Vector Operations
Learn how to add, subtract, and scale vectors
When you cook, you combine ingredients. A recipe might call for two cups of flour plus half a cup of sugar. You do not just dump everything together randomly; you add specific amounts of specific things to get the result you want. Vector operations work the same way. You can combine vectors by adding them, subtract one from another, or scale a vector up or down. These simple operations, as natural as combining ingredients, turn out to be the building blocks for everything else in linear algebra.
Think about walking around a city. If you walk three blocks north and then two blocks east, you end up at the same place as someone who walked two blocks east and then three blocks north. The order does not matter because you are combining displacements. This is vector addition in action: combining separate movements into a single net movement. And if someone tells you to “walk twice as far in that direction,” they are asking you to scale a vector, to stretch it by a factor of two while keeping its direction.
In this lesson, you will learn the precise rules for combining vectors. These rules are simple and intuitive when you think about them geometrically, and they become powerful computational tools when you work with components algebraically. By the end, you will understand how any vector can be built up from other vectors through addition and scaling, an idea that leads directly to the concept of linear combinations.
Core Concepts
Vector Addition: The Geometric View
Imagine you are giving someone directions. “Walk to the coffee shop, then from there, walk to the bookstore.” You are describing two displacements, and the person ends up in a final location that represents the combination of both movements. This is exactly what vector addition does.
Geometrically, we add two vectors using the tip-to-tail method:
- Draw the first vector $\vec{u}$
- Place the tail (starting point) of the second vector $\vec{v}$ at the tip (ending point) of $\vec{u}$
- The sum $\vec{u} + \vec{v}$ is the vector from the tail of $\vec{u}$ to the tip of $\vec{v}$
Think of it as chaining movements: first do one displacement, then do the other. The result is the direct route from where you started to where you ended up.
This is why vector addition makes sense for combining forces, velocities, or any quantities that have both magnitude and direction. If two people push on a box, one pushing east and one pushing north, the box moves in the combined direction, the vector sum of the two forces.
Vector Addition: The Algebraic View
While the geometric picture builds intuition, we need a computational method for actual calculations. The rule is remarkably simple: add corresponding components.
For 2D vectors:
$$\begin{pmatrix} u_1 \ u_2 \end{pmatrix} + \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \ u_2 + v_2 \end{pmatrix}$$
For 3D vectors:
$$\begin{pmatrix} u_1 \ u_2 \ u_3 \end{pmatrix} + \begin{pmatrix} v_1 \ v_2 \ v_3 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \ u_2 + v_2 \ u_3 + v_3 \end{pmatrix}$$
Each component of the result is just the sum of the corresponding components from the original vectors. This works because the $x$-component tells you the total horizontal movement, the $y$-component tells you the total vertical movement, and so on.
Important: You can only add vectors of the same dimension. Adding a 2D vector to a 3D vector is undefined, like trying to add apples and oranges.
The Parallelogram Rule
There is another way to visualize vector addition that reveals an important symmetry. If you place both vectors $\vec{u}$ and $\vec{v}$ with their tails at the same point, they form two sides of a parallelogram. The sum $\vec{u} + \vec{v}$ is the diagonal of this parallelogram, starting from the shared tail.
Why does this work? Because a parallelogram has opposite sides that are equal and parallel. If you go along $\vec{u}$ then $\vec{v}$, or along $\vec{v}$ then $\vec{u}$, you arrive at the same corner (the one diagonally opposite to where you started). The diagonal represents this final position.
The parallelogram rule is equivalent to the tip-to-tail method; it just offers a different way to visualize the same operation. It is particularly useful when you want to see both vectors starting from the same point.
