Linear-Algebra-Done-Right · Unit 1 · Video 2 · Interactive Practice

Lists and Fⁿ: Coordinate Addition and a First Proof

IKey Formulas

Formula	Name	What it says
$\mathbf{F}^n = \{(x_1, \dots, x_n) : x_k \in \mathbf{F} \text{ for } k = 1, \dots, n\}$	The space $\mathbf{F}^n$	All lists of length $n$ of scalars; $x_k$ is the $k^\text{th}$ coordinate
$(x_1, \dots, x_n) + (y_1, \dots, y_n) = (x_1 + y_1, \dots, x_n + y_n)$	Addition in $\mathbf{F}^n$	Add corresponding coordinates
$x + y = y + x$ for all $x, y \in \mathbf{F}^n$	Commutativity	Inherited from commutativity in $\mathbf{F}$
$0 = (0, \dots, 0)$ , $\ x + 0 = x$	Zero list	The additive identity of $\mathbf{F}^n$

Key Insight: A property of addition in $\mathbf{F}^n$ is proved one coordinate at a time, from the same property of addition in $\mathbf{F}$ .

IIVisualization 1 — Lists versus Sets

Two lists are equal only with the same length and the same entries in the same order; sets ignore both.

Example List A List B

💡 Every list has a finite length, so $(x_1, x_2, \dots)$ is not a list, while the empty list $(\,)$ of length $0$ is one.

IIIVisualization 2 — Adding Corresponding Coordinates

Addition in $\mathbf{F}^n$ is $n$ separate scalar sums, one per coordinate, so it works for every $n$ .

Length n

IVVisualization 3 — The Proof of Commutativity

Commutativity in $\mathbf{F}^n$ is commutativity in $\mathbf{F}$ , applied once in each coordinate.

Step 1

Line	General: $x, y \in \mathbf{F}^n$	Instance in $\mathbf{C}^2$ : $x = (2 + i,\ -3)$ , $y = (1 - 4i,\ 5)$
1	$x + y = (x_1, \dots, x_n) + (y_1, \dots, y_n)$	$x + y = (2 + i,\ -3) + (1 - 4i,\ 5)$
2	$= (x_1 + y_1, \dots, x_n + y_n)$	$= \big((2 + i) + (1 - 4i),\ -3 + 5\big) = (3 - 3i,\ 2)$
3	$= (y_1 + x_1, \dots, y_n + x_n)$	$= \big((1 - 4i) + (2 + i),\ 5 + (-3)\big) = (3 - 3i,\ 2)$
4	$= (y_1, \dots, y_n) + (x_1, \dots, x_n)$	$= (1 - 4i,\ 5) + (2 + i,\ -3)$
5	$= y + x \quad \blacksquare$	$= y + x$

VQuiz Questions

Problem 1 · Addition in R⁴

Given: $x = (1, -2, 0, 3)$ and $y = (4, 5, -1, -3)$ in $\mathbf{R}^4$ — find $x + y$ .

✅ Correct! Each coordinate of

x + y

is the sum of the matching coordinates:

1 + 4

-2 + 5

0 + (-1)

3 + (-3)

❌ Check the signs. One coordinate sum has a sign slip:

-2 + 5 = 3

and

3 + (-3) = 0

❌ Not quite.

7

is the total of all eight entries, a single number. The sum of two elements of

\mathbf{R}^4

is again a list of length

4

❌ Not quite. Add corresponding coordinates, one at a time.

Show solution

By the definition of addition in $\mathbf{R}^4$ :

x + y = (1 + 4,\ -2 + 5,\ 0 + (-1),\ 3 + (-3)) = (5, 3, -1, 0).

The answer is a list of length $4$ , not a number: addition in $\mathbf{F}^n$ never mixes different coordinates.

Problem 2 · Lists or Sets?

Which of these equalities is true?

✅ Correct! A set ignores order and repetition, so

\{7, 2, 7\}

and

\{2, 7\}

have exactly the same elements.

❌ Not quite. In a list order matters: the first coordinates are

3

and

5

, which differ.

