Continuity Of Linear Mappings In Finite Dimensional Spaces

In the realm of linear algebra, linear mappings, also known as linear transformations, serve as the cornerstone for understanding relationships between vector spaces. A linear mapping is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. When dealing with finite-dimensional spaces, a fascinating property emerges: every linear mapping is continuous. This article delves into the proof of this theorem, explores its implications, and highlights its significance within mathematics.

The central theorem we will explore states that any linear mapping from one finite-dimensional normed space to another is continuous. This implies that small changes in the input vector result in small changes in the output vector, a crucial property for many applications. To fully grasp this theorem, we first need to define some fundamental concepts.

• Linear Mapping: A function T: V → W between vector spaces V and W is a linear mapping if it satisfies the following two conditions:
  • T(u + v) = T(u) + T(v) for all vectors u, v in V.
  • T(αu) = αT(u) for all vectors u in V and all scalars α.
• Normed Space: A normed space is a vector space V equipped with a norm ||·||, which is a function that assigns a non-negative real number to each vector, representing its length or magnitude. The norm must satisfy the following properties:
  • ||u|| ≥ 0 for all u in V, and ||u|| = 0 if and only if u = 0.
  • ||αu|| = |α| ||u|| for all u in V and all scalars α.
  • ||u + v|| ≤ ||u|| + ||v|| for all u, v in V (the triangle inequality).
• Continuity: A mapping T: V → W between normed spaces V and W is continuous at a point v in V if for every ε > 0, there exists a δ > 0 such that if ||u - v|| < δ, then ||T(u) - T(v)|| < ε. A mapping is continuous if it is continuous at every point in its domain.

Proof of Continuity

Now, let's delve into the formal proof demonstrating that linear mappings between finite-dimensional normed spaces are inherently continuous. This proof leverages the concept of a basis for finite-dimensional spaces and the properties of norms.

1. Establish Finite Dimensional Normed Spaces: Assume we have two finite-dimensional normed spaces, V and W, with dimensions n and m, respectively. Let T: V → W be a linear mapping.
2. Choose a Basis for V: Let {v₁, v₂, ..., vₙ} be a basis for V. This means that any vector v in V can be uniquely expressed as a linear combination of these basis vectors: v = α₁v₁ + α₂v₂ + ... + αₙvₙ, where α₁, α₂, ..., αₙ are scalars.
3. Utilize Linearity of T: Due to the linearity of T, we can express the image of v under T as follows: T(v) = T(α₁v₁ + α₂v₂ + ... + αₙvₙ) = α₁T(v₁) + α₂T(v₂) + ... + αₙT(vₙ). This step is pivotal as it allows us to decompose the transformation of any vector into the transformation of basis vectors.
4. Apply the Triangle Inequality and Norm Properties: Now, let's consider the norm of T(v). Using the properties of norms and the triangle inequality, we have: ||T(v)|| = ||α₁T(v₁) + α₂T(v₂) + ... + αₙT(vₙ)|| ≤ |α₁| ||T(v₁)|| + |α₂| ||T(v₂)|| + ... + |αₙ| ||T(vₙ)||. This inequality provides an upper bound for the norm of the transformed vector.
5. Introduce a Constant M: Let M = max{||T(v₁)||, ||T(v₂)||, ..., ||T(vₙ)||}. This constant M represents the maximum norm among the transformed basis vectors. Then, we can further bound the inequality: ||T(v)|| ≤ M(|α₁| + |α₂| + ... + |αₙ|). This step simplifies the expression and highlights the role of the coefficients in the linear combination.
6. Equivalence of Norms in Finite Dimensions: A crucial fact is that in finite-dimensional spaces, all norms are equivalent. This means that for any two norms ||·||₁ and ||·||₂ on V, there exist constants C₁ and C₂ such that C₁||v||₁ ≤ ||v||₂ ≤ C₂||v||₁ for all v in V. We can introduce the norm ||v||∞ = max{|α₁|, |α₂|, ..., |αₙ|}. Then, there exists a constant K such that |α₁| + |α₂| + ... + |αₙ| ≤ K||v||. This equivalence of norms is a cornerstone of the proof, as it allows us to relate the coefficients of the linear combination to the norm of the vector v.
7. Combine Inequalities: Combining the previous inequalities, we get: ||T(v)|| ≤ M(|α₁| + |α₂| + ... + |αₙ|) ≤ M K ||v||. Let C = M K. Then, ||T(v)|| ≤ C||v||. This inequality is a pivotal result, demonstrating that the norm of the transformed vector is bounded by a constant multiple of the norm of the original vector.
8. Establish Continuity: Now, we can demonstrate continuity. Let u, v be vectors in V. Consider ||T(u) - T(v)||. Using the linearity of T, we have T(u) - T(v) = T(u - v). Applying the inequality we derived, we get: ||T(u) - T(v)|| = ||T(u - v)|| ≤ C||u - v||. This final inequality is the key to proving continuity. For any ε > 0, choose δ = ε/C. Then, if ||u - v|| < δ, we have ||T(u) - T(v)|| ≤ C||u - v|| < C(ε/C) = ε. This precisely matches the definition of continuity, establishing that T is continuous.

Implications of Continuity

The continuity of linear mappings on finite-dimensional spaces has several important implications across various mathematical fields.

1. Boundedness: The inequality ||T(v)|| ≤ C||v|| implies that T is a bounded linear operator. Boundedness is a critical property in functional analysis, ensuring that the operator does not