Vector space: properties – "Math for Non-Geeks"

In this chapter, we will consider some properties of vector spaces which can be derived directly from the definition of a vector space. So every vector space must satisfy these properties, no matter how abstract or high-dimensional.

Overview

In this section we use again the operation symbols "$⊞$" and "$⊡$" to distinguish the vector addition and the scalar multiplication with the field addition "$+$" and the field multiplication "$\cdot$".

We want to derive simple properties and rules from the eight axioms of the vector space. Since we have demanded in the axioms only the existence of the zero vector and the additive inverse, the following questions arise first: Is the zero vector unique or are there several zero vectors? Is the inverse element of addition unique or can there be more than one? The answer to both questions is:

• In every vector space there is exactly one zero vector $0_V\in V$. So nope, there cannot be more than one zero vector in a vector space.
• The inverse with respect to addition is unique. So for every vector $v\in V$ there is exactly one other vector $w\in V$ with $v+w=0$.

Further statements that we will prove in the following are:

• For every $v\in V$ we have that: $0_K⊡v=0_V$.
• For every $\rho \in K$ we have that: $\rho ⊡0_V=0_V$.
• From $\rho ⊡v=0_V$ it follows that $\rho =0_K$ or $v=0_V$.
• For all $\rho \in K$ and all $v\in V$ we have that: $(-\rho )⊡v=-(\rho ⊡v)=\rho ⊡(-v)$.

In the following, we denote by $V$ a vector space over a field $K$.

Uniqueness of the zero vector

Proving uniqueness of the zero vector (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Uniqueness of the zero vector)

In every vector space $V$, the zero vector $0_V$ is unique

Proof (Uniqueness of the zero vector)

Suppose we had two vectors $0_V$ and $0{\text{'}}_V$ with the zero vector property, i.e. $0_V$ and $0{\text{'}}_V$ satisfy for all vectors $v\in V$ the equation $v⊞0_V=v$ respectively.

If we take the first equation $v⊞0_V=v$ and plug in for $v$ the concrete vector $0{\text{'}}_V$, and if we analogously, in the second equation $v⊞0{\text{'}}_V=v$ plug in for $v$ the vector $0_V$, then we get $0{\text{'}}_V⊞0_V=0{\text{'}}_V$ and $0_V⊞0{\text{'}}_V=0_V$.

Because of the commutative law of vector addition $0_V⊞0{\text{'}}_V=0{\text{'}}_V⊞0_V$ and accordingly

$0_V=0_V⊞0{\text{'}}_V=0{\text{'}}_V⊞0_V=0{\text{'}}_V$

Thus $0_V=0{\text{'}}_V$.

We have shown in total that in a vector space $V$ two vectors with the null property are equal. Thus, the zero vector is unique.

Inverses are unique

Proving the uniqueness of inverses (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Inverses are unique)

The inverse with respect to addition is unique. That means, for every vector $v\in V$ there is exactly one vector $w\in V$ with $v⊞w=0_V$.

Proof (Inverses are unique)

Let $v,w,x\in V$. We assume that $v$ and $w$ are two additive inverses to $x$. Let thus $x⊞v=0_V$ and $x⊞w=0_V$. We now show that $v=w$ and that there can be only one additive inverse.

$\begin{array}{cccc}v& =v⊞0_V& & \\ & \downarrow \text{zero vector satisfies }x⊞w=0_V\\ & =v⊞(x⊞w)\\ & \downarrow \text{associative law of vector addition}\\ & =(v⊞x)⊞w\\ & \downarrow \text{commutative law of vector addition}\\ & =(x⊞v)⊞w\\ & \downarrow x⊞v=0_V\\ & =0_V⊞w=w\end{array}$

Thus, the two additive inverses $v$ and $w$ are identical.

Hint

In the proof above, we wrote $w$ and $v$, respectively, for the additive inverse of $x$. However, we have seen that there is exactly one additive inverse. This is normally denoted by $-x$ and we will use this notation for the additive inverse in the following sections.

Scaling by zero results in the zero vector

Scaling by zero results in the zero vector - proof (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Scaling by zero results in the zero vector)

For every $v\in V$ we have that: $0_K⊡v=0_V$.

Proof (Scaling by zero results in the zero vector)

$\begin{array}{cc}& 0_V\\ & \downarrow \text{vector addition by the inverse of }{0}_K⊡v\\ [0.3em]=& 0_K⊡v⊞(-0_K⊡v)\\ & \downarrow \text{replace }{0}_K\text{ by }{0}_K+0_K\\ =& (0_K+0_K)⊡v⊞(-0_K⊡v)\\ & \downarrow \text{scalar distributive law }\\ =& (0_K⊡v⊞0_K⊡v)⊞(-0_K⊡v)\\ & \downarrow \text{associative law of vector addition}\\ =& 0_K⊡v⊞(0_K⊡v⊞(-0_K⊡v))\\ & \downarrow \text{0 is its own additive inverse: }{0}_K⊡v⊞(-0_K⊡v)=0_V\\ =& 0_K⊡v\end{array}$

So $0_V=0_K⊡v$.

Question: Can a vector $\rho \cdot v\ne 0$ be multiplied with a scalar $\mu \in K$ in such a way that the original vector $v\ne 0$ is the result?

The answer is yes, because if we choose $\mu ={\rho }^{-1}$, then according to the associative law of scalar multiplication, we have

$\begin{array}{cc}{\rho }^{-1}⊡(\rho ⊡v)& =({\rho }^{-1}\cdot \rho )⊡v\\ & =1⊡v\\ & \downarrow \text{axiom: }1⊡v=v\\ & =v\end{array}$

Now ${\rho }^{-1}$ is only well-defined if $\rho \ne 0$. But this is always true, since $\rho \cdot v\ne 0$. If there was $\rho =0$ then $\rho \cdot v=0$ which is excluded by the premise $\rho \cdot v\ne 0$.

