2.2 Span, independence, basis, dimension

A basis is a set of building blocks with no waste. Every vector is built from it in exactly one way, and every basis of a space has the same size, its dimension.

Chapter 2.1 said what a vector space is. This chapter asks how to describe one economically. The answer is to pick a few vectors so that every vector in the space is built from them, and none of them is wasted. That set is a basis. The number of vectors in it is the dimension, which measures how big the space really is.

Before you start

2.1: vector space, subspace and the subspace test. For one proof below you also need induction: prove a statement for 1, then show that if it holds for $k - 1$ it holds for $k$ . Here are three quick checks on 2.1.

Quick checkWhich of these is a subspace of

\mathbb R^3

Quick checkIn a vector space, what is

0x

for a vector

x

Quick checkIs the set of polynomials of degree exactly 2 closed under addition?

The picture

Two arrows in space. Slide $a$ and $b$ : the point $au + bv$ moves, but it never leaves the blue plane. That plane is everything two arrows can build. Now pick “v is a multiple of u”. The plane collapses to a line, because the second arrow adds no new direction.

Questions this chapter answers: which vectors can a set build (the span)? When is a vector in the set wasted (dependence)? What is the smallest set that builds everything (a basis)? And how many vectors does that take (the dimension)?

Definitions and theorems

Building vectors: span

Definition T

Linear combination

Given vectors $x_1, \dots, x_k$ in $V$ and scalars $c_1, \dots, c_k$ , the vector

c_1 x_1 + c_2 x_2 + \dots + c_k x_k = \sum_{i=1}^k c_i x_i

is a linear combination of $x_1, \dots, x_k$ . The $c_i$ are its coefficients. It lies in $V$ by the closure axioms.

In one lineA sum of scalar multiples, $c_1 x_1 + c_2 x_2 + \dots + c_k x_k$ .

Why it existsIt is everything you can build from given vectors using only the two vector-space operations.

Definition T

Span

The span of a set $S = \{x_1, \dots, x_k\}$ is the set of all linear combinations of its elements:

L(S) = \left\{ \sum_{i=1}^k c_i x_i : c_1, \dots, c_k \in \mathbb R \right\}.

Apostol writes $L(S)$ , “the linear span”. Many books write $\operatorname{span}(S)$ . If $L(S) = V$ , we say $S$ spans $V$ .

In one lineThe set of all linear combinations of the given vectors: everything they can build.

Why it existsIt is the smallest subspace containing the vectors, so it answers 'what can these vectors reach?'

Spagnolie’s gloss: the span is “the space that can be reached by $S$ ”. In $\mathbb R^3$ with $x_1 = (1, 0, 0)$ and $x_2 = (1, 1, 0)$ :

c_1 x_1 + c_2 x_2 = (c_1 + c_2,\ c_2,\ 0).

Every vector with third entry 0 is reached: to get $(p, q, 0)$ , take $c_2 = q$ and $c_1 = p - q$ . So $L(S)$ is the $xy$ -plane. For example, $(0, 2, 0) = -2x_1 + 2x_2$ .

Your answerWith

x_1 = (1, 0, 0)

and

x_2 = (1, 1, 0)

as above,

(5, 3, 0) = c_1 x_1 + c_2 x_2

. What is

c_1

Quick checkWhat is the span of the single vector

(1, 2, 3)

\mathbb R^3

Worked example

(3, 5, 7)

in the span of

(1, 1, 1)

and

(0, 1, 2)

? Is

(3, 5, 8)

Solution.

Ask the question as an equation. Is there a pair $c_1, c_2$ with $c_1(1, 1, 1) + c_2(0, 1, 2) = (3, 5, 7)$ ? ‘In the span’ means ‘is some linear combination’, so we look for the coefficients.
Compare entries. $c_1 = 3$ , then $c_1 + c_2 = 5$ , then $c_1 + 2c_2 = 7$ . One vector equation in $\mathbb R^3$ is three ordinary equations.
Solve with the first two. $c_1 = 3$ and $c_2 = 5 - 3 = 2$ . Two equations fix the two unknowns.
Test the third equation. $c_1 + 2c_2 = 3 + 4 = 7$ . It holds. Three equations, two unknowns: the leftover equation is a check that can fail.
Repeat for $(3, 5, 8)$ . The first two entries still give $c_1 = 3$ , $c_2 = 2$ , but now the third needs $c_1 + 2c_2 = 8$ , and $3 + 4 = 7 \ne 8$ . The only candidate coefficients fail the third entry, so no combination works.

