# Another linear transformation question

#### KataKoniK

Q: In each case, show that T is not a linear transformation.

T[x y]^T = [0 y^2]^T

A: If X = [0 1]^T then T(2X) = [0 4]^T while 2T(X) = [0 2]^T

I don't quite understand this solution. What are we trying to accomplish here? So, since T(2X) = [0 4]^T while 2T(X) = [0 2]^T do not yeild the same answer, then it's not linear? If the answer was T(2X) = [0 4]^T while 2T(X) = [0 4]^T, then it would be linear?

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#### Muzza

One of the defining characteristics of a linear transformation f (with some given domain/codomain) is that f(kx) = k * f(x) (k is any scalar, x any vector). So if we have a function (like your "T") which /doesn't/ satisfy the previous condition, it can't be linear.

However, just because a function satisfies f(kx) = k * f(x) for some pair k, x doesn't mean it's linear. It may fail to satisfy f(kx) = k * f(x) for some other pair of k, x, or it might not satisfy f(x + y) = f(x) + f(y) (x, y vectors).

Last edited:

#### quasar987

Homework Helper
Gold Member
To check if a transformation is linear, you must verify that the two following properties are satisfied:

1. T(u+v)=T(u)+T(v)

2.T(cu)=cT(u)

That is the definition.

What the author does is that he gives a counter-exemple of 2. I.e. he gives a specific case (u = (0,1)) where 2. fails. And this is sufficient to conclude that T is not linear.

Now I hope you understand that simply verifying that "T(2X) = [0 4]^T while 2T(X) = [0 4]^T" wouldn't prove anything. To conclude that T is linear, you'd have to show that properties 1. AND 2. are satisfied for EVERY imaginable vector.

#### KataKoniK

Understood. Thanks a lot

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