This shows why it is useful to define vector spaces $V$ over a field $K$ and not something weaker (like a ring). This is because in fields there is a multiplicative inverse to every non-zero element. So there is an inverse $\rho \in K$ with $\rho \ne 0$ to every $\mu \in K$ with $\mu \cdot \rho =1$. Thus, the fact that $K$ is a field guarantees that every scaling (stretching) $\rho ⊡v$ can be scaled back by the inverse scaling (with the inverse stretching factors) $\mu$ such that $\mu ⊡(\rho ⊡v)=v$.

Scaling the zero vector again gives the zero vector

Scaling the zero vector again gives the zero vector - proof (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Scaling the zero vector again gives the zero vector)

For all $\rho \in K$ we have that: $\rho ⊡0_V=0_V$.

Proof (Scaling the zero vector again gives the zero vector)

Let $0_V$ be the neutral element of vector addition and $\rho \in K$ arbitrary. We have that:

$\begin{array}{cc}& 0_V\\ & \downarrow \text{vector addition by the inverse of }\rho ⊡0_V\\ =& \rho ⊡0_V⊞(-\rho ⊡0_V)\\ & \downarrow \text{replace the zero vector }{0}_V\text{ by }{0}_V⊞0_V\\ =& \rho ⊡(0_V⊞0_V)⊞(-\rho ⊡0_V)\\ & \downarrow \text{vectorial distributive law}\\ =& (\rho ⊡0_V⊞\rho ⊡0_V)⊞(-\rho ⊡0_V)\\ & \downarrow \text{associative law of vector addition}\\ =& \rho ⊡0_V⊞(\rho ⊡0_V⊞(-\rho ⊡0_V))\\ & \downarrow \text{0 can be written as }\rho ⊡0_V⊞(-\rho ⊡0_V)=0_V\\ =& \rho ⊡0_V\end{array}$

So $0_V=\rho ⊡0_V$.

Scalar multiplication leaves no zero divisors

Proof that scalar multiplication leaves no zero divisors (YouTube video (German) by the YouTube channel Maths CA).

Theorem

Let $\rho \in K$ and $v\in V$ be such that $\rho ⊡v=0_V$. Then, either $\rho =0_K$ or $v=0_V$ (or both).

In already proved theorems we have shown that in the case $\rho =0_K$ or $v=0_V$ the equation $\rho ⊡v=0_V$ is satisfied. Now we show conversely that it follows from $\rho ⊡v=0_V$ that $\rho =0_K$ or $v=0_V$.

Proof

Let $\rho ⊡v=0_V$. If $\rho \ne 0_K$, then there exists a ${\rho }^{-1}\in K$ with ${\rho }^{-1}\cdot \rho =1$. Thus:

$\begin{array}{cc}& v\\ & \downarrow \text{axiom: }1⊡v=v\\ =& 1⊡v\\ [0.3em]=& ({\rho }^{-1}\rho )⊡v\\ & \downarrow \text{associative law for scalars }\\ =& {\rho }^{-1}⊡(\rho ⊡v)\\ & \downarrow \rho ⊡v=0_V\\ =& {\rho }^{-1}⊡0_V\\ & \downarrow \text{theorem (already proven), that for all }\gamma \in K\text{ we have }\gamma ⊡0_V=0_V\\ =& 0_V\end{array}$

So from $\rho \ne 0_K$ we get $v=0_V$. Thus $\rho =0_K$ or $v=0_V$ must hold, and we cannot at the same time have $\rho \ne 0_K$ and $v\ne 0_V$.

Scaling by a negative scalar

Scaling a vector by a negative scalar (YouTube video (German) by the YouTube channel Maths CA).

Theorem (Scaling by a negative scalar)

For all $\rho \in K$ and all $v\in V$ we have that

$(-\rho )⊡v=-(\rho ⊡v)=\rho ⊡(-v)$

Proof (Scaling by a negative scalar)

By the scalar distributive law for scalar multiplication we have on one hand:

$\begin{array}{cc}& (\rho ⊡v)⊞((-\rho )⊡v)\\ & \downarrow (\lambda +\rho )⊡v=(\lambda ⊡v)⊞(\rho ⊡v)\\ =& (\rho +(-\rho ))⊡v\\ =& 0_K⊡v\\ =& 0_V\end{array}$

This shows that $(-\rho )⊡v$ is the additive inverse of $\rho ⊡v$. Thus $(-\rho )⊡v=-(\rho ⊡v)$. On the other hand, according to vectorial distributive law.

$\begin{array}{cc}& (\rho ⊡v)⊞(\rho ⊡(-v))\\ & \downarrow \lambda ⊡(v⊞w)=(\lambda ⊡v)⊞(\lambda ⊡w)\\ =& \rho ⊡(v⊞(-v))\\ =& \rho ⊡0_V\\ =& 0_V\end{array}$

This shows that $\rho ⊡(-v)$ is the additive inverse of $\rho ⊡v$, so it is equal to $-(\rho ⊡v)$. Thus, in total

$(-\rho )⊡v=-(\rho ⊡v)=\rho ⊡(-v)$

Hint

From the above theorem, we immediately get $(-1)⊡v=-v$. This follows from $(-1)⊡v=-(1⊡v)=-v$ (where we have used $1⊡v=v$).