stated earlier, (a+b)² ≠ a² + b², therefore

T(u + v) ≠ T(u) + T(v)......

Q.E.D.

~WOMBAT

Answer 2

In order for T to be linear you need to proof that T(u+v)=T(u)+T(v)...FOR ALL u,v in T's domain (among other things). Therefore, to proof that your "T" is not linear, you simply need to find vectors u,v such that T(u+v) <> T(u)+T(v). ( <> meaning "different than")

How to find them? Well, there no generic way to find them, just follow your intuition in this one. For example, consider u=v=(1,0) aswell. Then T(u+v)=T(2,0)=(4,0), but T(u)=T(v)=T(1,0)=(1,0). Hence T(u+v) <> T(u)+T(v), and then T is not linear.

Answer 3

Only For all value of y

Answer 4

That is correct, and your reasoning is sound.