Linear Transformations

Definition. Let X and Y be vector spaces over a field F. A function T : X → Y is a linear transformation if it preserves linear combinations, that is, ∀u, v ∈ X, c ∈ F:

1. T(u + v) = T(u) + T(v)
2. T(cu) = cT(u)

Or, alternatively, T(c₁u + c₂v) = c₁T(u) + c₂T(v). In general T is a linear transformation if it preserves linear combinations, T(c₁x₁ + ··· + c_nx_n) = c₁T(u) + ··· c_nT(x_n).

X and Y are called the domain and codomain of T, respectively; the image or range of T is the set of images, Img(T) = {T(x) | x ∈ X}; and the kernel of T is the set of vectors which map to the vector zero, that is, Ker(T) = {x | T(x) = 0}.

Examples:

• The derivate is a linear transformation, d/dx: X → X, X is the set of differential functions. Reclaim that $\frac{d}{dx}(c_1f(x)+c_2g(x))=c_1\frac{d}{dx}f(x)+c_2\frac{d}{dx}g(x)$. Futhermore, ker(d/dx) is the set of constant functions.
• Limits are linear operators, too. $\lim_{x \to a}(c_1f(x)+c_2g(x)) = c_1\lim_{x \to a}f(x)+c_2\lim_{x \to a}g(x)$. Besides, definite and indefinite integrals, series, antiderivatives, etc., are all linear operators on functions.
• T: ℝ³ → ℝ², T(x₁, x₂, x₃) = (x₁+x₃, x₂-x₁) is a linear transformation.

$[ T(c_1x+c_2y)= T(c_1\left(\begin{smallmatrix}x_1\\ x_2\\ x_3\end{smallmatrix} \right)+c_2\left(\begin{smallmatrix}y_1\\ y_2\\ y_3\end{smallmatrix} \right)) = T\left(\begin{smallmatrix}c_1x_1+c_2y_1\\ c_1x_2+c_2y_2\\ c_1x_3+c_2y_3\end{smallmatrix} \right) = \left(\begin{smallmatrix}c_1x_1+c_2y_1+c_1x_3+c_2y_3\\ c_1x_2+c_2y_2-c_1x_1-c_2y_1\end{smallmatrix} \right) = \left(\begin{smallmatrix}c_1x_1+c_1x_3\\ c_1x_2-c_1x_1\end{smallmatrix} \right) + \left(\begin{smallmatrix}c_2y_1+c_2y_3\\ c_2y_2-c_2y_1\end{smallmatrix} \right) ] = c_1\left(\begin{smallmatrix}x_1+x_3\\ x_2-x_1\end{smallmatrix} \right) + c_2\left(\begin{smallmatrix}y_1+y_3\\ y_2-y_1\end{smallmatrix} \right) = c_1T\left(\begin{smallmatrix}x_1\\ x_2\\ x_3\end{smallmatrix} \right)+c_2T\left(\begin{smallmatrix}y_1\\ y_2\\ y_3\end{smallmatrix} \right) ] $

Exercise. Calculate the kernel and image of T: ℝ³ → ℝ², T(x₁, x₂, x₃) = (x₁ + 2x₂, x₃ - 3x₂).

$T(x)=0↭ \left( \begin{smallmatrix}x_1 + 2x_2\\ x_3-3x_2\end{smallmatrix} \right) = \left( \begin{smallmatrix}0\\ 0\end{smallmatrix} \right)$

$\begin{cases} x_1 + 2x_2 = 0 \\ x_3 -3x_2 = 0 \end{cases}↭ x_3=3x_2,~x_1=-2x_2$

ker(T)={$\left(\begin{smallmatrix}-2t\\ t\\ 3t\end{smallmatrix}\right): t ∈ ℝ$} = {t$\left(\begin{smallmatrix}-2\\ 1\\ 3\end{smallmatrix}\right): t ∈ ℝ$}

$T(x)=\left( \begin{smallmatrix}b_1\\ b_2\end{smallmatrix} \right)↭ \left( \begin{smallmatrix}x_1 + 2x_2\\ x_3-3x_2\end{smallmatrix} \right) = \left( \begin{smallmatrix}b_1\\ b_2\end{smallmatrix} \right)$

$\begin{cases} x_1 + 2x_2 = b_1 \\ x_3 -3x_2 = b_2 \end{cases}$ In particular, x₂ = 0, and $T \left( \begin{smallmatrix}b_1\\ 0 \\ b_2\end{smallmatrix} \right) = \left( \begin{smallmatrix}b_1\\ b_2\end{smallmatrix} \right)$, that is, Img(T) = ℝ²

Definitions. A mapping T: X → Y is said to be…

• … onto or surjective if every element in the codomain (∀y ∈ Y) has an element in the domain (∃x ∈ X) that maps into it (T(x) = y) ↭ the codomain and range or image of T are the same. T is onto if and only if Img(T) = Y, e.g., the latest example is onto.
• … one to one or injective if there’s no instance of a vector in the codomain that gets mapped more than one time, that is, T(u) = T(v) ⇒ u = v. T is one to one if and only if Ker(T) = {0}, e.g., the latest example is not one-to-one because its ker(T)= {t$\left(\begin{smallmatrix}-2\\ 1\\ 3\end{smallmatrix}\right): t ∈ ℝ$} ≠ {0}
• … bijective or invertible if T is one-to-one and onto. Futhermore, if T is bijective, there exists another linear transformation, an inverse map of T (S and T undo each other under composition), S: Y → X: S∘T = Id_X and T:S=Id_Y.

Example. T: ℤ₂³ → ℤ₂³, $T(x_1,x_2,x_3)=(x_1+x_2, x_2, x_1+x_2+x_3)$

$T(x)=0↭ \left( \begin{smallmatrix}x_1 + x_2\\ x_2\\ x_1+x_2+x_3\end{smallmatrix} \right) = \left( \begin{smallmatrix}0\\ 0\\ 0\end{smallmatrix} \right)$

$\begin{cases} x_1 + x_2 = 0 \\ x_2 = 0 \\ x_1 +x_2 +x_3 = 0 \end{cases}↭ Ker(T)=\left( \begin{smallmatrix}0\\ 0\\ 0\end{smallmatrix} \right)$ T is one to one.

$T(x)=b↭ \left( \begin{smallmatrix}x_1 + x_2\\ x_2\\ x_1+x_2+x_3\end{smallmatrix} \right) = \left( \begin{smallmatrix}b_1\\ b_2\\ b_3\end{smallmatrix} \right)$

$\begin{cases} x_1 + x_2 = b_1 \\ x_2 = b_2 \\ x_1 +x_2 +x_3 = b_3 \end{cases}$ ↭ $\begin{cases} x_1 = b_1 + b_2 \\ x_2 = b_2 \\ x_3 = b_1 + b_3 \end{cases}$

Therefore, T is onto and bijective.