1. Linear map equation:

Let T:Rⁿ→R^m be a linear transformation and b∈R^m . Then equation T(X)=b is called a linear map equation. The linear map equation T(X)=b is called consistent if it has a solution, i.e. if there exists X∈Rⁿ such that T(X)=b. If it is not consistent, then it is called inconsistent. Note that for X∈Rⁿ , X≡(x₁ , x₂ , x₃ , …, x_n ) and T(X)∈R^m ; where x₁ , x₂ , x₃ , …, x_n ∈R. For n=2, X≡( x₁ , x₂ ) or (x, y), where x₁ , x₂ , x, y∈R.

2. Examples:

(i) Let T:R² →R³ be the linear transformation defined by T(x, y)=(x+y, x-y, 0), where x, y∈R. Consider the linear map equation T(X)=b, where b=(1, 1, 0). Then we get (x+y, x-y, 0)=(1, 1, 0). This implies that x+y=1and x-y=1. Thus x=1 and y=0. Clearly T(1, 0)=(1, 1, 0) which means that solution of linear map equation exists. That is, linear map equation is consistent.
(ii) Let T:R² →R³ be the linear transformation defined by T(x, y)=(x+y, 2y, 2x), where x, y∈R. Consider the linear map equation T(X)=b, where b=(1, 4, 0). Then we get (x+y, 2y, 2x)=(1, 4, 0). This implies that x+y=1, 2y=4, 2x=0. By solving equations (ii) and (iii) we get, y=2 and x=0 which do not satisfy the equation x+y=1, a contradiction. This means that solution of linear map equation does not exist. That is, linear map equation is inconsistent.

3. Solution:

Consider a linear map equation
T(X)=b ……. (i)
where T:Rⁿ→R^m is a linear map and b∈R ^m . Note that in particular, if b=0, then the linear map equation (i) reduces to
T(X)=0 ……. (ii)

3.1. Condition for existence:

By the definition of range T, linear map equation (i) is consistent if and only if b∈Range of T. However, the linear map equation (ii) is always consistent, since T being linear T(0) =0.

3.2. Solution set:

(a.) Assume X_o to be a solution of (i).Then the solution set, i.e. the collection of all solutions of (i) is X₀+ker(T)≡ {X_o+a: a∈ker(T)}.
Reason: If a∈ker(T), then T(X_o+a)=T(X_o)+T(a)=b+0=0, so that X_o+a is a solution of linear map equation of (i). Conversely, let X be a solution of (i). Then T(X)=b. Since X= X_o+(X-X_o) and (X-X_o)∈ker(T), hence X∈X0+ker(T).
(b.) From 3.1, 0 is a solution of (ii). Clearly ker(T) is its solution set.

3.3 Uniqueness of solution:

(a.) Let solution of (i) exist. Then solution is unique if and only if T is one-to-one by the definition of one-to-one map.
(b.) Solution of (ii) is unique if and only if T is one-to-one by definition. [Note that solution of (ii) always exists.]

3.4. Summary:

(a.) Solution of (i) may or may not exist.
(b.) Solution of (i) is unique if and only if T is one-to-one, provided solution exists.
(c.) Solution of (ii) always exists.
(d.) Solution of (ii) is unique if and only if T is one-to-one.

4. Remark:

4.1.

If T:Rⁿ→R^m is a one-to-one linear transformation, then it does not imply that solution of (i) or (ii) exist, in general [this is shown in the example 5(ii) below].

4.2.

If T:Rⁿ→R^m is a one-to-one linear transformation and n=m, then T is onto and hence solution of (i) exists. Further, T being one-to-one solution is unique.

5. Examples:

(i) Let T:R²→R² be the linear transformation defined by T(x, y)=(y, x), where x, y∈R. Consider the linear map equation T(X)=b, where b=(1, 4). Then solution of linear map equation exists because b∈Range of T. Further the solution is unique, since T is one-to-one.
(ii) Let T:R²→R³ be the linear transformation defined by T(x, y)=(x-y, 0), where x, y∈R. Consider the linear map equation T(X)=b, where b=(1, 4, 1). Notice that this linear map equation is not consistent and T is one-to-one. This shows that solution of linear map equation may not exist even if T is one-to-one.