Linear Operator

Definition. A map $φ : L \to L$ of a linear space into itself is called a linear operator if, for all $\overline{x}, \overline{y} \in L$ and all $λ \in R$ , two conditions hold:

additivity: $φ (\overline{x} + \overline{y}) = φ (\overline{x}) + φ (\overline{y})$ ;
homogeneity: $φ (λ \overline{x}) = λ φ (\overline{x})$ .

The vector $φ (\overline{x})$ is called the image of the vector $\overline{x}$ , and $\overline{x}$ is called its preimage.

Remark. Both conditions are essential. By itself, "a correspondence assigning to every vector $\overline{x}$ some vector $φ (\overline{x})$ " defines an arbitrary map; it is precisely additivity and homogeneity that make it linear. They can be combined into a single condition: $φ (λ \overline{x} + μ \overline{y}) = λ φ (\overline{x}) + μ φ (\overline{y})$ .

Examples: the identity operator $φ (\overline{x}) = \overline{x}$ ; the zero operator $φ (\overline{x}) = \overline{0}$ ; scaling $φ (\overline{x}) = k \overline{x}$ ; rotation of the plane about the origin.

The Matrix of a Linear Operator

Let $L$ be an $n$ -dimensional space with basis $\overline{e}_{1}, \dots, \overline{e}_{n}$ . Expand the images of the basis vectors in this same basis:

φ (\overline{e}_{j}) = a_{1 j} \overline{e}_{1} + a_{2 j} \overline{e}_{2} + \dots + a_{nj} \overline{e}_{n}, j = 1, \dots, n .

The coefficients of these expansions, written by columns, form the matrix of the linear operator in this basis:

A = a_{11} a_{21} ⋮ a_{n 1} a_{12} a_{22} ⋮ a_{n 2} \dots \dots ⋱ \dots a_{1 n} a_{2 n} ⋮ a_{nn} .

(Column $j$ holds the coordinates of the image $φ (\overline{e}_{j})$ .)

If $\overline{x} = x_{1} \overline{e}_{1} + \dots + x_{n} \overline{e}_{n}$ , then by linearity $φ (\overline{x}) = \sum_{j} x_{j} φ (\overline{e}_{j})$ ; equating coordinates shows that in coordinates the operator acts by multiplication by the matrix:

x_{1}^{'} ⋮ x_{n}^{'} = A x_{1} ⋮ x_{n}, where φ (\overline{x}) = x_{1}^{'} \overline{e}_{1} + \dots + x_{n}^{'} \overline{e}_{n} .

Remark. The converse also holds: if a basis is fixed in $L_{n}$ , then any square matrix of order $n$ is the matrix of some linear operator. The operator $φ$ is nonsingular $⟺ det A \neq = 0$ .

Example

Let $\overline{e}_{1}, \overline{e}_{2}, \overline{e}_{3}, \overline{e}_{4}$ be a basis, and let an operator be given by the images:

φ (\overline{e}_{1}) = \overline{e}_{1} - \overline{e}_{2} + \overline{e}_{4}, φ (\overline{e}_{2}) = \overline{e}_{2} + 2 \overline{e}_{3}, φ (\overline{e}_{3}) = 2 \overline{e}_{1} - \overline{e}_{4}, φ (\overline{e}_{4}) = - \overline{e}_{1} - \overline{e}_{2} .

Writing the coordinates of the images by columns, we obtain the matrix

A = 1 - 1 01 0120 200 - 1 - 1 - 1 00 .

Let's check nonsingularity by expanding along the third row $(0, 2, 0, 0)$ :

det A = 2 \cdot (- 1)^{3 + 2} 1 - 1 1 20 - 1 - 1 - 1 0 = 2 \cdot (- 1) \cdot (- 4) = 8 \neq = 0,

so the operator $φ$ is nonsingular.

Change of the Operator's Matrix Under a Change of Basis

Theorem. Let $A$ and $B$ be the matrices of the same linear operator $φ$ in the "old" basis $\overline{e}_{1}, \dots, \overline{e}_{n}$ and the "new" basis $\overline{e}_{1}^{'}, \dots, \overline{e}_{n}^{'}$ , and let $C$ be the change-of-basis matrix from the old basis to the new one. Then

B = C^{- 1} A C .

Proof. The coordinates of any vector in the old and new bases are related by $\overline{x} = C \overline{x}^{'}$ . Let $\overline{y} = φ (\overline{x})$ ; then in the old basis $\overline{y} = A \overline{x}$ , in the new one $\overline{y}^{'} = B \overline{x}^{'}$ , and likewise $\overline{y} = C \overline{y}^{'}$ . Substituting $\overline{x} = C \overline{x}^{'}$ , $\overline{y} = C \overline{y}^{'}$ into $\overline{y} = A \overline{x}$ :

C \overline{y}^{'} = A C \overline{x}^{'} multiply by C^{- 1} on the left \overline{y}^{'} = C^{- 1} A C \overline{x}^{'} .

Comparing with $\overline{y}^{'} = B \overline{x}^{'}$ , we get $B = C^{- 1} A C$ .

Corollary. The determinant of the matrix of a linear operator does not depend on the choice of basis:

det B = det (C^{- 1} A C) = \frac{1}{det C} \cdot det A \cdot det C = det A .

(Matrices $A$ and $B = C^{- 1} A C$ are called similar.)

For a linear operator, its invariant directions — vectors that it only stretches — matter a great deal. These are eigenvectors and eigenvalues, which we turn to in the next article.

Linear Operator

The Matrix of a Linear Operator

Example

Change of the Operator's Matrix Under a Change of Basis

Next

Related works