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Chapter 3: Problem 19

Beweisen Sie folgendes (au脽erordentlich n眉tzliche!) Unterraumkriterium: Sei \(U\) eine Teilmenge eines \(K\)-Vektorraums \(V\). Wenn gilt \- \(U \neq \varnothing\) \- f眉r alle \(k \in K\) und \(u \in U\) gilt \(k \cdot u \in U\), \- f眉r alle \(u, u^{\prime} \in U\) ist \(u-u^{\prime} \in U\), dann ist \(U\) ein Unterraum von \(V\).

Short Answer

Expert verified
\(U\) is a subspace of \(V\) because it is non-empty, closed under scalar multiplication and addition.

Step by step solution

01

Identify the Problem

We need to prove that a subset \(U\) of a \(K\)-vector space \(V\) is a subspace. Given conditions are: \(U eq \varnothing\), for any scalar \(k\) in \(K\) and element \(u\) in \(U\), \(k \cdot u\) belongs to \(U\), and for any elements \(u, u^{\prime}\) in \(U\), the difference \(u - u^{\prime}\) is in \(U\).
02

Check Non-emptiness

Since \(U eq \varnothing\), it has at least one element. This is the first condition needed to ensure that \(U\) can be a subspace.
03

Verify Closure under Scalar Multiplication

The condition that for all \(k \in K\) and \(u \in U\), \(k \cdot u \in U\) confirms that \(U\) is closed under scalar multiplication. This means multiplying any element of \(U\) by any scalar from \(K\) keeps the result within \(U\), which is a key property of subspaces.
04

Verify Closure under Addition

To be closed under addition, we use the given condition: for all \(u, u^\prime \in U\), \(u - u^{\prime} \in U\). Replace \(u^\prime\) with \(-v\) where \(v\) is also in \(U\) (note that subtraction \(u - (-v) = u + v\)). Hence, this condition implies \(u + v \in U\), proving closure under addition.
05

Conclusion of the Proof

Having proven the subset is non-empty, closed under scalar multiplication, and addition, \(U\) fulfills all conditions required to be a subspace of \(V\). Thus, by the subspace criterion, \(U\) is a subspace.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Spaces
Vector spaces form the foundation of linear algebra. Understanding them is crucial when working with subspaces, matrices, and linear transformations. A vector space over a field \( K \) is a set \( V \), along with two operations that follow specific axioms: vector addition and scalar multiplication.
  • Vector Addition: For any two vectors \( u, v \in V \), there exists a vector \( u + v \) that is also in \( V \).
  • Scalar Multiplication: For any vector \( v \in V \) and scalar \( k \in K \), there exists a vector \( k \cdot v \) that is in \( V \).
These operations must satisfy 10 axioms, such as associativity, distributive properties, and the existence of a zero vector. It is this structured framework that supports the study of more intricate concepts, like subspaces.
Scalar Multiplication
Scalar multiplication is a defining operation in vector spaces. It involves taking a scalar \( k \) from the field \( K \) and multiplying it by a vector \( u \) from the vector space \( V \), resulting in another vector in \( V \). The operation must satisfy certain properties to ensure the structure of the vector space is maintained.
The properties include:
  • Associativity with Field Multiplication: \((kl)v = k(lv)\) for any scalars \( k, l \in K \) and vector \( v \in V \).
  • Distributive Property: \(k(u + v) = ku + kv\) and \((k + l)v = kv + lv\).
  • Multiplicative Identity: The scalar 1 satisfies \(1 \cdot v = v\) for any vector \( v \in V \).
These properties guarantee that multiplying a vector by a scalar from the field doesn't lead you outside of the vector space, preserving its structure.
Closure Properties
In the context of vector spaces, closure properties are essential for identifying subspaces. These properties ensure that operations performed within a vector space result in vectors that still reside within that space. For a subset \( U \) of a vector space \( V \) to be a subspace, it must be closed under two main operations:
  • Closure under Addition: If \( u, v \in U \), then their sum \( u + v \) must also be in \( U \).
  • Closure under Scalar Multiplication: If \( u \in U \) and \( k \in K \), then \( k \cdot u \) must be in \( U \) as well.
These criteria are integral to the subspace criterion we analyzed earlier. If a subset is closed under both addition and scalar multiplication, as well as non-empty, it fulfills the necessary conditions to be a subspace of \( V \). This ensures that any linear combination of vectors from \( U \) remains within \( U \), maintaining the subset's integrity.
Linear Algebra Proofs
Proving statements in linear algebra often involves a structured approach, breaking down the problem into smaller, manageable pieces. The subspace criterion proof is an excellent example of this methodology. The proof requires you to show that a subset \( U \) of a vector space \( V \) meets the conditions of being a subspace:
  • Non-emptiness: \( U eq \varnothing \), ensuring at least one element is present in \( U \).
  • Closure under Scalar Multiplication: If \( k \in K \) and \( u \in U \), then \( k \cdot u \in U \).
  • Closure under Addition: From the condition \( u - u^{'} \in U \), derive that \( u + v \in U \) (since \( u - (-v) = u + v \)).
Proving these conditions systematically confirms that \( U \) is indeed a subspace. In linear algebra, breaking down the problem and addressing each criterion step-by-step not only simplifies the proof but also deepens your understanding of the concepts involved.

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Most popular questions from this chapter

Zeigen Sie: (a) Ist \(g\) eine Gerade der euklidischen Ebene \(\mathbf{R}^{2}\) durch den Nullpunkt, so ist \(g\) ein Vektorraum. (b) Sei \(E\) eine Ebene des \(\mathbf{R}^{3}\) durch den Nullpunkt. Dann ist \(E\) ein Vektorraum.

Beweisen Sie folgende Aussagen. Sei \(V\) ein \(K\)-Vektorraum. (a) F眉r alle \(k \in K, v \in V\) gilt $$ k \cdot o=o \quad \text { und } \quad(-k) \cdot v=-(k \cdot v)=k \cdot(-v) $$ [脺berlegen Sie zun盲chst, in welchen Strukturen das Minuszeichen was bedeutet.] (b) F眉r alle \(k_{1}, k_{2} \mathrm{~K} \in\) und alle \(v \in V\) mit \(v \neq o\) gilt: $$ k_{1} \cdot v=k_{2} \cdot v \Leftrightarrow k_{1}=k_{2}. $$

Zeigen Sie: Je drei Vektoren der Menge \(\left\\{\left(1, x, x^{2}\right) \mid x \in K\right\\}\) von \(K^{3}\) sind linear unabh盲ngig.

Seien \(U\) und \(U^{\prime}\) Unterr盲ume eines \(K\)-Vektorraums \(V\). (a) Zeigen Sie: Die Menge $$ U+U^{\prime}:=\left\\{u+u^{\prime} \mid u \in U, u^{\prime} \in U^{\prime}\right\\} $$ (die Summe von \(U\) und \(U^{\prime}\) ) ist ein Unterraum von \(V\). (b) Gilt \(U+U^{\prime}=\left\langle U, U^{\prime}\right\rangle\) ?

Zeigen Sie, dass das Erzeugnis \(\left\langle v_{1}, \ldots, v_{n}\right\rangle\) von Vektoren \(v_{1}, \ldots, v_{n}\) eines Vektorraums \(V\) ein Unterraum von \(V\) ist.
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