❌ Not quite. In a list repetition has meaning: these lists have lengths

2

and

3

, so they are not equal. (The sets

\{4, 4\}

and

\{4, 4, 4\}

are equal.)

❌ Not quite.

(\,)

has length

0

, while

(0)

is a list of length

1

whose only entry is the number

0

❌ Not quite. Lists are equal exactly when they have the same length and the same elements in the same order.

Show solution

(a) False: same length $2$ , but the first coordinates $3 \ne 5$ .
(b) False: lengths $2 \ne 3$ .
(c) False: lengths $0 \ne 1$ .
(d) True: both sets have exactly the elements $2$ and $7$ ; writing $7$ twice adds nothing.

Problem 3 · Sum and Zero in C²

Given: $x = (2 + i,\ 3i)$ and $y = (1 - i,\ 4)$ in $\mathbf{C}^2$ — find $x + y$ , and the element $w \in \mathbf{C}^2$ with $x + w = x$ .

What is x + y?

Which w satisfies x + w = x?

✅ Correct!

x + y = (3,\ 4 + 3i)

, and the zero list

(0, 0)

is the additive identity:

x + (0, 0) = (2 + i + 0,\ 3i + 0) = x

❌ Check the first coordinate.

(2 + i) + (1 - i) = 3 + (i - i) = 3

❌ Not quite.

3i + 4

is the complex number

4 + 3i

; a real part and an imaginary part do not combine into

7i

❌ Not quite. That adds all four entries into one complex number. The sum of two elements of

\mathbf{C}^2

is a list of length

2

❌ Not quite. Adding

w

must leave every coordinate of

x

unchanged.

❌ Not quite. The sum of an element of

\mathbf{C}^2

and a number has not been defined, so

x + 0

only makes sense with

0

the zero list

(0, 0)

❌ Not quite. That

w

-x

: it gives

x + w = (0, 0)

, not

x

Show solution

Sum. Add corresponding coordinates, using complex addition in each:

x + y = \big((2 + i) + (1 - i),\ 3i + 4\big) = (3,\ 4 + 3i).

Identity. Write $w = (w_1, w_2)$ . Then $x + w = (2 + i + w_1,\ 3i + w_2)$ , which equals $x$ exactly when $2 + i + w_1 = 2 + i$ and $3i + w_2 = 3i$ , that is $w_1 = w_2 = 0$ . So $w = (0, 0)$ , the zero list of $\mathbf{C}^2$ .

Problem 4 · Transfer: Associativity

Given: the proof that $(x + y) + z = x + (y + z)$ for $x, y, z \in \mathbf{F}^n$ , written like the commutativity proof:

\begin{aligned}(x + y) + z &= \big((x_1 + y_1) + z_1, \dots, (x_n + y_n) + z_n\big)\\ &= \big(x_1 + (y_1 + z_1), \dots, x_n + (y_n + z_n)\big)\\ &= x + (y + z).\end{aligned}

Which fact justifies the middle equality?

✅ Correct! Each coordinate changes from

(x_k + y_k) + z_k

x_k + (y_k + z_k)

, which is associativity of scalars, and lists with equal coordinates are equal.

❌ Not quite. No two terms change places in the middle line; only the grouping of the parentheses changes.

❌ Not quite. That is the statement being proved, so using it would be circular.

❌ Not quite. The definition of addition in

\mathbf{F}^n

produces the first and last equalities. The middle one changes what happens inside each coordinate.

❌ Not quite. Look at what changes inside a single coordinate.

Show solution

The first equality applies the definition of addition in $\mathbf{F}^n$ twice (first $x + y$ , then adding $z$ ). In the middle equality, coordinate $k$ changes from $(x_k + y_k) + z_k$ to $x_k + (y_k + z_k)$ , which is associativity of addition in $\mathbf{F}$ . It is used $n$ times, once per coordinate, and lists whose coordinates agree are equal. The last equality reads the definition of addition in $\mathbf{F}^n$ from right to left.

The pattern is the same as in the commutativity proof: the property in $\mathbf{F}^n$ is inherited coordinate by coordinate from the same property in $\mathbf{F}$ .

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