Is this method of finding linear transformation correct?

Question

The method of linear transformation I have been using is:

\begin{aligned} & \text { (The Question I will be using to show my method:) } T(x, y)=(x+1, y) \\ & T(u+v)=T\left(x_1+x_2, y_1+y_2\right) \\ & =\left(x_1+x_2+1+1\right),\left(y_1+y_2\right) \\ & =\left(x_1+1, y_1\right),\left(x_2+1, y_2\right) \\ & T(u)+T(v) \\ & T\left(x_1, y_1\right)+T\left(x_2, y_2\right) \\ & =\left[x_1+1, y_1\right]+\left[x_2+1, y_2\right] \\ & \end{aligned}

What I have done here is I found out the value for T(u+v) and then proved it to T(u) + T(v)

Q1: Is my method correct & do there needs to be some adjustments ?

Q2: Every question that i solve using this method comes out to be a linear transformation. So , when do i not get a linear transformation ? It would be great if u could give a example

Answer 1

Note that a linear transformation $T: V \to W$ must satisfy the following:

1. $T(x + y) = T(x) + T(y)$ for each $x, y \in V.$
2. $T(\lambda x) = \lambda T(x)$ for each $\lambda \in \mathbb{F},$ where $\mathbb{F}$ is the parent field of the vector space $V.$

With these, we have the fact that $T(0) = 0,$ that is, zeros map to zeros.

The map $T(x, y) = (x+1, y)$ is additive but is certainly not homogeneous, for $T(\lambda(x, y)) = (\lambda x + 1, \lambda y) \neq \lambda(x + 1, y).$ So, this map is not linear.

You can also quickly check that it fails to be linear since $T((0, 0)) = (1, 0).$ I remark that this map $T$ is an affine map, but not linear.

Answer 2

Let's consider your function $T$, where $T(x,y) = (x+1,\, y)$. This definition implicitly tells us that both the domain and codomain of $T$ are two-dimensional. For concreteness, let's suppose the field of coefficients is the real numbers (but any field will do). A common notation for this situation is

$$T: \mathbb{R}^2 \to \mathbb{R}^2$$ $$ (x, y) \mapsto (x + 1, y).$$

This function is not a linear transformation. By definition, every linear transformation is required to exhibit two properties:

1. $T(u + v) = T(u) + T(v)$ for all vectors $u$ and $v$
2. $T(\lambda u) = \lambda T(u)$ for all vectors $u$ and scalars $\lambda$

Let us say that a function that satisfies property 1. is called additive. Let us say further that a function that satisfies property 2. is called homogeneous. A linear transformation must be both. Your function $T$ is neither.

In your question, you give a flawed proof of additivity. Let $u = (x_1, y_1)$ and let $v = (x_2, y_2)$ be any two vectors in $\mathbb{R}^2$. Then $T(u + v) = $ $$T((x_1, y_1) + (x_2, y_2)) = T(x_1 + x_2, y_1 + y_2) = (x_1 + x_2 + 1, y_1 + y_2).$$

On the other hand, $T(u) + T(v) = $ $$T(x_1, y_1) + T(x_2, y_2) = (x_1 + 1, y_1) + (x_2 + 1, y_2) = (x_1 + x_2 + 2, y_1 + y_2).$$

Your function $T$ is also not homogeneous. For suppose $\lambda$ is a real number and $u = (x, y)$ is a vector. Then $T(\lambda u) = $

$$T(\lambda x, \lambda y) = (\lambda x + 1, \lambda y).$$

On the other hand, $\lambda T(u) = $ $$\lambda T(x, y) = \lambda (x + 1, y) = (\lambda x + \lambda, \lambda y).$$

With experience, a function like $T$ can immediately be classified as non-linear. The giveaway is the "+ 1" in the definition. If a function $T: \mathbb{R}^2 \to \mathbb{R}^2$ is linear, then there exist real numbers $a, b, c, d$ such that for every vector $(x, y)$, $T(x, y) = (ax + by, cx + dy)$. See how both components of the output are linear combinations of the components of the input? That's why we call these things linear transformations.