Determining Whether a Set is a Vector Space

# Determining Whether a Set is a Vector Space

We have looked at a variety of different vector spaces so far including:

We will now look at some contrived examples of sets under specified operations of addition and scalar multiplication and determine whether or not they are vector spaces.

## Example 1

Let $V = \mathbb{R}^2$, and let $u, v \in \mathbb{V}$ such that $u = (u_1, u_2)$ and $v = (v_1, v_2)$. Define addition component-wise, that is $u + v = (u_1 + v_1, u_2 + v_2)$ and define scalar multiplication by $a \in \mathbb{F}$ to be $au = (au_1, 0)$. Determine whether or not this set under these operations is a vector space.

We first note that addition of vectors $u + v$ is defined as standard addition, however, multiplication is not defined standardly. Let $u$ be a vector whose second component is nonzero, that is $u_2 \neq 0$, and consider the existence of a multiplicative identity $1$ such that $1 u = u$. By the definition of scalar multiplication it follows that:

(1)
\begin{align} 1u = 1(u_1, u_2) = (1u_1, 0) \neq u \end{align}

Since $u_2 \neq 0$, it follows that $1u \neq u$ under the operation of scalar multiplication, and so axiom 6 (the existence of a multiplicative identity) does not hold, so $V$ is not a vector space under these prescribed operations.

## Example 2

Let $V = \mathbb{R}^2$, and let $u, v, w \in \mathbb{V}$ such that $u = (u_1, u_2)$ and $v = (v_1, v_2)$. Define addition as $u + v = (u_1v_1, u_2v_2)$ and define scalar multiplication by $a \in \mathbb{F}$ to be $au = (au_1, au_2 + 1)$. Determine whether or not this set under these operations is a vector space.

In this example both addition and scalar multiplication are not standard. We will attempt to verify that all ten axioms hold, and will stop verifying if one axiom fails.

• 1. $u + v = (u_1v_1, u_2v_2) = (v_1u_1, v_2u_2) = v + u$.
• 2. $u + (v + w) = (u_1[v_1w_1], u_2[v_2w_2]) = (u_1v_1w_1, u_2v_2w_2) = ([u_1v_1]w_1,[u_2v_2]w_2) = (u + v) + w$.
• 3. The zero vector is $1 = (1, 1)$, that is $1 + u = (1u_1, 1u_2) = (u_1, u_2) = u$.
• 4. This is where $V$ under the specified operations fails. Notice that for any vector $u$, that additive inverse would have to be $-u = \left ( \frac{1}{u_1}, \frac{1}{u_2} \right)$. However notice that for any vector that contains at least one component that is zero, for example the vector $m = (2, 0)$, no such additive inverse exists since $-m = \left ( \frac{1}{2}, \frac{1}{0} \right)$ is undefined.

Therefore $V$ is not a vector space under the prescribed operations.

## Example 3

Let $V = \mathbb{R}^3$, and let $u, v, w \in \mathbb{V}$ such that $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$. Define addition as $u + v = (u_1^2 + v_1^2, u_2^2 + v_2^2, u_3^2 + v_3^2)$ and define scalar multiplication by $a \in \mathbb{F}$ to be $au = (au_1, au_2 + 1)$. Determine whether or not this set under these operations is a vector space.

Once again, we will attempt to verify all ten axioms, and we will stop if at least one axiom fails.

• 1. $u + v = (u_1^2 + v_1^2, u_2^2 + v_2^2, u_3^2 + v_3^2) = (v_1^2 + u_1^2, v_2^2 + u_2^2, v_3^2 + u_3^2)$.
• 2. $u + (v + w) = (u_1^2 + [v_1^2 + w_1^2], u_2^2 + [v_2^2 + w_2^2], u_3^2 + [v_3^2 + w_3^2]) = ([u_1^2 + v_1^2] + w_1^2, [u_2^2 + v_2^2] + w_2^2, [u_3^2 + v_3^2] + w_3^2) = (u + v) + w$.
• 3. No such zero vector exists. Suppose that one did and call it $z$. Then $u + z = (u_1^2 + z_1^2, u_2^2 + z_2^2 + u_3^2 + z_3^2) = (u_1, u_2, u_3)$, or in other words, $u_i^2 + z_i^2 = u_i$ for $i = 1, 2, 3$ and so $z_i = \sqrt{u_i - u_i^2}$, and then $z = \left ( \sqrt{u_1 - u_1^2},\sqrt{u_2 - u_2^2},\sqrt{u_3 - u_3^2} \right)$. But this zero vector only holds for $u$. If $u \neq v$, then $v$ has a different zero vector.