Vector Subtraction
Subtraction is closely related to addition. To subtract $\vec{v}$ from $\vec{u}$, you add the negative of $\vec{v}$:
$$\vec{u} - \vec{v} = \vec{u} + (-\vec{v})$$
The negative of a vector $-\vec{v}$ has the same magnitude as $\vec{v}$ but points in the opposite direction. In component form:
$$-\begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} -v_1 \ -v_2 \end{pmatrix}$$
So subtraction becomes straightforward: subtract corresponding components.
$$\begin{pmatrix} u_1 \ u_2 \end{pmatrix} - \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} u_1 - v_1 \ u_2 - v_2 \end{pmatrix}$$
Geometric interpretation of subtraction: The vector $\vec{u} - \vec{v}$ points from the tip of $\vec{v}$ to the tip of $\vec{u}$ (when both vectors are drawn from the same starting point). This is exactly the displacement you would need to get from “where $\vec{v}$ ends” to “where $\vec{u}$ ends.”
This explains why, in the previous lesson, we found the displacement vector from point $A$ to point $B$ by subtracting: the position of $B$ minus the position of $A$ gives you the vector pointing from $A$ to $B$.
Scalar Multiplication: Stretching and Shrinking
A scalar is just a number (as opposed to a vector). When you multiply a vector by a scalar, you scale the vector, hence the name.
Scalar multiplication multiplies each component by the scalar:
$$c \cdot \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} c \cdot v_1 \ c \cdot v_2 \end{pmatrix}$$
The geometric effect depends on the value of $c$:
- If $c > 1$: The vector gets longer (stretched) in the same direction
- If $0 < c < 1$: The vector gets shorter (shrunk) but stays in the same direction
- If $c = 1$: The vector is unchanged
- If $c = 0$: The result is the zero vector (all movement eliminated)
- If $c < 0$: The vector reverses direction and scales by $|c|$
For example:
- $2\vec{v}$ is twice as long as $\vec{v}$, pointing in the same direction
- $\frac{1}{2}\vec{v}$ is half as long as $\vec{v}$, pointing in the same direction
- $-1\vec{v} = -\vec{v}$ has the same length as $\vec{v}$ but points in the opposite direction
- $-3\vec{v}$ is three times as long as $\vec{v}$ and points in the opposite direction
Parallel Vectors
Two vectors are parallel if one is a scalar multiple of the other. This means they point in either the same direction or exactly opposite directions.
$$\vec{u} \text{ and } \vec{v} \text{ are parallel if } \vec{u} = c\vec{v} \text{ for some scalar } c$$
For example:
- $\begin{pmatrix} 2 \ 4 \end{pmatrix}$ and $\begin{pmatrix} 1 \ 2 \end{pmatrix}$ are parallel because $\begin{pmatrix} 2 \ 4 \end{pmatrix} = 2\begin{pmatrix} 1 \ 2 \end{pmatrix}$
- $\begin{pmatrix} 3 \ -6 \end{pmatrix}$ and $\begin{pmatrix} -1 \ 2 \end{pmatrix}$ are parallel because $\begin{pmatrix} 3 \ -6 \end{pmatrix} = -3\begin{pmatrix} -1 \ 2 \end{pmatrix}$
To check if two vectors are parallel, see if the ratios of corresponding components are equal. For $\vec{u} = \begin{pmatrix} u_1 \ u_2 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} v_1 \ v_2 \end{pmatrix}$, they are parallel if $\frac{u_1}{v_1} = \frac{u_2}{v_2}$ (assuming no division by zero).
Linear Combinations
Now we come to one of the most important concepts in linear algebra. A linear combination of vectors is what you get when you scale some vectors and then add them together.
If $\vec{v}_1, \vec{v}_2, \ldots, \vec{v}_n$ are vectors and $c_1, c_2, \ldots, c_n$ are scalars, then:
$$c_1\vec{v}_1 + c_2\vec{v}_2 + \cdots + c_n\vec{v}_n$$
is a linear combination of those vectors. The scalars $c_1, c_2, \ldots, c_n$ are called the coefficients or weights.
