Linear-Algebra-Done-Right · Unit 1 · Video 3 · Interactive Practice

Vectors in R², Scalar Multiplication, and Fields

IKey Formulas

Formula	Name	Meaning
$-x = (-x_1, \dots, -x_n)$ , $x + (-x) = 0$	Additive inverse	Negate every coordinate; in $\mathbf{R}^2$ the same length, opposite direction
$\lambda(x_1, \dots, x_n) = (\lambda x_1, \dots, \lambda x_n)$	Scalar multiplication	A scalar $\lambda \in \mathbf{F}$ times a vector in $\mathbf{F}^n$ gives a vector in $\mathbf{F}^n$
length of $\lambda x$ $=$ $\|\lambda\|$ times the length of $x$	Geometry in $\mathbf{R}^2$	Same direction if $\lambda \gt 0$ , opposite if $\lambda \lt 0$
$1 + 1 = 0$ in $\{0, 1\}$	Two-element field	$-1 = 1$ and $1^{-1} = 1$

Key Insight: The arrows live only in $\mathbf{R}^2$ and $\mathbf{R}^3$ ; the algebra works coordinate by coordinate in every $\mathbf{F}^n$ , and the scalars $\mathbf{F}$ may be any field.

IIAddition Tip to Tail

Adding coordinates is the same as placing $v$ at the tip of $u$ : the sum runs from the tail of $u$ to the tip of the moved $v$ .

IIIScalar Multiplication Stretches, Shrinks and Flips

Multiplying by $\lambda$ multiplies the length by $|\lambda|$ , and a negative $\lambda$ reverses the direction.

Scalar λ 1/2

IVWhen Is $\{0, 1\}$ a Field?

Which addition and multiplication tables on $\{0, 1\}$ satisfy all six properties of complex arithmetic?

+	0	1
0
1

·	0	1
0
1

💡 Because $1 + 1 = 0$ in this field, it fails the hypothesis $1 + 1 \ne 0$ that a few later results require; $\mathbf{Q}$ , $\mathbf{R}$ and $\mathbf{C}$ all satisfy it.

VQuiz Questions

Problem 1 · Scalar Multiplication in $\mathbf{R}^4$

Compute $-2\,\big(3, -1, \tfrac{1}{2}, 0\big)$ .

✅ Correct! Every coordinate is multiplied by

-2

, including the negative one:

(-2)(-1) = 2

❌ Not quite. That adds

-2

to each coordinate. Scalar multiplication multiplies each coordinate by

\lambda

❌ Check the signs. Multiply each coordinate by

-2

: a positive coordinate becomes negative and a negative coordinate becomes positive.

❌ Not quite. Use

\lambda(x_1, \dots, x_n) = (\lambda x_1, \dots, \lambda x_n)

with

\lambda = -2

Show solution

By the definition of scalar multiplication, each coordinate is multiplied by $-2$ :

-2\,\big(3, -1, \tfrac{1}{2}, 0\big) = \big((-2)(3),\ (-2)(-1),\ (-2)\tfrac{1}{2},\ (-2)(0)\big) = (-6, 2, -1, 0).

Problem 2 · Direction and Length of $\lambda x$

Given: $x = (4, 3) \in \mathbf{R}^2$ and $\lambda = -\tfrac{3}{2}$ . Describe the arrow $\lambda x$ .

Direction of λx

Length of λx

✅ Correct!

\lambda x = (-6, -\tfrac{9}{2})

points opposite to

x

and is

\tfrac{3}{2}

times as long:

\tfrac{3}{2} \cdot 5 = \tfrac{15}{2}

❌ Check the direction. A negative scalar reverses the direction;

\lambda x

still lies on the line through

0

and

x

❌ A length is never negative. The sign of

\lambda

sets the direction; the length is

|\lambda|

times the length of

x

❌ Check the length. The length of

x

\sqrt{4^2 + 3^2} = 5

, and the length of

\lambda x

|\lambda|

times that.

