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. — GMA Steals & Deals: Find The Best Savings Now!
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. — Free Stuff Craigslist Dallas: Your Ultimate Guide
- 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. — Iowa State Game Guide: Tickets, Sports & How To Attend
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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
