Definition

A map $T$ from a vector space $V$ to a vector space $W$, $T: V \rightarrow W$ is linear for all $v_1, v_2 \in V$ and $c \in \mathbf{R}$ if

$T(v_1+v_2) = T(v_1) + T(v_2))$
$T(cv_1) = cT(v_1)$

Remark: The two conditions can be combined together and so $T: V \rightarrow W$ is linear above if and only if $T(v_1 + cv_2) = T(v_1) + cT(v_2)$ for all $v_1, v_2 \in V$ and $c \in \mathbf{R}$.

Proof:
$\Rightarrow$: Assume $T$ is linear. Then,

$$ \begin{align*} T(v_1 + cv_2) &= T(v_1 + cv_2) \\ &= T(v_1) + T(cv_2) \text{ (By property (1))} \\ &= T(v_1) + cT(v_2) \text{ (By property (2))} \end{align*} $$

$\Leftarrow$: Assume $T(v_1 + cv_2) = T(v_1) + cT(v_2)$. Then to show that property (1) is true, notice that

$$ \begin{align*} T(v_1 + v_2) &= T(v_1 + (1)v_2) \\ &= T(v_1) + (1)T(v_2) \\ &= T(v_1) + T(v_2). \end{align*} $$

And to see that property (2) is true, notice that

$$ \begin{align*} T(cv_1) &= T(bar{0}_V + cv_1) \\ &= T(\bar{0}_V) + cT(v_1) \end{align*} $$

To finish the proof we want to additionally show that $T(\bar{0}_V) = \bar{0}_W$. How do we do this? We can only use the assumption that $T(v_1 + cv_2) = T(v_1) + cT(v_2)$. To do this notice that,

$$ \begin{align*} \bar{0}_W + T(\bar{0}_V) &= T(\bar{0}_V) \text{ (We're just adding the zero vector)} \\ &= T(\bar{0}_V + (1)\bar{0}_V) \text{ (using (c))} \\ &= T(\bar{0}_V) + (1)T(\bar{0}_V) \text{ (using (c))} \\ &= T(\bar{0}_V) + T(\bar{0}_V) \\ \bar{0}_W &= T(\bar{0}_V) \end{align*} $$

Another way to do this is the following

$$ \begin{align*} T(\bar{0}_V) &= T(v_1 - v_1) \\ &= T(v_1 + (-1)v_1) \\ &= T(v_1) + (-1)T(v_1) \\ &= \bar{0}_W \end{align*} $$

Remark 2: Suppose we have a linear transformation $T: V \rightarrow W$

$$ \begin{align*} T(a_1u_1 + ... + a_ku_k) &= T(a_1u_1 + ... + a_{k-1}u_{k-1}) + a_kT(u_k) \text{ Using property (3)}\\ &= a_1T(u_1) + ... + a_kT(u_k) \end{align*} $$

This is crucial because this says that the image of a linear combination with coefficients $a_1, ... a_k$ is again a linear combination in the new vector space with coefficients $a_1,...a_k$ except that it’s a linear combination of the image of the original vectors.

Example 1

For $V, W$, the map

$$ \begin{align*} T_0: \ &V \rightarrow W \\ &u \rightarrow \bar{0}_W. \end{align*} $$

is linear.

We need to verify that it is linear by verifying $T(v_1 + cv_2) = T(v_1) + cT(v_2)$. This is easy because for any vectors $v_1, v_2$,

$$ \begin{align*} T_0(v_1 + cv_2) &= \bar{0}_W. \end{align*} $$

Moreover, we also have

$$ \begin{align*} T_0(v_1) + T_0(cv_2) &= \bar{0}_W + c\bar{0}_W \\ &= \bar{0}_W \end{align*} $$

The two sides are equal and so $T_0$ is linear.

Example 2

The map

$$ \begin{align*} I_V: \ &V \rightarrow V \\ &u \rightarrow u. \end{align*} $$

is linear as well.

Example 3

The map

$$ \begin{align*} T: \ &\mathbf{R}^2 \rightarrow \mathbf{R}^2 \\ &(x,y) \rightarrow (-y,x) \end{align*} $$

is linear as well. (a rotation by 90 degrees, todo: add pic). To see why it’s linear, notice that

$$ \begin{align*} T((x_1,y_1) + c(x_2,y_2)) &= T((x_1 + cx_2, y_1 + cy_2)) \text{ (add the two vectors to get one vector)}\\ &= (-y_1-cy_2, x_1+cx_2). \end{align*} $$

Moreover, notice that

$$ \begin{align*} T((x_1,y_1)) + T((cx_2, cy_2)) &= (-y_1, x_1) + (-cy_2, cx_2) \\ &= (-y_1-cy_2, x_1 + cx_2). \end{align*} $$

Both sides are equal and so the transformation is linear.

Example 4

The map

$$ \begin{align*} T: \ &P \rightarrow P \\ &f(x) \rightarrow f'(x). \end{align*} $$

is linear. Note here that the map

$$ \begin{align*} T: \ &P \rightarrow P \\ &f(x) \rightarrow f'(x). \end{align*} $$

is different because the domain and codomain are different here! this is crucial.

Example 6

Let $A \in M_{m \times n}$. The map

$$ \begin{align*} L_A: \ &\mathbf{R}^n \rightarrow \mathbf{R}^m \\ &\bar{x} \rightarrow A\bar{x}. \end{align*} $$

is linear. Remember here $A\bar{x}$ are the linear combinations of the column vectors of $A$ with the coefficients being the entries of $\bar{x}$. The crucial thing here is that if $V$ and $W$ are both finite dimensional, then the map can be represented with this kind of transformation (matrix).

Example 6

For $a < b$, define the map

$$ \begin{align*} T_a^b: \ &C^0(\mathbf{R}) \rightarrow \mathbf{R} \\ &f \rightarrow \int_a^b f(x)dx. \end{align*} $$

($C^0$ is the set of continuous functions on $\mathbf{R})$).

Recall the dimension of $\mathbf{R}$ is 1 and the dimension of $C^0$ is infinte because the set of all polynomials (which has dimension infinity) is a subset of the set of continuous functions. Therefore, the set of continuous function has dimension infinity as well. This mapping goes from an infinite dimensional space to a finite dimensional space.

To prove that this mapping in linear, we notice that

$$ \begin{align*} T_a^b(f + cg) &= \int_a^b (f(x) + cg(x))dx \\ &= \int_a^b f(x)dx + c \int_a^b g(x)dx \ \text{ (By Calculus)} \\ &= T_a^b(f) + cT_a^b(g). \end{align*} $$

References:

Math416 by Ely Kerman