For example, with two vectors $\vec{u}$ and $\vec{v}$:
- $3\vec{u} + 2\vec{v}$ is a linear combination
- $\vec{u} - 4\vec{v} = 1\vec{u} + (-4)\vec{v}$ is a linear combination
- $\frac{1}{2}\vec{u} + \frac{1}{2}\vec{v}$ is a linear combination (the midpoint of the two vectors, in a sense)
Linear combinations are everywhere:
- The position vector $\begin{pmatrix} 3 \ 4 \end{pmatrix}$ can be written as $3\begin{pmatrix} 1 \ 0 \end{pmatrix} + 4\begin{pmatrix} 0 \ 1 \end{pmatrix}$, a linear combination of the standard basis vectors
- Any movement in a room can be described as a linear combination of “move right,” “move forward,” and “move up”
- Any color on your screen is a linear combination of red, green, and blue
A fundamental question in linear algebra is: given some vectors, which other vectors can be expressed as their linear combinations? This question leads to concepts like span, linear independence, and dimension.
Properties of Vector Operations
Vector addition and scalar multiplication obey a set of rules that make them behave “nicely.” These properties are not just abstract formalism; they reflect the intuitive behavior you would expect from combining displacements.
Properties of Vector Addition:
-
Commutativity: $\vec{u} + \vec{v} = \vec{v} + \vec{u}$
The order does not matter. Walking north then east gets you to the same place as walking east then north.
-
Associativity: $(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$
When adding three or more vectors, grouping does not matter. You can add them in any order.
-
Additive Identity: $\vec{v} + \vec{0} = \vec{v}$
Adding the zero vector changes nothing. “Go somewhere, then go nowhere” leaves you where you were.
-
Additive Inverse: $\vec{v} + (-\vec{v}) = \vec{0}$
Every vector has an opposite that cancels it out. Going somewhere and then going back leaves you where you started.
Properties of Scalar Multiplication:
-
Associativity: $a(b\vec{v}) = (ab)\vec{v}$
Scaling by $a$ and then by $b$ is the same as scaling by $ab$ once.
-
Multiplicative Identity: $1\vec{v} = \vec{v}$
Scaling by 1 does nothing.
-
Distributivity over Vector Addition: $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
Scaling a sum is the same as scaling each vector and then adding.
-
Distributivity over Scalar Addition: $(a + b)\vec{v} = a\vec{v} + b\vec{v}$
Scaling by a sum of scalars is the same as scaling by each scalar and then adding the results.
These eight properties are so important that any mathematical structure satisfying them is called a vector space. Vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$ are just the most familiar examples of vector spaces.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| $\vec{u} + \vec{v}$ | Vector addition | $\begin{pmatrix} 1 \ 2 \end{pmatrix} + \begin{pmatrix} 3 \ 1 \end{pmatrix} = \begin{pmatrix} 4 \ 3 \end{pmatrix}$ |
| $c\vec{v}$ | Scalar multiplication | $2\begin{pmatrix} 3 \ 4 \end{pmatrix} = \begin{pmatrix} 6 \ 8 \end{pmatrix}$ |
| Linear combination | Sum of scaled vectors | $a\vec{u} + b\vec{v}$ |
| Parallel vectors | Same or opposite direction | $\vec{v}$ and $2\vec{v}$ are parallel |
| $-\vec{v}$ | Negative of a vector (opposite direction) | $-\begin{pmatrix} 3 \ -2 \end{pmatrix} = \begin{pmatrix} -3 \ 2 \end{pmatrix}$ |
| $\vec{u} - \vec{v}$ | Vector subtraction | $\begin{pmatrix} 5 \ 3 \end{pmatrix} - \begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \end{pmatrix}$ |
| Tip-to-tail method | Geometric method for adding vectors | Place tail of second at tip of first |
| Parallelogram rule | Alternative visualization of vector sum | Sum is the diagonal |
| Coefficient | Scalar multiplier in a linear combination | In $3\vec{u} + 2\vec{v}$, the coefficients are 3 and 2 |
Examples
Compute $\begin{pmatrix} 1 \ 3 \end{pmatrix} + \begin{pmatrix} 2 \ -1 \end{pmatrix}$.
Solution:
Add corresponding components.
