Skip to content

Latest commit

 

History

History
293 lines (253 loc) · 18.2 KB

linear_algebra-linear_transformations.md

File metadata and controls

293 lines (253 loc) · 18.2 KB
Raw

Linear transformations

Erika Duan 2022-09-16

Vector transformation notation

A key focus of linear algebra is the linear transformations of vector spaces.

A linear transformation can be described as:

  • A function that maps a vector \vec x in \mathbb{R}^n to a vector \vec w in \mathbb{R}^m, where T(\vec x) = \vec w.
  • This is denoted by T: \mathbb{R}^n \to \mathbb{R}^m respectively.
  • The domain of T(\vec x) is \mathbb{R}^n.
  • The co-domain of T(\vec x) can be \mathbb{R}^m respectively.
  • The image of \vec x under T is the set \{\vec w \in \mathbb{R}^m | \vec w = T(\vec x)\} where \vec x \in \mathbb{R}^n.
  • The range of T(\vec x) also describes the set \{\vec w \in \mathbb{R}^m | \vec w = T(\vec x)\} where \vec x \in \mathbb{R}^n.

A linear transformation can also be described as a matrix transformation, where A_T is the standard matrix for the linear transformation T: \mathbb{R}^n \to \mathbb{R}^m and T(\vec x) = A\vec x where A_T = \begin{bmatrix}T(\vec e_1) & T(\vec e_2) & \cdots & T(\vec e_n) \end{bmatrix}.

A linear transformation T: \mathbb{R}^n \to \mathbb{R}^m must satisfy the following two properties:

  • For vectors \vec u, \vec v \in \mathbb{R}^n, T(\vec u + \vec v) = T(\vec u) + T(\vec v).
  • Let c be a scalar, T(c \vec u) = cT(\vec u).

Examples of linear transformations include projections onto lower dimensions, sheering transformations, scaling transformations and rotations around the point of origin.

Linear transformation compositions

If function f(x) maps element A to B and function g(x) maps element B to C, then the composition of f then g, denoted as g \circ f, is the function which maps element A to C and (g \circ f)(a) = g(f(a)).

Similarly, if T_1: \mathbb{R}^n \to \mathbb{R}^m and T_2: \mathbb{R}^m \to \mathbb{R}^p, the co-domain of T_1 equals the domain of T_2 and the composition T_2 \circ T_1 maps \mathbb{R}^n to \mathbb{R}^p.

The linear transformation composition T_2 \circ T_1 also satisfies the following two properties:

  • (T_2 \circ T_1)(\vec u + \vec v) = T_2(T_1(\vec u + \vec v)) = \cdots = (T_2 \circ T_1)(\vec u) + (T_2 \circ T_1)(\vec v).
  • (T_2 \circ T_1)(c\vec u) = T_2(T_1(c\vec u) = T_2(cT_1(\vec u) = cT_2(T_1(\vec u) = c(T_2 \circ T_1)(\vec u).

Note: In the example above, even though the sequence of transformations (T_2 \circ T_1)(\vec x) and (T_1 \circ T_2)(\vec x) produce the same grid lines in the 2D plane, the position of the basis vectors \hat i and \hat j are different.

Injective linear transformations

A linear transformation T: \mathbb{R}^n \to \mathbb{R}^m is injective (or one-to-one) if:

  • Every vector \vec b \in \mathbb{R}^m is the image of at most one vector \vec x in \mathbb{R}^n.
  • Different vectors \in \mathbb{R}^n have different images in \in \mathbb{R}^m.
  • If T(\vec u) = T(\vec v), then \vec u = \vec v.

Another way of thinking about this is that A_T must contain a set of independent vectors \{\vec v_1, \cdots, \vec v_p\} which spans a p-dimensional space in \in \mathbb{R}^m. Therefore a unique set of coordinates \{c_1, \cdots, c_p \} must exist which scales \{\vec v_1, \cdots, \vec v_p\} to form T(\vec x) = \vec b and T(\vec x) = \vec 0 only contains the trivial solution.

By extension, a linear transformation T: \mathbb{R}^n \to \mathbb{R}^n is only injective if A_T contains a basis for \mathbb{R}^n i.e. a set of independent vectors \{\vec v_1, \cdots, \vec v_n\} which span \mathbb{R}^n. The matrix rank, or dimensions of ColA, must be n for A_T to be injective when T: \mathbb{R}^n \to \mathbb{R}^n.

Surjective linear transformations

A linear transformation T: \mathbb{R}^n \to \mathbb{R}^m is surjective (or onto) if:

  • The range of T(\vec x), \vec b, spans \mathbb{R}^m for T(\vec x) = \vec b.
  • The equation A(\vec x) = \vec b has a solution for all \vec b \in \mathbb{R}^m.
  • The column space of A must span the co-domain \in \mathbb{R}^m i.e. the dimensions of the basis for ColA must be m.

Another way of thinking about this is that A_T must span \mathbb{R}^m i.e. the range and co-domain of T(\vec x) must both be \in \mathbb{R}^m. By definition, A_T \in \mathbb{R}^m if it contains a set of linearly independent vectors \{\vec v_1, \cdots, \vec v_m\}. Therefore, ColA \in \mathbb{R}^m for a surjective linear transformation T: \mathbb{R}^n \to \mathbb{R}^m.

Note: The set of vectors \{\vec v_1, \cdots, \vec v_m, \cdots, \vec v_p \} in A_T does not need to be linearly independent for surjective linear transformations where T: \mathbb{R}^n \to \mathbb{R}^m.

By extension, a linear transformation T: \mathbb{R}^n \to \mathbb{R}^n is only surjective if A_T contains a basis for \mathbb{R}^n i.e. the image of T(\vec x) is also in \mathbb{R}^n.

Bijective linear transformations

A linear transformation T: \mathbb{R}^n \to \mathbb{R}^n is therefore bijective (one-to-one and unto) if:

  • A_T contains a linearly independent set of vectors {\vec v_1, \cdots, \vec v_n} and a unique set of coordinates scales {\vec v_1, \cdots, \vec v_n} to form a different \vec b for each unique \vec x, where T(\vec x) = \vec b.
  • As A_T contains a basis with n dimensions, the range of T(\vec x) is therefore equal to the co-domain i.e. A_T \in \mathbb{R}^n.

Bijective linear transformations are an example of the rank and nullity theorem.

Given a bijective linear transformation T: \mathbb{R}^n \to \mathbb{R}^n where A_T has dimensions n \times n, the rank of T is the column space of A_T, which is n. The nullity of T is the null space of A_T, which is 0. Rank\,T + Nullity\,T = n + 0 = n.

Resources

  • Great YouTube videos on 2D and 3D linear transformations by 3Blue1Brown.