Lec 04-16-2026: Scalar Multiplication & Vector Subtraction | MATH 245

In our last lecture, we ended off with vector addition (tail-to-tip). Before we discuss vector subtraction, we’ll discuss scalar multiplication.

What does it mean to multiply a vector by a number? Intuitively, it should stretch or shrink the vector - and possibly flip its direction.

Def: Given a vector $\vec{u} = \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2$, and a real number $c$, we define the scalar multiple of $\vec{u}$ by $c$, denoted by $c\vec{u}$, to be

$$c\vec{u} = \begin{bmatrix} cx \\ cy \end{bmatrix}$$

Ex: Suppose $\vec{u} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Find $2\vec{u}$, $3\vec{u}$, $\frac{1}{2}\vec{u}$, $-\vec{u}$.

$$2\vec{u} = 2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$$ $$3\vec{u} = 3\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix}$$

$$\tfrac{1}{2}\vec{u} = \tfrac{1}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} \tfrac{1}{2} \\[4pt] \tfrac{1}{2} \end{bmatrix}$$ $$-\vec{u} = -1\vec{u} = -1\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix}$$

Geometrically, what does $c\vec{u}$ do to $\vec{u}$? Let’s plot the vectors above.

Geometrically, multiplying by a positive scalar $c > 0$ stretches or shrinks $\vec{u}$ while keeping it on the same ray - all of $\vec{u}$, $2\vec{u}$, $3\vec{u}$, $\frac{1}{2}\vec{u}$ point in the same direction. Multiplying by $c < 0$ reverses the direction and scales by $|c|$, so $-\vec{u}$ points exactly opposite to $\vec{u}$ with the same length.

Vector Arithmetic

Since vector subtraction is defined as $\vec{u} - \vec{v} = \vec{u} + (-1)\vec{v}$, we can compute it by combining scalar multiplication and vector addition. Here is a worked example that mixes both operations.

Ex: Given $\vec{u} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$ & $\vec{v} = \begin{bmatrix} 2 \\ -5 \end{bmatrix}$, find $4\vec{u} - 3\vec{v}$.

Sol 1: (Long Method)

$$\begin{align*} 4\vec{u} - 3\vec{v} &= 4\vec{u} + (-3)\vec{v} \\ &= 4\begin{bmatrix} 1 \\ -2 \end{bmatrix} + (-3)\begin{bmatrix} 2 \\ -5 \end{bmatrix} \\ &= \begin{bmatrix} 4 \\ -8 \end{bmatrix} + \begin{bmatrix} -6 \\ 15 \end{bmatrix} = \begin{bmatrix} -2 \\ 7 \end{bmatrix} \end{align*}$$

Sol 2: (Component-wise Subtraction)

$$\begin{align*} 4\vec{u} - 3\vec{v} &= 4\begin{bmatrix} 1 \\ -2 \end{bmatrix} - 3\begin{bmatrix} 2 \\ -5 \end{bmatrix} \\ &= \begin{bmatrix} 4 \\ -8 \end{bmatrix} - \begin{bmatrix} 6 \\ -15 \end{bmatrix} = \begin{bmatrix} -2 \\ 7 \end{bmatrix} \end{align*}$$

Geometrically, we understand vector addition (Tail-to-Tip method from last time). Since $\vec{u} - \vec{v} = \vec{u} + (-\vec{v})$, vector subtraction is: negate $\vec{v}$ (flip its direction), then add tail-to-tip.

Ex: Geometrically, what is vector subtraction?

Addition

$\vec{u} + \vec{v}$

Subtraction

$\vec{u} - \vec{v} = \vec{u} + (-\vec{v})$

Taking both diagrams together, we get a cleaner geometric interpretation of $\vec{u} - \vec{v}$:

If we have two vectors $\vec{u}$ & $\vec{v}$ whose initial point is the origin, then $\vec{u} - \vec{v}$ is the vector whose initial point is the tip of $\vec{v}$ & terminal point is the tip of $\vec{u}$. In other words, $\vec{u} - \vec{v}$ points from the tip of $\vec{v}$ to the tip of $\vec{u}$.

Vectors in $\mathbb{R}^n$

Vectors in $\mathbb{R}^n$ behave similarly.

$$\mathbb{R}^n = \left\{ \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \;\middle|\; x_1, x_2, \ldots, x_n \in \mathbb{R} \right\}$$

Def: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$ where $\vec{u} = \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix}$ & $\vec{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}$. Then,

$$\vec{u} + \vec{v} = \begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ \vdots \\ u_n + v_n \end{bmatrix} \qquad \& \qquad c\vec{u} = c\begin{bmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{bmatrix} = \begin{bmatrix} cu_1 \\ cu_2 \\ \vdots \\ cu_n \end{bmatrix} \quad \text{where } c \in \mathbb{R}$$

These vector operations satisfy the same familiar rules as regular number arithmetic - the same properties you relied on in algebra still hold.

Thm: For all vectors $\vec{u}$, $\vec{v}$ & $\vec{w}$ in $\mathbb{R}^n$ & for all scalars $c$, $d$, we have:

• $\vec{u} + \vec{v} = \vec{v} + \vec{u}$
• $(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$
• $\vec{u} + \vec{0} = \vec{u}$; $\vec{0}$ is the vector whose entries are all 0’s.
• $\vec{u} + (-\vec{u}) = \vec{0}$
• $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
• $(c + d)\vec{u} = c\vec{u} + d\vec{u}$
• $c(d\vec{u}) = (cd)\vec{u}$
• $1\vec{u} = \vec{u}$

Norms (Magnitude / Length)

The length (or norm or magnitude) of a vector is an intuitive idea. Let’s show the progression from $\mathbb{R}^2 \to \mathbb{R}^3 \to \mathbb{R}^n$.