Step 1: Add the first components (the $x$-components). $$1 + 2 = 3$$
Step 2: Add the second components (the $y$-components). $$3 + (-1) = 2$$
Step 3: Write the result as a vector.
$$\begin{pmatrix} 1 \ 3 \end{pmatrix} + \begin{pmatrix} 2 \ -1 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \end{pmatrix}$$
Geometric interpretation: If you walk 1 unit right and 3 units up, then walk 2 units right and 1 unit down, you end up 3 units right and 2 units up from where you started.
Compute $3\begin{pmatrix} 2 \ -4 \end{pmatrix}$.
Solution:
Multiply each component by the scalar.
Step 1: Multiply the first component by 3. $$3 \cdot 2 = 6$$
Step 2: Multiply the second component by 3. $$3 \cdot (-4) = -12$$
Step 3: Write the result as a vector.
$$3\begin{pmatrix} 2 \ -4 \end{pmatrix} = \begin{pmatrix} 6 \ -12 \end{pmatrix}$$
Geometric interpretation: The resulting vector points in the same direction as $\begin{pmatrix} 2 \ -4 \end{pmatrix}$ but is three times as long. If the original vector represented moving 2 units right and 4 units down, the scaled vector represents moving 6 units right and 12 units down.
Find $2\vec{u} - 3\vec{v}$ where $\vec{u} = \begin{pmatrix} 1 \ 4 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \ 1 \end{pmatrix}$.
Solution:
This expression combines scalar multiplication and subtraction. We will compute each scaled vector first, then subtract.
Step 1: Compute $2\vec{u}$. $$2\vec{u} = 2\begin{pmatrix} 1 \ 4 \end{pmatrix} = \begin{pmatrix} 2 \ 8 \end{pmatrix}$$
Step 2: Compute $3\vec{v}$. $$3\vec{v} = 3\begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 6 \ 3 \end{pmatrix}$$
Step 3: Subtract (which means subtracting corresponding components). $$2\vec{u} - 3\vec{v} = \begin{pmatrix} 2 \ 8 \end{pmatrix} - \begin{pmatrix} 6 \ 3 \end{pmatrix} = \begin{pmatrix} 2 - 6 \ 8 - 3 \end{pmatrix} = \begin{pmatrix} -4 \ 5 \end{pmatrix}$$
Answer: $2\vec{u} - 3\vec{v} = \begin{pmatrix} -4 \ 5 \end{pmatrix}$
Alternative method: You can also do this in one step by applying the scalars directly to each component: $$2\vec{u} - 3\vec{v} = \begin{pmatrix} 2(1) - 3(2) \ 2(4) - 3(1) \end{pmatrix} = \begin{pmatrix} 2 - 6 \ 8 - 3 \end{pmatrix} = \begin{pmatrix} -4 \ 5 \end{pmatrix}$$
Write $\begin{pmatrix} 7 \ 1 \end{pmatrix}$ as a linear combination of $\begin{pmatrix} 1 \ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \ 1 \end{pmatrix}$.
Solution:
We want to find scalars $a$ and $b$ such that:
$$a\begin{pmatrix} 1 \ 0 \end{pmatrix} + b\begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 7 \ 1 \end{pmatrix}$$
Step 1: Compute the left side in terms of $a$ and $b$. $$a\begin{pmatrix} 1 \ 0 \end{pmatrix} + b\begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} a \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ b \end{pmatrix} = \begin{pmatrix} a \ b \end{pmatrix}$$
Step 2: Match components with the target vector. $$\begin{pmatrix} a \ b \end{pmatrix} = \begin{pmatrix} 7 \ 1 \end{pmatrix}$$
This gives us $a = 7$ and $b = 1$.
