Problem 10
Question
Let \(V\) be a vector space over \(F\). Prove that \(-(\alpha v)=(-\alpha) v=\alpha(-v)\) for all \(\alpha \in F\) and all \(v \in V\).
Step-by-Step Solution
Verified Answer
Question: Prove the identity \(-(\alpha v)=(-\alpha) v=\alpha(-v)\) for all scalars \(\alpha\) and all vectors \(v\).
1Step 1: Understand the notation and definitions
Here, \(V\) represents a vector space over the field \(F\), and \(\alpha \in F\) stands for a scalar value, while \(v \in V\) represents a vector. In vector spaces, there are certain properties that the elements must satisfy such as the existence of additive inverses for every element. In this context, the additive inverse of a scalar multiple of a vector (\(\alpha v\)) must also satisfy these properties.
2Step 2: Prove \(-(\alpha v)\) is the additive inverse of \(\alpha v\)
We will first show that \(-(\alpha v)\) is the additive inverse of \(\alpha v\). By definition, an additive inverse must satisfy the following property:
\(v + (-v) = 0\), where \(0\) is the zero vector in the vector space \(V\).
Now, we have \((\alpha v) + (-(\alpha v))\). By the distributive property of scalar multiplication over vector addition, we can write this as:
\(\alpha(v + (-v)) = \alpha0=0\)
Since \(v + (-v) = 0\), the equation simplifies to \(\alpha0=0\). Therefore, we have shown that \(-(\alpha v)\) is the additive inverse of \(\alpha v\).
3Step 3: Prove \((-\alpha)v\) is the additive inverse of \(\alpha v\)
Next, we will show that \((-\alpha)v\) is the additive inverse of \(\alpha v\). We proceed similarly to the previous step, considering the sum \((\alpha v) + ((-\alpha)v)\). By the distributive property of scalar multiplication over vector addition, we can write this as:
\((\alpha + (-\alpha))v = 0v\)
Since \(\alpha + (-\alpha) = 0\), the equation simplifies to \(0v=0\). Therefore, we have shown that \((-\alpha)v\) is also the additive inverse of \(\alpha v\).
4Step 4: Prove \(\alpha(-v)\) is the additive inverse of \(\alpha v\)
Finally, we will show that \(\alpha(-v)\) is the additive inverse of \(\alpha v\). We proceed similarly to the previous steps, considering the sum \((\alpha v) + (\alpha(-v))\). By the distributive property of scalar multiplication over vector addition, we can write this as:
\(\alpha(v + (-v)) = \alpha0=0\)
Since \(v + (-v) = 0\), the equation simplifies to \(\alpha0=0\). Therefore, we have shown that \(\alpha(-v)\) is also the additive inverse of \(\alpha v\).
5Step 5: Establish the identity
Since all three expressions \(-(\alpha v)\), \((-\alpha)v\), and \(\alpha(-v)\) are additive inverses of \(\alpha v\), they must be equal to each other according to the uniqueness of additive inverses in a vector space. Thus, we have established the identity:
\(-(\alpha v)=(-\alpha)v=\alpha(-v)\) for all \(\alpha \in F\) and all \(v \in V\).
Key Concepts
Vector AdditionScalar MultiplicationAdditive InverseZero Vector
Vector Addition
Vector addition is a fundamental operation in vector spaces. It involves taking two vectors and combining them to produce a third vector. You can think of this like adding points or directions together.
For example, if you have two vectors, such as \( \mathbf{a} \) and \( \mathbf{b} \), their vector addition \( \mathbf{a} + \mathbf{b} \) will result in a new vector.
For example, if you have two vectors, such as \( \mathbf{a} \) and \( \mathbf{b} \), their vector addition \( \mathbf{a} + \mathbf{b} \) will result in a new vector.
- This operation is commutative, meaning \( \mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \).
- It's also associative, so \( (\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c}) \).
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar, which is a single real number from the field. This operation changes the magnitude of the vector without affecting its direction. For example, if you have a vector \( \mathbf{v} \) and a scalar \( k \), scalar multiplication is represented as \( k \mathbf{v} \).
- One important property of scalar multiplication is that it distributes over vector addition: \( k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v} \).
- Scalar multiplication is also associative with respect to the scalar values: \( (cd)\mathbf{v} = c(d\mathbf{v}) \).
Additive Inverse
The additive inverse is a property that states for every vector \( \mathbf{v} \) in a vector space, there exists another vector \( -\mathbf{v} \) such that their sum equals the zero vector: \( \mathbf{v} + (-\mathbf{v}) = \mathbf{0} \). This concept ensures that every vector has a counterpart that can "cancel" it out.
- The additivity of this inverse makes it crucial for simplifying expressions and solving vector equations.
- The identity \( \mathbf{v} + \mathbf{0} = \mathbf{v} \) is always maintained where \( \mathbf{0} \) is the zero vector.
Zero Vector
The zero vector is a special vector in a vector space, denoted often as \( \mathbf{0} \). It plays a crucial role as it acts as the identity element for addition. No matter which vector it is added to, the result is always that vector: \( \mathbf{v} + \mathbf{0} = \mathbf{v} \).
- The zero vector also satisfies the condition \( 0\mathbf{v} = \mathbf{0} \) for any vector \( \mathbf{v} \), representing the scalar multiplication effect.
- In any vector space, we also find that adding the additive inverse \( -\mathbf{v} \) results in the zero vector again \( \mathbf{v} + (-\mathbf{v}) = \mathbf{0} \).
Other exercises in this chapter
Problem 8
Show that the set of all possible solutions \((x, y, z) \in \mathbb{R}^{3}\) of the equations $$ \begin{array}{l} A x+B y+C z=0 \\ D x+E y+C z=0 \end{array} $$
View solution Problem 9
Let \(W\) be the subset of continuous functions on [0,1] such that \(f(0)=0\). Prove that \(W\) is a subspace of \(C[0,1]\)
View solution Problem 11
Let \(V\) be a vector space of dimension \(n .\) Prove each of the following statements. (a) If \(S=\left\\{v_{1}, \ldots, v_{n}\right\\}\) is a set of linearly
View solution Problem 12
Prove that any set of vectors containing \(\mathbf{0}\) is linearly dependent.
View solution