Matrix Representation Theorem

We noted that multiplication by an m x n matrix defines a transformation from 𝔽ⁿ into 𝔽^m. Matrix transformations are linear transformations since the properties in Definition 2.1.1 are simply restatement of the linearity properties for matrix multiplication. The matrix representation theorem says that every linear transformation from 𝔽ⁿ into 𝔽^m is a matrix transformation.
Just like every vector in a finite-dimensional vector space can be associated with a vector in 𝔽ⁿ, every linear transformation between vector spaces can be associated with a matrix in M_m,n(𝔽):

Theorem 3.3.1. (Matrix Representation Theorem) Let V be an n-dimensional vector space over 𝔽 and W be m-dimensional vector space over 𝔽. Let B = {u₁, u₂, ..., u_n} and B' = {w₁, w₂, ..., w_m} be ordered bases for V and W respectively. Let T: V ⟶ W be any linear transformation. Then there is a unique matrix A such that T(x) = Ax for all X ∈ V. Then there exists a natural isomorpshim J: L(V,W) ⟶ M_m,n(𝔽) given by J(T) = A.

Proof. Let x ∈ V then x = x₁ u₁ + x₂ u₂ + ... + x_nu_n, where u_j is the jth element of the basis of V.
Applying the additivity and scalar properties of T, we obtain

The vectors T(u_j) ∈ W, can be written as linear combination of the vectors of the basis B' of W:

Clearly these mn scalar A_ij, 1 ≤ i ≤ m, 1 ≤ j ≤ n determine T completely.

The m x n matrix A = (A_ij) whose j-th column is the coordinate matrix of T(u_j) relative to ordered basis B' is called the matrix of T relative to the pair of ordered basis B and B'. It is denoted by [T]_B,B'. Thus

$$[T{]}_{B,B^{\prime }}=\left(\begin{array}{cccc}{A}_{11}& A_{12}& \dots & A_{1n}\\ A_{21}& A_{22}& \dots & A_{2n}\\ ⋮& ⋮& ⋮& ⋮\\ A_{m1}& A_{m2}& \dots & A_{mn}\end{array}\right)$$

Then we can express the linear transformation as T(x) = [T]_B,B [x]_B = Ax, since

$$\begin{gathered} T(\mathbf{x})=T(\sum _{j=1}^n{x}_j{\mathbf{u}}_j)=\sum _{j=1}^n{x}_{j}T({\mathbf{u}}_j)=\sum _{j=1}^n{x}_j\sum _{i=1}^m{A}_{ij}{\mathbf{w}}_i \\ =\sum _{j=1}^n\sum _{i=1}^m{A}_{ij}{x}_j{\mathbf{w}}_i=\sum _{i=1}^m(\sum _{j=1}^n{A}_{ij}{x}_j){\mathbf{w}}_i \end{gathered}$$

from which is evident that the scalar components of L(x), with respect to the basis B' are given by

$$y_i=\sum _{j=1}^n{A}_{ij}{x}_j$$

or in matrix form as

$$\left(\begin{array}{c}{y}_1\\ y_2\\ ⋮\\ y_m\end{array}\right)=A\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right)$$

Conversely suppose that A = (A_ij) be any given m x n matrix. Define T: V ⟶ W by

$$T(\sum _{j=1}^n{x}_j{\mathbf{u}}_{\mathbf{j}})=\sum _{i=1}^n(\sum _{j=1}^n{A}_{ij}{x}_j){\mathbf{w}}_{\mathbf{i}}$$

Then T, is a linear transformation such that the properties in definition 2.1.1 hold:

1. Let a ∈ F then

   $$T(\mathbf{ax})=\sum _{i=1}^n(a(\sum _{j=1}^n{A}_{ij}{x}_j)){\mathbf{w}}_{\mathbf{i}}$$

   Hence

   a[T]_B' = aT_B,B' [x]_B

2. Let S be any other linear transformation then

   [(T + S)(x)]_B' = ([T_B,B'] + [S_B,B']) [x]_B

This completely proves the theorem.  □

Example 3.3.2. Consider the linear transformation T: ℝ³ ⟶ ℝ² given by

Consider B = {(1,0,0),(0,1,0),(0,0,1)} and B' = {(1,0), (0,1)} be standard basis of ℝ³ and ℝ² respectively. Then

Hence [T(1,0,0)]_B = (1 0)^T,   [T(0,1,0)]_B = (1 −1)^T,   [T(0,0,1)]_B = (1 0)^T. Thus the matrix T relative to B and B' is

$$[T{]}_{B,B^{\prime }}=\left(\begin{array}{ccc}1& 1& 1\\ 0& -1& 0\end{array}\right)\text{\hspace{1em}■}$$

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