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. \(\blacksquare\)