---
title: "Product of Vector Space"
source: https://www.jemoka.com/posts/kbhproduct_of_vector_spaces/
---

A product of vector spaces is a vector space formed by putting an element from each space into an element of the vector.
constituents Suppose \(V_1 \dots V_{m}\) are vector spaces over the same field \(\mathbb{F}\)
requirements Product between \(V_1 \dots V_{m}\) is defined:
\begin{equation} V_1 \times \dots \times V_{m} = \{(v_1, \dots, v_{m}): v_1 \in V_1 \dots v_{m} \in V_{m}\} \end{equation}
&ldquo;chain an element from each space into another vector&rdquo;
additional information operations on Product of Vector Spaces The operations on the product of vector spaces are defined in the usual way.
Addition: \((u_1, \dots, u_{m})+(v_1, \dots, v_{m}) = (u_1+v_1, \dots, u_{m}+v_{m})\)
Scalar multiplication: \(\lambda (v_1 \dots v_{m}) = (\lambda v_1, \dots, \lambda v_{m})\)
Product of Vector Spaces is a vector space The operations defined above inherits closure from their respective vector spaces.
additive identity: \((0, \dots, 0)\), taking the zero from each vector space additive inverse: \((-v_1, \dots, -v_{m})\), taking the additive inverse from each vector space scalar multiplicative identity: \(1\) operations: commutativity, associativity, distributivity &mdash; inheriting from vector spaces \(\blacksquare\)
dimension of the Product of Vector Spaces is the sum of the spaces&rsquo; dimension Proof:
Take each \(V_{j}\); construct a list such that, for each basis vector in the basis of \(V_{j}\), we have an element of the list such that we have that basis vector in the \(j^{th}\) slot and \(0\) in all others.
This list is linearly independent; and, a linear combination thereof span all of \(V_1 \times \dots \times V_{m}\). The length of this is the sum of the number of basis vectors of each space, as desired. \(\blacksquare\)
product summation map See: product summation map