Show solution

Compute the vector: $\lambda x = -\tfrac{3}{2}(4, 3) = \big(-6, -\tfrac{9}{2}\big)$ . Each coordinate has the opposite sign from the matching coordinate of $x$ , so $\lambda x$ points in the opposite direction, because $\lambda \lt 0$ .

The length of $x$ is $\sqrt{16 + 9} = 5$ . The length of $\lambda x$ is

\sqrt{36 + \tfrac{81}{4}} = \sqrt{\tfrac{225}{4}} = \tfrac{15}{2} = \big|-\tfrac{3}{2}\big| \cdot 5.

Problem 3 · Solve for a Vector

Find the $x \in \mathbf{R}^3$ with $2x + (3, -1, 4) = (7, 5, -2)$ .

✅ Correct! Check:

2(2, 3, -3) + (3, -1, 4) = (4, 6, -6) + (3, -1, 4) = (7, 5, -2)

❌ Not quite. That adds

(3, -1, 4)

to the right side. To cancel it, add its additive inverse

(-3, 1, -4)

to both sides.

❌ Close, but…

(4, 6, -6)

2x

, not

x

. Multiply by the scalar

\tfrac{1}{2}

to finish.

❌ Not quite. That halves

(7, 5, -2)

but not

(3, -1, 4)

. Cancel

(3, -1, 4)

first, then multiply by

\tfrac{1}{2}

❌ Not quite. Add

-(3, -1, 4)

to both sides, then multiply by

\tfrac{1}{2}

Show solution

Step 1: cancel the known vector. Its additive inverse is $-(3, -1, 4) = (-3, 1, -4)$ . Adding it to both sides gives

2x = (7, 5, -2) + (-3, 1, -4) = (4, 6, -6).

Step 2: multiply by the scalar $\tfrac{1}{2}$ .

x = \tfrac{1}{2}(4, 6, -6) = (2, 3, -3).

Check: $2(2, 3, -3) + (3, -1, 4) = (7, 5, -2)$ ✓

Problem 4 · Arithmetic over the Two-Element Field

Let $\mathbf{F} = \{0, 1\}$ with $1 + 1 = 0$ , and work in $\mathbf{F}^3$ . Compute $x + x$ for $x = (1, 0, 1)$ , and the additive inverse $-(1, 1, 0)$ .

x + x for x = (1, 0, 1)

The additive inverse −(1, 1, 0)

✅ Correct! Since

1 + 1 = 0

, every vector in

\mathbf{F}^3

satisfies

x + x = 0

, so every vector is its own additive inverse.

❌ Not quite. The coordinates must be elements of

\mathbf{F} = \{0, 1\}

; neither

2

nor

-1

is. In this field

1 + 1 = 0

and

-1 = 1

❌ Not quite. Swapping

0

and

1

in each coordinate is not negation:

(1, 1, 0) + (0, 0, 1) = (1, 1, 1) \ne 0

❌ Check the sum. Add coordinate by coordinate using

1 + 1 = 0

and

0 + 0 = 0

❌ Check the inverse. You need the vector

y

with

(1, 1, 0) + y = (0, 0, 0)

; in this field

-0 = 0

and

-1 = 1

Show solution

The sum. Coordinate by coordinate, $(1, 0, 1) + (1, 0, 1) = (1 + 1,\ 0 + 0,\ 1 + 1) = (0, 0, 0)$ , because $1 + 1 = 0$ in $\mathbf{F}$ .

The inverse. $-(1, 1, 0) = (-1, -1, -0)$ . In $\mathbf{F}$ the additive inverse of $1$ is $1$ (since $1 + 1 = 0$ ) and the additive inverse of $0$ is $0$ , so $-(1, 1, 0) = (1, 1, 0)$ .

Check: $(1, 1, 0) + (1, 1, 0) = (0, 0, 0)$ ✓

Solved: 0 / 4

Vectors in R², Scalar Multiplication, and Fields

IKey Formulas

IIAddition Tip to Tail

IIIScalar Multiplication Stretches, Shrinks and Flips

IVWhen Is {0,1}\{0, 1\}{0,1} a Field?

VQuiz Questions

IVWhen Is $\{0, 1\}$ a Field?