Properties of Vector Spaces

# Properties of Vector Spaces

We will now look at some important properties of vector spaces and provide what may seem like trivial proofs. Please review the Vector Spaces page first nevertheless.

Theorem 1: If $V$ is a vector space and $x \in V$ then there exists only one additive identity $0 \in V$ such that $x + 0 = 0 + x = x$ $\forall x \in V$. |

**Proof:**Suppose that there exists two additive identities, that is $0$ and $0'$ are both additive identities for all $x \in V$. Then it follows that:

\begin{align} 0 = 0 + 0' = 0' \implies 0 = 0' \end{align}

- We note that these equalities are true since if $0$ and $0'$ are both additive identities, then $0 = 0 + 0'$ and $0' = 0' + 0$. $\blacksquare$

Theorem 2: If $V$ is a vector space then $\forall x \in V$, $\exists (-x) \in V$ where $(-x)$ is a unique additive inverse to $x$. |

**Proof:**Suppose that $x \in V$ and that both $-x$ and $-x'$ are additive inverses to $x$. Then it follows that:

\begin{align} \quad -x =(-x) + 0 = (-x) + (x + (-x')) = (-x + x) + (-x') = 0 + (-x') = -x' \implies -x = -x' \end{align}

- Note that we made two substitutions in this proof, that is $0 = (x + (-x'))$ which is true if $-x'$ is an additive inverse to $x$, and $(-x + x) = 0$ which is true if $-x$ is an additive inverse to $x$. $\blacksquare$

Theorem 3: The product of the $0$ scalar and a vector $x \in V$ is equal to the zero vector, that is $0x = 0$. |

**Proof:**Let $x \in V$, and therefore:

\begin{align} \quad 0x = (0 + 0)x = 0x + 0x \implies 0x = 0x + 0x \implies 0x + (-0x) = 0x + 0x + (-0x) \implies 0 = 0x \quad \blacksquare \end{align}

Theorem 4: The product of any scalar $k$ and the zero vector $0 \in V$ is equal to the zero vector, that is $k0 = 0$. |

**Proof:**Let $k$ be a scalar and let $0 \in V$ be the zero vector. Therefore:

\begin{align} \quad k0 = k(0 + 0) = k0 + k0 \implies k0 = k0 + k0 \implies k0 + (-k0) = k0 + k0 + (-k0) \implies 0 = k0 \quad \blacksquare \end{align}

Theorem 5: The scalar $-1$ multiplied by the vector $x \in V$ produced the additive inverse of $x$, that is $(-1)x = (-x)$. |

**Proof:**Consider the vectors $x$ and $(-1)x$, and suppose we add them together as follows:

\begin{align} \quad x + (-1)x = 1x + (-1)x = (1 + (-1))x = 0x = 0 \implies x + (-1)x = 0 \end{align}

- Therefore $(-1)x$ is the additive inverse of $x$. $\blacksquare$

Theorem 6: Let $a, b \in \mathbb{F}$. Then the equation $a + x = b$ has a unique solution $x \in \mathbb{F}$. |

**Proof:**Suppose that both $x$ and $y$ are solutions to the equation $a + x = b$. Then $a + x = b$ and $a + y = b$, and so $x = b + (-a)$ and $y = b + (-a)$, which implies that $x = y$ and so the solution is unique. $\blacksquare$

Theorem 7: Let $a, b \in \mathbb{F}$. Then the equation $ax = b$ has a unique solution $x \in \mathbb{F}$. |

**Proof:**Suppose that both $x$ and $y$ are solutions to the equation $ax = b$. Then $ax = b$ and $ay = b$, and so $x = a^{-1}b$ and $y = a^{-1}b$, which implies that $x = y$ and so the solution is unique. $\blacksquare$