$\mathbb{R}^2$

Given $\vec{v} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$, let’s denote the length of $\vec{v}$ by $\|\vec{v}\|$. Can we find $\|\vec{v}\|$?

By the Pythagorean Theorem:

$$3^2 + 4^2 = \|\vec{v}\|^2$$ $$\|\vec{v}\| = \sqrt{3^2 + 4^2}$$ $$9 + 16 = \|\vec{v}\|^2$$ $$25 = \|\vec{v}\|^2$$ $$\|\vec{v}\| = 5$$

In general, if $\vec{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$, then $\|\vec{v}\| = \sqrt{v_1^2 + v_2^2}$.

$\mathbb{R}^3$

If $\vec{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$, then $\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}$.

Ex: Given $\vec{v} = \begin{bmatrix} 2 \\ 3 \\ 6 \end{bmatrix}$, find $\|\vec{v}\|$.

The strategy is to apply the Pythagorean Theorem twice. First, look straight down (bird’s eye view) at the $xy$-plane and find $\ell$, the length of the horizontal leg:

$$2^2 + 3^2 = \ell^2 \implies \ell = \sqrt{13}$$

Now treat the full vector as the hypotenuse of a right triangle with legs $\ell$ and $6$ (the $z$-component), and apply the Pythagorean Theorem again:

$$\begin{align*} \ell^2 + 6^2 &= \|\vec{v}\|^2 \\ 13 + 6^2 &= \|\vec{v}\|^2 \\ 2^2 + 3^2 + 6^2 &= \|\vec{v}\|^2 \\ \|\vec{v}\| &= \sqrt{2^2 + 3^2 + 6^2} \end{align*}$$

$\mathbb{R}^n$

The pattern is clear: in $\mathbb{R}^2$ we summed 2 squared components; in $\mathbb{R}^3$ we summed 3. In $\mathbb{R}^n$ the pattern continues to $n$ components:

If $\vec{w} = \begin{bmatrix} w_1 \\ \vdots \\ w_n \end{bmatrix}$, then $\|\vec{w}\| = \sqrt{w_1^2 + \cdots + w_n^2}$ (called the Euclidean norm).

Q: Given a vector $\vec{v}$, what is the length of $c\vec{v}$?

$$\begin{align*} \|c\vec{v}\| &= \left\| c \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \right\| = \left\| \begin{bmatrix} cv_1 \\ cv_2 \\ \vdots \\ cv_n \end{bmatrix} \right\| = \sqrt{(cv_1)^2 + (cv_2)^2 + \cdots + (cv_n)^2} \\ &= \sqrt{c^2(v_1^2 + v_2^2 + \cdots + v_n^2)} = \sqrt{c^2}\,\sqrt{v_1^2 + v_2^2 + \cdots + v_n^2} = |c|\,\|\vec{v}\| \end{align*}$$

So $\|c\vec{v}\| = |c|\,\|\vec{v}\|$ - scaling a vector by $c$ scales its length by $|c|$. This confirms our earlier geometric intuition: when $c > 0$, the vector stretches or shrinks by a factor of $c$; when $c < 0$, the direction flips and the length scales by $|c|$.

Unit Vectors

Sometimes we care only about the direction of a vector, not its magnitude. To isolate direction, we can normalize $\vec{v}$: scale it down to length 1 while preserving its direction. The result is called a unit vector, denoted $\hat{v}$.

Ex: Given $\vec{v} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$, normalize $\vec{v}$.

Recall $\|\vec{v}\| = 5$. We want to find a scalar $c$ such that $\|c\vec{v}\| = 1$. Using the result from above, $\|c\vec{v}\| = |c|\,\|\vec{v}\|$, so we need $|c| \cdot 5 = 1$, giving $c = \frac{1}{5} = \frac{1}{\|\vec{v}\|}$. Applying this:

$$\hat{v} = \frac{1}{\|\vec{v}\|}\vec{v} = \frac{1}{5}\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix}$$

This is the unit vector of $\vec{v}$: it points in the same direction as $\vec{v}$, but has length 1.

Angles & the Dot Product

Goal: Given two vectors $\vec{u}$ & $\vec{v}$, find a formula for the angle $\theta$ between them directly from their components, without needing to draw a picture.

To do this, we will use the vector $\vec{u} - \vec{v}$ as the third side of the triangle formed by $\vec{u}$ and $\vec{v}$, then apply the Law of Cosines. Recall given vectors $\vec{u}$ & $\vec{v}$, the vector going from the tip of $\vec{v}$ to the tip of $\vec{u}$ is $\vec{u} - \vec{v}$.

We will now apply the Law of Cosines to the triangle formed by $\vec{u}$, $\vec{v}$, and $\vec{u} - \vec{v}$, using the three side lengths $\|\vec{u}\|$, $\|\vec{v}\|$, and $\|\vec{u} - \vec{v}\|$.

Law of Cosines

In any triangle $ABC$, we have the following:

$$a^2 = b^2 + c^2 - 2bc\cos(A)$$

Proof

We want to express $a^2$ in terms of $b$, $c$, and the angle $A$. The trick is to drop a perpendicular (altitude) from vertex $C$ down to side $AB$, which splits the triangle into two right triangles. Calling the foot of the altitude $H$, let $x = AH$ (the horizontal distance from $A$ to $H$) and $h = CH$ (the height). Then we can apply the Pythagorean Theorem to each right triangle separately:

(Acute triangle)

From the left right triangle (legs $x$ and $h$, hypotenuse $b$):

$$x^2 + h^2 = b^2 \implies h^2 = b^2 - x^2$$

From the right right triangle (legs $c - x$ and $h$, hypotenuse $a$):

$$h^2 + (c - x)^2 = a^2$$