Question

Let \(V\) be a vector spacc, and define \(V^{n}\) to be the set of all \(n\) -tuples \(\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right)\) of \(n\) vectors \(\mathbf{v}_{i}\), each belonging to \(V\). Define addition and scalar multiplication in \(V^{n}\) as follows: $$\begin{array}{r} \left(\mathbf{u}_{1}, \mathbf{u}_{2}, \ldots, \mathbf{u}_{n}\right)+\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right) \\ =\left(\mathbf{u}_{1}+\mathbf{v}_{1}, \mathbf{u}_{2}+\mathbf{v}_{2}, \ldots, \mathbf{u}_{n}+\mathbf{v}_{n}\right) \\ a\left(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\right)=\left(a \mathbf{v}_{1}, a \mathbf{v}_{2}, \ldots, a \mathbf{v}_{n}\right) \end{array}$$ Show that \(V^{n}\) is a vector space.

Short answer

\(V^n\) is a vector space by satisfying all 10 vector space properties.

Step-by-step solution

Closure Under Addition

To show that the set \(V^n\) is closed under addition, take any two elements \(\left( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n \right)\) and \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\) in \(V^n\). The sum \(\left( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n \right) + \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) = \left( \mathbf{u}_1 + \mathbf{v}_1, \mathbf{u}_2 + \mathbf{v}_2, \ldots, \mathbf{u}_n + \mathbf{v}_n \right) \) is still in \(V^n\) because each component \(\mathbf{u}_i + \mathbf{v}_i\) is in \(V\), and since each individual vector space \(V\) is closed under addition, the property holds for all components.

Associativity of Addition

Show that addition is associative: for any \(\left( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n \right)\), \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\), and \(\left( \mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n \right)\) in \(V^n\), we have:\[\left[ \left( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n \right) + \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) \right] + \left( \mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n \right) = \left( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n \right) + \left[ \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) + \left( \mathbf{w}_1, \mathbf{w}_2, \ldots, \mathbf{w}_n \right) \right]\]This works because addition of each component vector is associative in \(V\).

Identity Element of Addition

Confirm that there is an identity element for addition in \(V^n\). This element is \(\left( \mathbf{0}_1, \mathbf{0}_2, \ldots, \mathbf{0}_n \right)\), where each \(\mathbf{0}_i\) is the zero vector in \(V\). For any \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) \in V^n\), the equation \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) + \left( \mathbf{0}_1, \mathbf{0}_2, \ldots, \mathbf{0}_n \right) = \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\) confirms that each component vector has an additive identity in \(V\).

Existence of Additive Inverses

Show that each element \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\) has an additive inverse, which is \(\left( -\mathbf{v}_1, -\mathbf{v}_2, \ldots, -\mathbf{v}_n \right)\). The sum \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) + \left( -\mathbf{v}_1, -\mathbf{v}_2, \ldots, -\mathbf{v}_n \right) = \left( \mathbf{0}_1, \mathbf{0}_2, \ldots, \mathbf{0}_n \right)\) shows the existence of an additive inverse because each vector \(\mathbf{v}_i\) has an inverse in \(V\).

Closure Under Scalar Multiplication

Demonstrate that \(V^n\) is closed under scalar multiplication by taking any scalar \(a\) and any vector \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\) in \(V^n\). The product \(a \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) = \left( a\mathbf{v}_1, a\mathbf{v}_2, \ldots, a\mathbf{v}_n \right)\) remains in \(V^n\) because scalar multiplication of each vector is still in \(V\).

Distributive Properties

Verify distributive properties:1. For any scalars \(a\) and vectors \(\left( \mathbf{u}_1, \ldots, \mathbf{u}_n \right)\), \(\left( \mathbf{v}_1, \ldots, \mathbf{v}_n \right)\), show that:\[a \left[ \left( \mathbf{u}_1, \ldots, \mathbf{u}_n \right) + \left( \mathbf{v}_1, \ldots, \mathbf{v}_n \right) \right] = a \left( \mathbf{u}_1 + \mathbf{v}_1, \ldots, \mathbf{u}_n + \mathbf{v}_n \right) = \left( a\mathbf{u}_1 + a\mathbf{v}_1, \ldots, a\mathbf{u}_n + a\mathbf{v}_n \right)\]2. Show.\[(a + b) \left( \mathbf{v}_1, \ldots, \mathbf{v}_n \right) = \left( (a+b)\mathbf{v}_1, \ldots, (a+b)\mathbf{v}_n \right) = a\left( \mathbf{v}_1, \ldots, \mathbf{v}_n \right) + b\left( \mathbf{v}_1, \ldots, \mathbf{v}_n \right)\]Both are valid as elements in \(V\) satisfy the distributive property.