Answer. $(3, 5, 7) = 3(1,1,1) + 2(0,1,2)$ is in the span. $(3, 5, 8)$ is not: the span of two independent vectors is a plane in $\mathbb R^3$ , and this point lies off it.

Theorem T Apostol 1.6

The span is a subspace

For any finite set $S = \{x_1, \dots, x_k\}$ in a vector space $V$ , the span $L(S)$ is a subspace of $V$ .

In one lineFor any finite set $S$ in $V$ , the span $L(S)$ is a subspace of $V$ .

Why it is trueA sum of two combinations is a combination (add coefficients); a multiple of a combination is a combination (multiply coefficients).

ProofSpagnolie Lecture 2

The idea: Use the subspace test. Combinations of the same vectors add by adding their coefficients, and scale by scaling their coefficients.

O = 0x_1 + \dots + 0x_k \in L(S)

Choosing every coefficient to be 0 gives the zero vector, by 'Scaling by zero' from 2.1.

Let

v = \sum c_i x_i

and

w = \sum b_i x_i

be in

L(S)

Two arbitrary elements of

L(S)

. Each is SOME combination, so each has its own coefficients.

v + w = \sum (c_i + b_i)\, x_i

, a combination, so

v + w \in L(S)

Axioms 3 and 4 let us regroup the terms; axiom 9 combines

c_i x_i + b_i x_i

into

(c_i + b_i) x_i

av = \sum (a c_i)\, x_i \in L(S)

Axiom 8 distributes

a

over the sum; axiom 7 turns

a(c_i x_i)

into

(ac_i)x_i

L(S)

is a subspace of

V

All three parts of the subspace test hold.

Wasted vectors: dependence

Spagnolie’s motivation: “It is very useful to describe a space using only a few building blocks. But redundancy is not great.” Take $x_1 = (1, 0)$ , $x_2 = (1, 1)$ , $x_3 = (-1, 3)$ in $\mathbb R^2$ . Then $x_3 = -4x_1 + 3x_2$ . The third vector is wasted, because anything built with it could be built without it. Moving everything to one side gives

-4x_1 + 3x_2 - 1x_3 = O,

a combination equal to zero with coefficients that are not all zero. This form is the definition, because it does not single out which vector is the wasted one.

Taking every coefficient 0 always gives $O$ , for any vectors at all, so that tells us nothing. Dependence asks whether there is a second way to get $O$ .

Definition T

Linearly dependent

A set $\{x_1, \dots, x_k\}$ is linearly dependent if there are scalars $c_1, \dots, c_k$ , not all zero, with

c_1 x_1 + \dots + c_k x_k = O.

In one lineSome combination of the vectors equals zero without all coefficients being zero; equivalently, one of them is a combination of the others.

Why it existsIt detects waste: a dependent set has a vector that adds nothing to the span.

Definition T

Linearly independent

A set $\{x_1, \dots, x_k\}$ is linearly independent if it is not dependent. That is, the equation

c_1 x_1 + \dots + c_k x_k = O

has only the solution $c_1 = c_2 = \dots = c_k = 0$ .

In one lineThe only combination of the vectors equal to zero is the one with every coefficient zero.

Why it existsIndependent vectors carry no waste, so each vector they build is built in only one way.

Why· Why the two forms say the same thing

Suppose $c_1 x_1 + \dots + c_k x_k = O$ with some $c_j \ne 0$ . Move the other terms across and divide by $c_j$ :

x_j = -\frac{1}{c_j} \sum_{i \ne j} c_i x_i.

So $x_j$ is a combination of the others. Conversely, if $x_j = \sum_{i \ne j} d_i x_i$ , then $\sum_{i \ne j} d_i x_i - 1 \cdot x_j = O$ , and the coefficient $-1$ is not zero. Dependence means some vector is a combination of the rest.

Quick checkThree vectors satisfy

x_1 - 2x_2 + x_3 = O

. Are they dependent or independent?

Your answerFor which value of

t

are the vectors

(1, 2)

and

(3, t)

dependent?