Answer: $$\begin{pmatrix} 7 \ 1 \end{pmatrix} = 7\begin{pmatrix} 1 \ 0 \end{pmatrix} + 1\begin{pmatrix} 0 \ 1 \end{pmatrix}$$
Insight: The vectors $\begin{pmatrix} 1 \ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \ 1 \end{pmatrix}$ are the standard basis vectors $\hat{i}$ and $\hat{j}$. Every 2D vector can be written as a linear combination of these two vectors, and the coefficients are simply the components of the vector. This is why component notation is so natural: the components tell you exactly how much of each basis direction you need.
Can $\begin{pmatrix} 5 \ 3 \end{pmatrix}$ be written as a linear combination of $\begin{pmatrix} 1 \ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \ 1 \end{pmatrix}$?
Solution:
We need to determine if there exist scalars $a$ and $b$ such that:
$$a\begin{pmatrix} 1 \ 2 \end{pmatrix} + b\begin{pmatrix} 3 \ 1 \end{pmatrix} = \begin{pmatrix} 5 \ 3 \end{pmatrix}$$
Step 1: Expand the left side. $$\begin{pmatrix} a \ 2a \end{pmatrix} + \begin{pmatrix} 3b \ b \end{pmatrix} = \begin{pmatrix} a + 3b \ 2a + b \end{pmatrix}$$
Step 2: Set up equations by matching components.
Comparing with $\begin{pmatrix} 5 \ 3 \end{pmatrix}$:
- First component: $a + 3b = 5$
- Second component: $2a + b = 3$
Step 3: Solve the system of equations.
From the first equation: $a = 5 - 3b$
Substitute into the second equation: $$2(5 - 3b) + b = 3$$ $$10 - 6b + b = 3$$ $$10 - 5b = 3$$ $$-5b = -7$$ $$b = \frac{7}{5}$$
Now find $a$: $$a = 5 - 3\left(\frac{7}{5}\right) = 5 - \frac{21}{5} = \frac{25 - 21}{5} = \frac{4}{5}$$
Step 4: Verify the solution. $$\frac{4}{5}\begin{pmatrix} 1 \ 2 \end{pmatrix} + \frac{7}{5}\begin{pmatrix} 3 \ 1 \end{pmatrix} = \begin{pmatrix} 4/5 \ 8/5 \end{pmatrix} + \begin{pmatrix} 21/5 \ 7/5 \end{pmatrix} = \begin{pmatrix} 25/5 \ 15/5 \end{pmatrix} = \begin{pmatrix} 5 \ 3 \end{pmatrix} \quad \checkmark$$
Answer: Yes, $\begin{pmatrix} 5 \ 3 \end{pmatrix}$ can be written as a linear combination:
$$\begin{pmatrix} 5 \ 3 \end{pmatrix} = \frac{4}{5}\begin{pmatrix} 1 \ 2 \end{pmatrix} + \frac{7}{5}\begin{pmatrix} 3 \ 1 \end{pmatrix}$$
Deeper insight: Since the two given vectors $\begin{pmatrix} 1 \ 2 \end{pmatrix}$ and $\begin{pmatrix} 3 \ 1 \end{pmatrix}$ are not parallel (one is not a scalar multiple of the other), they can be used to reach any point in the 2D plane through linear combinations. Any non-parallel pair of 2D vectors spans all of $\mathbb{R}^2$. This is a preview of ideas about span and basis that you will explore later.
