Basic Properties of Vectors

# Basic Properties of Vectors

We will now look at some important definitions and properties of vectors in Euclidean n-space. All proofs given are rather straightforward from the definition of the operations we are about to look at, and most explanation will be omitted.

 Definition: Two vectors $\vec{u}$ and $\vec{v}$ are said to be Equal if and only if each of their respective components are equal, that is $u_{i} = v_{i}$ for every vector component $i$.

For example, the following vectors $\vec{u} = (1, 3)$ and $\vec{v} = (1, 3)$ are equal since $u_1 = v_1 = 1$ and $u_2 = v_2 = 3$.

 Definition: If $\vec{u}, \vec{v} \in \mathbb{R}^n$ are vectors, then Vector addition is defined by adding corresponding components to each other, that is $\vec{u} + \vec{v} = (u_1, u_2, ..., u_n) + (v_1, v_2, ..., v_n) = (u_1 + v_1, u_2 + v_2, ..., u_n + v_n)$. Furthermore, vector subtraction is defined by $\vec{u} - \vec{v} = (u_1, u_2, ..., u_n) - (v_1, v_2, ..., v_n) = (u_1 - v_1, u_2 - v_2, ..., u_n - v_n)$.

We will now look at the important commutativity property of vector addition.

 Theorem 1 (Commutativity of Vector Addition): If $\vec{u}, \vec{v} \in \mathbb{R}^n$, then $\vec{u} + \vec{v} = \vec{v} + \vec{u}$.
• Proof: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$ The addition of vectors is defined such that each respective component of vector the first vector will be added to the second vector, that is:
(1)
\begin{align} \vec{u} + \vec{v} = (u_{1}, u_{2}, ..., u_{n}) + (v_{1}, v_{2}, ..., v_{n}) \\ \vec{u} + \vec{v} = (u_{1} + v_{1}, u_{2} + u_{2}, ..., u_{n} + v_{n}) \\ \vec{u} + \vec{v} = (v_{1} + u_{1}, v_{2} + u_{2}, ..., v_{n} + u_{n}) \\ \vec{u} + \vec{v} = \vec{v} + \vec{u} \\ \blacksquare \end{align}

For example, if we wanted to find the sum $\vec{u} + \vec{v}$ given that $\vec{u} = (1, 2)$ and $\vec{v} = (4, 2)$, then:

(2)
\begin{align} \vec{u} + \vec{v} = (1, 2) + (4, 2) \\ \vec{u} + \vec{v} = (5, 4) \end{align}

Furthermore, if we instead calculate $\vec{v} + \vec{u}$, we get the same result:

(3)
\begin{align} \vec{v} + \vec{u} = (4, 2) + (1, 2) \\ \vec{v} + \vec{u} = (5, 4) \end{align}
 Definition: The Zero Vector in Euclidean n-space denoted $\vec{0} = \underbrace{(0, 0, ..., 0)}_{\mathrm{n-times}}$ has all of its components as zeroes.
 Theorem 2 (Zero Vector Addition Property): If $\vec{u}, \vec{0} \in \mathbb{R}^n$, then $\vec{u} + \vec{0} = \vec{0} + \vec{u} = \vec{u}$.
• Proof: From Theorem 1, we have that $\vec{u} + \vec{0} = \vec{0} + \vec{u}$ already, so we only need to show that $\vec{u} + \vec{0} = \vec{u}$.
(4)
\begin{align} \vec{u} + \vec{0} = (u_1, u_2, ..., u_n) + (0, 0, ..., 0) \\ \vec{u} + \vec{0} = (u_1 + 0, u_2 + 0, ..., u_n + 0) = \vec{u}\\ \vec{u} + \vec{0} = (0 + u_1, 0 + u_2, ..., 0 + u_n) \\ \vec{u} + \vec{0} = (u_1, u_2, ..., u_n) \\ \vec{u} + \vec{0} = \vec{u} \quad \blacksquare \end{align}
 Theorem 3 (Associativity of Vector Addition): If $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$, then it follows that $\vec{u} + ( \vec{v} + \vec{w} ) = ( \vec{u} + \vec{v} ) + \vec{w}$.
• Proof:
(5)
\begin{align} \quad \vec{u} + (\vec{v} + \vec{w}) = (u_1, u_2, ..., u_n) + (v_1 + w_1, v_2 + w_2, ..., v_n + w_n) \\ \quad \vec{u} + (\vec{v} + \vec{w}) = (u_1 + v_1 + w_1, u_2 + v_2 + w_2, ..., u_n + v_n + w_n) \\ \quad \vec{u} + (\vec{v} + \vec{w}) = (u_1 + v_1, u_2 + v_2, ..., u_n + v_n) + (w_1, w_2, ..., w_n) \\ \quad \vec{u} + (\vec{v} + \vec{w}) = (\vec{u} + \vec{v}) + \vec{w} \\ \blacksquare \end{align}

Recall that a scalar is defined as a quantity that has only magnitude and no direction. We will now look at our next definition.

 Definition: If $\vec{u} \in \mathbb{R}^n$, and $k \in \mathbb{R}$, then $k\vec{u}$ is defined to be a Scalar Multiple of the Vector $\vec{u}$ that has a length that is $k$ times that of $\vec{u}$, and $k\vec{u} = k(u_1, u_2, ..., u_n) = (ku_1, ku_2, ..., ku_n)$. If $k < 0$, the direction of the vector changes to be in the position opposite to $\vec{u}$.
 Theorem 4: If $\vec{u} \in \mathbb{R}^n$ and $k, l$ are scalars such that $k, l \in \mathbb{R}$, then $k(l\vec{u}) = (kl)\vec{u}$.
• Proof:
(6)
\begin{align} k(l\vec{u}) = k(l(u_1, u_2, ..., u_n)) \\ k(l\vec{u}) = k(lu_1, lu_2, ..., lu_n) \\ k(l\vec{u}) = (klu_1, klu_2, ..., klu_n) \\ k(l\vec{u}) = kl(u_1, u_2, ..., u_n) \\ k(l\vec{u}) = (kl)\vec{u} \\ \blacksquare \end{align}
 Theorem 5: If $\vec{u} \in \mathbb{R}^n$ and $k, l$ are scalars such that $k, l \in \mathbb{R}$, then $(k + l)\vec{u} = k\vec{u} + l\vec{u}$.
• Proof:
(7)
\begin{align} (k + l)\vec{u} = (k + l)(u_1, u_2, ..., u_n) \\ (k + l)\vec{u} = ((k+l)u_1, (k+l)u_2, ..., (k+l)u_n) \\ (k + l)\vec{u} =(ku_1 + lu_1, ku_2 + lu_2, ..., ku_n + lu_n) \\ \quad (k + l)\vec{u} =(ku_1, ku_2, ..., ku_n) + (lu_1, lu_2, ..., lu_n) \\ (k + l)\vec{u} =k(u_1, u_2, ..., u_n) + l(u_1, u_2, ..., u_n) \\ (k + l)\vec{u} =k\vec{u} + l\vec{u} \\ \blacksquare \end{align}