# Vector space

## Redirected from Vectorspace

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A vector space is an algebraic structure consisting of an additive Abelian group $V$ (elements of which are called vectors, and are denoted in bold), a field $F$ (elements of which are called scalars), and a scalar multiplication function $\times : F \times V \rightarrow V$ following these properties:

• Distributive property of scalar multiplication over vector addition: For all $a \in F$ and $\mathbf v, \mathbf w \in V$, $a \left(\mathbf v + \mathbf w \right) = a \mathbf v + a \mathbf w$.
• Distributive property of scalar multiplication over field addition: For all $a , b \in F$ and $\mathbf v \in V$, $\left(a + b\right) \mathbf v = a \mathbf v + b \mathbf v$.
• Associative law of combined scalar and field multiplication: For all $a , b \in F$ and $\mathbf v \in V$, $a \left(b \mathbf v\right) = \left(ab\right)\mathbf v$.
• Scalar multiplication identity: With 1 as the field multiplicative identity, for all v ∈ V, we have $1 \mathbf v = \mathbf v$.

As both $V$ and $F$ each have their own respective additive identites, we will denote boldface $\mathbf 0$ to represent the additive identity in $V$. We then say that $V$ is a vector space over the field $F$.

## Definitions

• A subset $S \subseteq V$ is:
• Linearly dependent if there exist (distinct) vectors $\mathbf v_1,\mathbf v_2,\dots,\mathbf v_n \in S$ and scalars $a_1,a_2,\dots,a_n \in F$, with at least one of these scalars non-zero, such that $\sum_{k=1}^n a_k\mathbf v_k = a_1\mathbf v_1 + a_2\mathbf v_2+\dots+a_n\mathbf v_n = \mathbf 0$. Otherwise, we say $S$ is linearly independent.
• A spanning set if, for any $\mathbf v \in V$ there exist vectors $\mathbf v_1,\mathbf v_2,\dots,\mathbf v_n \in S$ and scalars $a_1,a_2,\dots,a_n \in F$ such that $\sum_{k=1}^n a_k\mathbf v_k = a_1\mathbf v_1 + a_2\mathbf v_2+\dots+a_n\mathbf v_n = \mathbf v$. That is to say, any vector in $V$ is a linear combination of vectors in $S$.
• A basis for $V$ if $S$ is linearly independent and a spanning set.
• A subspace for $V$ if $S$ is also a subgroup and is closed under scalar multiplication: For any vector $\mathbf w \in S$ and scalar $a \in F$, $a\mathbf w \in S$.
• Given two vector spaces $V$ and $W$ over a field $F$, a function $T:V \rightarrow W$ is a linear transformation if for any $\mathbf u,\mathbf v, \in V$ and $a \in F$, $T\left( \mathbf u + \mathbf v \right) = T\left( \mathbf u \right) + T\left(\mathbf v \right)$ and $T\left(a \mathbf u \right) = aT\left( \mathbf u \right)$. In abstract algebra, this is known as a homomorphism.

## Theorems

• If $\mathcal B$ and $\mathcal C$ are bases for $V$, then they have the same cardinality. That is, there exists a bijective function $f:\mathcal B \rightarrow \mathcal C$ (proof). As such, we may define the dimension of $V$ as the cardinality of any basis for $V$.
• Given a basis $\mathcal B = \left\{\mathbf v_1, \mathbf v_2, \mathbf v_3, \dots \right\}$, and linear combinations $\sum_{k=1}^n a_k\mathbf v_k$ and $\sum_{k=1}^n b_k\mathbf v_k$, if $\sum_{k=1}^n a_k\mathbf v_k = \sum_{k=1}^n b_k\mathbf v_k$, then $a_k = b_k$ for each applicable $k$. This is to say any vector in $V$ is a unique linear combination of vectors in $\mathcal B$.