Key Properties and Rules
Vector Addition Rules
- Component-wise addition: $\begin{pmatrix} u_1 \ u_2 \end{pmatrix} + \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \ u_2 + v_2 \end{pmatrix}$
- Commutativity: $\vec{u} + \vec{v} = \vec{v} + \vec{u}$
- Associativity: $(\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})$
- Zero vector identity: $\vec{v} + \vec{0} = \vec{v}$
- Additive inverse: $\vec{v} + (-\vec{v}) = \vec{0}$
Scalar Multiplication Rules
- Component-wise scaling: $c\begin{pmatrix} v_1 \ v_2 \end{pmatrix} = \begin{pmatrix} cv_1 \ cv_2 \end{pmatrix}$
- Associativity: $a(b\vec{v}) = (ab)\vec{v}$
- Identity: $1\vec{v} = \vec{v}$
- Zero scalar: $0\vec{v} = \vec{0}$
- Scaling zero vector: $c\vec{0} = \vec{0}$
Distributive Rules
- Scalar distributes over vector addition: $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$
- Vector distributes scalar addition: $(a + b)\vec{v} = a\vec{v} + b\vec{v}$
Effects of Scalar Multiplication on Magnitude and Direction
| Scalar $c$ | Effect on Magnitude | Effect on Direction |
|---|---|---|
| $c > 1$ | Increases by factor of $c$ | Same direction |
| $0 < c < 1$ | Decreases by factor of $c$ | Same direction |
| $c = 1$ | No change | Same direction |
| $c = 0$ | Becomes 0 (zero vector) | No direction |
| $-1 < c < 0$ | Decreases by factor of $ | c |
| $c < -1$ | Increases by factor of $ | c |
| $c = -1$ | No change | Opposite direction |
Magnitude and Scalar Multiplication
The magnitude of a scaled vector satisfies:
$$|c\vec{v}| = |c| \cdot |\vec{v}|$$
The absolute value of $c$ appears because magnitude is always positive, even if $c$ is negative.
Real-World Applications
Combining Forces in Physics
When multiple forces act on an object, the net effect is the resultant force, which is the vector sum of all individual forces. If you push a heavy box with a force of 50 newtons to the right while your friend pushes with 30 newtons upward, the box accelerates in the direction of:
$$\vec{F}_{\text{total}} = \begin{pmatrix} 50 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 30 \end{pmatrix} = \begin{pmatrix} 50 \ 30 \end{pmatrix}$$
The magnitude of this resultant force is $\sqrt{50^2 + 30^2} = \sqrt{3400} \approx 58.3$ newtons, and it points diagonally up and to the right. Engineers use this principle constantly when designing structures, analyzing loads, and predicting how objects will move.
Swimming Across a River (Net Velocity)
Suppose you are swimming across a river. You can swim at 2 m/s, and you aim directly toward the opposite bank. But the river has a current flowing at 1 m/s downstream. Your velocity relative to the riverbank is the sum of your swimming velocity and the current:
$$\vec{v}{\text{actual}} = \vec{v}{\text{swim}} + \vec{v}_{\text{current}} = \begin{pmatrix} 2 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 2 \ 1 \end{pmatrix}$$
You end up moving diagonally, drifting downstream as you cross. Pilots face the same situation with wind; they must calculate the vector sum of their airspeed and wind velocity to determine their actual ground track.
Mixing Colors in RGB Space
Computer displays create colors by combining red, green, and blue light. Each color can be represented as a 3D vector where the components indicate the intensity of red, green, and blue (typically from 0 to 255). Mixing colors is essentially vector addition and scalar multiplication:
- Pure red: $\begin{pmatrix} 255 \ 0 \ 0 \end{pmatrix}$
- Pure green: $\begin{pmatrix} 0 \ 255 \ 0 \end{pmatrix}$
- Yellow (red + green): $\begin{pmatrix} 255 \ 0 \ 0 \end{pmatrix} + \begin{pmatrix} 0 \ 255 \ 0 \end{pmatrix} = \begin{pmatrix} 255 \ 255 \ 0 \end{pmatrix}$
- Dimmed red (half intensity): $0.5 \cdot \begin{pmatrix} 255 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 127.5 \ 0 \ 0 \end{pmatrix}$
Every color you see on a screen is a linear combination of red, green, and blue basis colors.
Portfolio Combinations in Finance
In finance, an investment portfolio can be viewed as a linear combination of individual assets. If you invest in two stocks, with weights (percentages of your total investment) $w_1$ and $w_2$, and each stock has a return vector (returns over different time periods or scenarios), your portfolio return is:
$$\vec{r}_{\text{portfolio}} = w_1\vec{r}_1 + w_2\vec{r}_2$$
Portfolio optimization is essentially finding the best linear combination of asset return vectors to maximize expected return while controlling risk. The mathematics of diversification is the mathematics of linear combinations.
