Question

Let \(V \stackrel{T}{\rightarrow} W \stackrel{S}{\rightarrow} V\) be functions such that \(T S=1_{W}\) and \(S T=1_{V}\). If \(T\) is linear, show that \(S\) is also linear.

Short answer

S is linear because it satisfies additivity and homogeneity.

Step-by-step solution

Understand the Given Problem

We are given functions where \(T: V \to W\) and \(S: W \to V\). These functions satisfy \(TS = 1_W\) and \(ST = 1_V\), meaning they are inverse functions to each other, and hence bijective. We need to show that if \(T\) is linear, then \(S\) is also linear.

Recall Definitions of Linear Functions

A function \(T\) is linear if it satisfies \(T(a + b) = T(a) + T(b)\) and \(T(\lambda a) = \lambda T(a)\) for all vectors \(a, b\) and scalars \(\lambda\). We need to establish that \(S\) follows these same properties.

Use the Relationship TS = 1_W

Because \(TS = 1_W\), for any \(w \in W\), applying \(T\) after \(S(w)\) results in \(w\). Thus, \(T(S(w)) = w\). We will use this to demonstrate that \(S\) is linear by exploiting the linearity of \(T\).

Demonstrate Additivity of S

Take any \(w_1, w_2 \in W\). We need to show \(S(w_1 + w_2) = S(w_1) + S(w_2)\). Using \(T\), we have \(T(S(w_1 + w_2)) = w_1 + w_2\). Knowing that \(T\) is linear, \(T(S(w_1) + S(w_2)) = T(S(w_1)) + T(S(w_2)) = w_1 + w_2\). Since \(T\) is injective (because \(ST = 1_V\)), we deduce that \(S(w_1 + w_2) = S(w_1) + S(w_2)\).

Demonstrate Homogeneity of S

Take any \(w \in W\) and scalar \(\lambda\). We need to show \(S(\lambda w) = \lambda S(w)\). By linearity of \(T\), \(T(S(\lambda w)) = \lambda w\). Also, \(T(\lambda S(w)) = \lambda T(S(w)) = \lambda w\). Since \(T\) is injective, \(S(\lambda w) = \lambda S(w)\).

Conclusion

Since \(S\) satisfies both additivity and homogeneity, \(S\) is linear.

Key concepts

Inverse Functions

In mathematics, inverse functions are essential as they essentially "undo" each other. If we have a function, say \( f(x) \), its inverse \( f^{-1}(x) \) will reverse the effect of \( f \). That means applying \( f \) and then \( f^{-1} \) will return you to the original input. This general property is denoted as \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).
In the context of linear transformations, if \( T: V \rightarrow W \) and \( S: W \rightarrow V \) are such that \( T \) and \( S \) are inverses, then \( TS = 1_W \) and \( ST = 1_V \). This guarantees that these functions are indeed undoing each other's operations.
For our specific problem, showing that if \( T \) is linear, \( S \) must also be linear hinges on the linear nature of these transformations and their inverses. This means that the properties of linearity (additivity and homogeneity) should be preserved when moving from \( T \) to \( S \).

Bijective Functions

A function is bijective if it is both injective (one-to-one) and surjective (onto). Injectivity ensures that different inputs map to different outputs, while surjectivity guarantees that every possible output is covered by the function. Together, these qualities ensure that there is a perfect "pairing" between sets.
In terms of linear transformations, bijective functions mean that every element in the range has a unique pre-image in the domain. For the functions \( T: V \rightarrow W \) and \( S: W \rightarrow V \), their mutual bijectivity comes from their inverse relationship, \( TS = 1_W \) and \( ST = 1_V \).
This bijective nature is vital in proving that if \( T \) is linear, then \( S \) must be linear too. This is because the bijective quality implies that functions are transformations that are reversible, thereby preserving structures like linearity.

Linearity Criteria

To determine if a function is linear, it must satisfy two main conditions: additivity and homogeneity. This means for a function \( T \), it must hold true that:

• Additivity: \( T(a + b) = T(a) + T(b) \) for any vectors \( a \) and \( b \) in the domain.
• Homogeneity: \( T(\lambda a) = \lambda T(a) \) for any scalar \( \lambda \) and vector \( a \) in the domain.

By applying these criteria to \( T \), and knowing that \( S \) is its inverse, we start to establish these same properties for \( S \).
Through properties like \( TS = 1_W \), linearity ensures that the structure of the mathematical space is maintained, making it possible to generalize these properties from \( T \) to \( S \). Therefore, these criteria form the backbone of analyzing and proving the linear nature of these functions.

Injective Functions

Injective functions, also known as one-to-one functions, map distinct elements from the domain to distinct elements in the codomain. This means that if \( f(a) = f(b) \), then \( a = b \). Having this property ensures that no information is lost during the transformation, as each input has a unique output.
For linear transformations, proving the injectivity of a function \( T \) or \( S \) is a critical step in establishing linearity across inverse functions. In our exercise, since \( ST = 1_V \), \( S \) must be injective. Any perturbation is preserved, maintaining the distinctiveness of each vector.
This injective nature, combined with the concept of inverses and bijections, further solidifies why \( S \) being the inverse of a linear \( T \), must also satisfy linearity principles as a natural consequence. Thus, injectivity is an essential foundation for understanding why such results hold in linear algebra.