Compatibility of Scalar Multiplication with Field Multiplication

Show that scalar multiplication is compatible with field multiplication: for any scalars \(a\) and \(b\) and vector \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\) in \(V^n\), the equation \(a(b\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)) = (ab)\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\) holds since for each component \(a(b\mathbf{v}_i)= (ab)\mathbf{v}_i\).

Identity Element of Scalar Multiplication

Verify the identity element of scalar multiplication: for any element \(\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) \in V^n\), multiplying by 1 gives \(1 \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) = \left( 1\mathbf{v}_1, 1\mathbf{v}_2, \ldots, 1\mathbf{v}_n \right) = \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\). Since each component satisfies this condition in \(V\), it holds for the tuple.

Key concepts

Closure Under Addition

To understand what closure under addition means for a vector space like \( V^n \), we first need to grasp that when we "add" two elements of this set, the result must also be an element of the same set. In mathematical terms, this implies for any two vectors \( \left( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n \right) \)and \( \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) \) from \( V^n \), their sum given by\[\left( \mathbf{u}_1 + \mathbf{v}_1, \mathbf{u}_2 + \mathbf{v}_2, \ldots, \mathbf{u}_n + \mathbf{v}_n \right) \]must also belong to \( V^n \). Each vector \( \mathbf{u}_i + \mathbf{v}_i \) is a sum of vectors in \( V \), and since \( V \) is a vector space, it is closed under addition. This means every sum \( \mathbf{u}_i + \mathbf{v}_i \) remains in \( V \), and therefore \( V^n \) is closed under addition as it satisfies this for each component.
Understanding closure under addition is vital. It's similar to being able to freely combine items in a group and still have something that belongs to the same group.
This is foundational to many operations in mathematics and helps us verify and ensure that certain operations don't break the rules of the structure we're working with.

Scalar Multiplication

Scalar multiplication in the context of vector spaces deals with multiplying elements of a vector by a scalar (a real number or any field element). For the vector space \( V^n \), this property ensures that for any vector \( \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) \)and any scalar \( a \), the result after scalar multiplication is still inside the vector space.
Mathematically, this is seen as:\[a \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) = \left( a\mathbf{v}_1, a\mathbf{v}_2, \ldots, a\mathbf{v}_n \right)\]Each component \( a\mathbf{v}_i \) must be within \( V \). Since \( V \) itself is closed under scalar multiplication, each \( a \mathbf{v}_i \) remains in \( V \), thereby keeping the entire tuple in \( V^n \).

• Scalars are like amplification factors—they change the size or direction of vectors but do not remove them from the vector space.
• This behavior ensures consistency in scaling operations, crucial in fields like physics and computer graphics.
• Scalar multiplication allows us to stretch or shrink vectors according to our needs while maintaining the rules of the space.

These concepts are inherent to ensuring the operations within a vector space stay valid and predictable, a desirable trait in mathematical analysis.

Additive Identity

The concept of an additive identity in a vector space means there exists a vector which, when added to any other vector in the space, leaves it unchanged. In the vector space \( V^n \), the additive identity is the zero vector \( (\mathbf{0}_1, \mathbf{0}_2, \ldots, \mathbf{0}_n) \), where each \( \mathbf{0}_i \) is the zero vector of the original space \( V \).
When you add this zero vector to any vector \( \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) \)from \( V^n \), the result is:\[\left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right) + \left( \mathbf{0}_1, \mathbf{0}_2, \ldots, \mathbf{0}_n \right) = \left( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \right)\]
This operation illustrates the stability and consistency of addition in a vector space.

• Additive identity acts as the neutral element in addition, a concept analogous to zero in arithmetic.
• It ensures the integrity of vector operations, maintaining vectors' properties without altering them.
• The additive identity forms a base case in many vector space proofs and computations.

Staying Consistent

The notion of an additive identity provides reliability and coherence in mathematical structures, laying down one of the cornerstone principles that uphold the nature of vector spaces.