Self-Test Problems
Problem 1: Compute $\begin{pmatrix} 4 \ -2 \end{pmatrix} + \begin{pmatrix} -1 \ 5 \end{pmatrix}$.
Show Answer
Add corresponding components:
$$\begin{pmatrix} 4 \ -2 \end{pmatrix} + \begin{pmatrix} -1 \ 5 \end{pmatrix} = \begin{pmatrix} 4 + (-1) \ -2 + 5 \end{pmatrix} = \begin{pmatrix} 3 \ 3 \end{pmatrix}$$
Problem 2: Compute $-4\begin{pmatrix} 2 \ -3 \end{pmatrix}$.
Show Answer
Multiply each component by $-4$:
$$-4\begin{pmatrix} 2 \ -3 \end{pmatrix} = \begin{pmatrix} -4 \cdot 2 \ -4 \cdot (-3) \end{pmatrix} = \begin{pmatrix} -8 \ 12 \end{pmatrix}$$
The resulting vector is 4 times as long as the original and points in the opposite direction.
Problem 3: Find $\vec{u} - \vec{v}$ where $\vec{u} = \begin{pmatrix} 5 \ 1 \ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 2 \ -1 \ 4 \end{pmatrix}$.
Show Answer
Subtract corresponding components:
$$\vec{u} - \vec{v} = \begin{pmatrix} 5 \ 1 \ 3 \end{pmatrix} - \begin{pmatrix} 2 \ -1 \ 4 \end{pmatrix} = \begin{pmatrix} 5 - 2 \ 1 - (-1) \ 3 - 4 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \ -1 \end{pmatrix}$$
Problem 4: Are the vectors $\begin{pmatrix} 6 \ -9 \end{pmatrix}$ and $\begin{pmatrix} -2 \ 3 \end{pmatrix}$ parallel? If so, find the scalar $c$ such that one equals $c$ times the other.
Show Answer
Check if the ratios of corresponding components are equal:
$$\frac{6}{-2} = -3 \quad \text{and} \quad \frac{-9}{3} = -3$$
The ratios are equal, so the vectors are parallel.
We have $\begin{pmatrix} 6 \ -9 \end{pmatrix} = -3 \begin{pmatrix} -2 \ 3 \end{pmatrix}$.
Verification: $-3 \cdot (-2) = 6$ and $-3 \cdot 3 = -9$. Correct.
Problem 5: Compute $3\vec{u} + 2\vec{v}$ where $\vec{u} = \begin{pmatrix} 1 \ -2 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 4 \ 1 \end{pmatrix}$.
Show Answer
Step 1: Compute $3\vec{u}$: $$3\vec{u} = 3\begin{pmatrix} 1 \ -2 \end{pmatrix} = \begin{pmatrix} 3 \ -6 \end{pmatrix}$$
Step 2: Compute $2\vec{v}$: $$2\vec{v} = 2\begin{pmatrix} 4 \ 1 \end{pmatrix} = \begin{pmatrix} 8 \ 2 \end{pmatrix}$$
Step 3: Add the results: $$3\vec{u} + 2\vec{v} = \begin{pmatrix} 3 \ -6 \end{pmatrix} + \begin{pmatrix} 8 \ 2 \end{pmatrix} = \begin{pmatrix} 11 \ -4 \end{pmatrix}$$
Problem 6: Write $\begin{pmatrix} 10 \ 4 \end{pmatrix}$ as a linear combination of $\begin{pmatrix} 2 \ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \ 2 \end{pmatrix}$.
Show Answer
We want to find $a$ and $b$ such that:
$$a\begin{pmatrix} 2 \ 0 \end{pmatrix} + b\begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 10 \ 4 \end{pmatrix}$$
This gives us: $$\begin{pmatrix} 2a + b \ 2b \end{pmatrix} = \begin{pmatrix} 10 \ 4 \end{pmatrix}$$
From the second component: $2b = 4$, so $b = 2$.
