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}

“chain an element from each space into another vector”

## 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 — inheriting from vector spaces

\(\blacksquare\)

### dimension of the Product of Vector Spaces is the sum of the spaces’ 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. \(\blacksquare\)