Abhijeet Mulgund's Personal Webpage

Search

Search IconIcon to open search

Vector Space

Last updated Nov 1, 2022

# Definition

Suppose $V$ is a Nonempty Set, $F$ is a Field, and $+: V \times V \to V$ and $*: F \times V \to V$ are Operations. Then $(V, F, +, *)$ is a Vector Space if

  1. $(V, +)$ is an Abelian Group:
    1. Additive Identity: There exists $\mathbf{0} \in V$ so that for all $v \in V$ we have that $\mathbf{0} + v = v$
    2. Associativity of $+$: For all $x,y,z \in V$, $(x + y) + z = x + (y + z)$
    3. Commutativity of $+$: For all $x, y \in V$, $x + y = y + x$
    4. Additive Inverses: If $x \in V$, then there exists $-x \in V$ so that $x + (-x) = \mathbf{0}$
  2. Compatibility with Scalar Multiplication ($*$):
    1. Multiplicative Identity: For all $v \in V$ we have that $1 * v = v$ (where $1$ is the Multiplicative Identity of $F$).
    2. Associativity: For all $c,d \in F$, $v \in V$, we have that $(cd)v = c(d*v)$
    3. Distributivity of $$ over $+$: For all $u, v \in V$ and $c \in F$, $c(u + v) = cu + cv$
    4. Distributivity of $*$ over $+{F}$: For all $u \in V$ and $c, d \in F$, $(c +{F} d) * u = c * u + d * u$

# Examples

  1. Functions to a Field form a Vector Space
  2. N-tuples over a Field form a Vector Space

# n-tuples

# Properties

# 1. Scalar multiplication by $0$ is $\mathbf{0}$

Suppose $v \in V$. Then $0v = \mathbf{0}$.

# Proof

$$\begin{align*} &0v = (0 + 0)v = 0v + 0v\\ &\Rightarrow \mathbf{0} = 0v. \end{align*}$$ $\blacksquare$

# 2. Scalar multiplication with $\mathbf{0}$ is $\mathbf{0}$

Suppose $c \in F$. Then $c \mathbf{0} = \mathbf{0}$.

# Proof

$$\begin{align*} c \mathbf{0} &= c (\mathbf{0} + \mathbf{0}) = c \mathbf{0} + c \mathbf{0}\\ \Rightarrow &\mathbf{0} = c \mathbf{0} \end{align*}$$ $\blacksquare$

# 3. If scalar product is $\mathbf{0}$, then either scalar is $0$ or vector is $\mathbf{0}$

Suppose $v \in V$, $c \in F$ and that $cv = \mathbf{0}$. Then either $c = 0$ or $v = \mathbf{0}$.

# Proof

If $c = 0$, then we are done (by (1) above). Otherwise, $c$ has Multiplicative Inverse $c^{-1}$. Then $$\begin{align*} v = 1v = (c^{-1}c)v = c^{-1}(cv) = c^{-1} \mathbf{0} = \mathbf{0}.\\ \end{align*}$$ $\blacksquare$

#