Substituting into the first component: $2a + 2 = 10$, so $2a = 8$, and $a = 4$.
Answer: $\begin{pmatrix} 10 \ 4 \end{pmatrix} = 4\begin{pmatrix} 2 \ 0 \end{pmatrix} + 2\begin{pmatrix} 1 \ 2 \end{pmatrix}$
Verification: $4\begin{pmatrix} 2 \ 0 \end{pmatrix} + 2\begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 8 \ 0 \end{pmatrix} + \begin{pmatrix} 2 \ 4 \end{pmatrix} = \begin{pmatrix} 10 \ 4 \end{pmatrix}$ (Correct)
Problem 7: A boat can travel at 5 m/s in still water. If the boat heads directly north but a current flows east at 3 m/s, what is the boat’s actual velocity vector? What is the magnitude of this velocity?
Show Answer
Set up coordinates with east as positive $x$ and north as positive $y$.
- Boat’s velocity in still water (heading north): $\vec{v}_{\text{boat}} = \begin{pmatrix} 0 \ 5 \end{pmatrix}$
- Current velocity (flowing east): $\vec{v}_{\text{current}} = \begin{pmatrix} 3 \ 0 \end{pmatrix}$
Actual velocity: $$\vec{v}{\text{actual}} = \vec{v}{\text{boat}} + \vec{v}_{\text{current}} = \begin{pmatrix} 0 \ 5 \end{pmatrix} + \begin{pmatrix} 3 \ 0 \end{pmatrix} = \begin{pmatrix} 3 \ 5 \end{pmatrix}$$
Magnitude: $$|\vec{v}_{\text{actual}}| = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \text{ m/s}$$
The boat actually moves northeast at about 5.83 m/s.
Problem 8: Using the properties of vector operations, simplify $2(\vec{u} + \vec{v}) - 2\vec{u}$.
Show Answer
Apply the distributive property to expand: $$2(\vec{u} + \vec{v}) - 2\vec{u} = 2\vec{u} + 2\vec{v} - 2\vec{u}$$
Use commutativity and associativity to group: $$= (2\vec{u} - 2\vec{u}) + 2\vec{v}$$
Since $2\vec{u} - 2\vec{u} = \vec{0}$: $$= \vec{0} + 2\vec{v} = 2\vec{v}$$
Answer: $2(\vec{u} + \vec{v}) - 2\vec{u} = 2\vec{v}$
Summary
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Vector addition combines vectors component-wise: add corresponding components. Geometrically, this is the tip-to-tail method or the parallelogram rule.
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Vector subtraction $\vec{u} - \vec{v}$ is the same as adding $\vec{u} + (-\vec{v})$. Subtract corresponding components.
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Scalar multiplication $c\vec{v}$ multiplies each component by $c$. This stretches or shrinks the vector, and reverses direction if $c < 0$.
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Two vectors are parallel if one is a scalar multiple of the other.
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A linear combination of vectors $\vec{v}_1, \vec{v}_2, \ldots$ with scalars $c_1, c_2, \ldots$ is $c_1\vec{v}_1 + c_2\vec{v}_2 + \cdots$. Linear combinations are central to linear algebra.
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Commutativity ($\vec{u} + \vec{v} = \vec{v} + \vec{u}$) and associativity hold for vector addition. Order and grouping do not matter.
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Distributive laws connect scalar multiplication with addition: $c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}$ and $(a + b)\vec{v} = a\vec{v} + b\vec{v}$.
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The zero vector $\vec{0}$ is the additive identity: $\vec{v} + \vec{0} = \vec{v}$. Every vector has an additive inverse $-\vec{v}$.
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The magnitude of a scaled vector satisfies $|c\vec{v}| = |c| \cdot |\vec{v}|$.
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Real-world applications include combining forces, calculating net velocities (boats, planes), mixing colors in RGB, and building investment portfolios.
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These operations form the foundation for understanding vector spaces, span, linear independence, and the rest of linear algebra.