Quick ways to be dependent, from Lecture 3:

one vector is a multiple of another;
the zero vector is in the set, since $1 \cdot O = O$ uses a non-zero coefficient;
too many vectors. Three vectors in the plane are always dependent, and the theorem below proves it.

Worked example

Are

(1, 1, 0)

(0, 1, 1)

and

(1, 0, 1)

independent in

\mathbb R^3

Solution.

Write the equation that defines independence. $c_1(1, 1, 0) + c_2(0, 1, 1) + c_3(1, 0, 1) = (0, 0, 0)$ . The set is independent exactly when the only solution is $c_1 = c_2 = c_3 = 0$ .
Compare entries. The first entries give $c_1 + c_3 = 0$ , the second give $c_1 + c_2 = 0$ , and the third give $c_2 + c_3 = 0$ . One vector equation in $\mathbb R^3$ is three ordinary equations, one per entry.
Solve. The first equation gives $c_3 = -c_1$ and the second gives $c_2 = -c_1$ . Put both into the third: $-c_1 - c_1 = 0$ , so $c_1 = 0$ . Then $c_2 = 0$ and $c_3 = 0$ . Substituting leaves one equation in one unknown, which has only one solution.
Read the result. The all-zero solution is the only one. That is the definition of independent.

Answer. Independent. In the step-through above a free unknown appeared, and that is what a dependence looks like. Here no unknown stayed free.

Worked example

Prove: if

u

and

v

are independent, then

u + v

and

u - v

are independent.

Solution. The plan: take any combination of $u + v$ and $u - v$ that equals $O$ , rewrite it as a combination of $u$ and $v$ , and use the independence of $u$ and $v$ .

Start with a combination equal to zero. $a(u + v) + b(u - v) = O$ . To show a set is independent, take an arbitrary combination equal to $O$ and show every coefficient is 0. We name the coefficients $a$ and $b$ and assume nothing else about them.
Regroup by $u$ and $v$ . $(a + b)u + (a - b)v = O$ . Axioms 8 and 9 multiply out the brackets, and axioms 3 and 4 let us collect the $u$ terms and the $v$ terms.
Use the hypothesis. $a + b = 0$ and $a - b = 0$ . $u$ and $v$ are independent, so a combination of them equal to $O$ must have both coefficients 0.
Solve the two equations. Adding them gives $2a = 0$ , so $a = 0$ . Then $b = 0$ . Two equations in two unknowns.

Answer. The only combination of $u + v$ and $u - v$ that equals $O$ has $a = b = 0$ , so they are independent. The same pattern proves most statements about independence: name a combination equal to $O$ , rewrite it, and read off the coefficients.

The key theorem: too many vectors are always dependent

This theorem is what makes “dimension” make sense. The course stated it and left the proof to Apostol. We prove it in full here, because the proof is the same elimination you just did in the step-through. We will call it the $k + 1$ theorem.

Theorem T Apostol Thm 1.5

More vectors than a spanning set means dependent

Let $x_1, \dots, x_k$ be any vectors and $L = L(\{x_1, \dots, x_k\})$ . Then every set of $k + 1$ vectors in $L$ is linearly dependent.

In one lineIf $k$ vectors span a space, then any $k + 1$ vectors in that space are linearly dependent.

Why it is trueEliminate one building block at a time. Each round uses up one of the $k+1$ vectors and one building block, and the vectors outlast the blocks.

Two small points. The case $k = 0$ is trivial: the span of no vectors is $\{O\}$ , so a single vector in it is $O$ , and $1 \cdot O = O$ is a dependence. And we treat a ‘set’ of vectors as a list, so a list with a repeated vector counts as dependent ( $1 \cdot y - 1 \cdot y = O$ ).

Why· The picture behind the proof

Take $k = 2$ . The two arrows $u$ and $v$ in the picture below are the building blocks, and their span is the blue plane. Now pick any three vectors $y_1, y_2, y_3$ in that plane. The theorem says they are dependent, and the picture shows why. Each $y_i$ is a mix of “some $u$ ” and “some $v$ ”, so a vector in the plane is described by two numbers, and three vectors carry three pairs of numbers.

The proof removes one building block at a time. Subtract the right multiple of $y_1$ from $y_2$ and from $y_3$ so that their $u$ -parts cancel. The two new vectors $z_2$ and $z_3$ then have no $u$ in them, so both lie along $v$ alone. Switch the picture to “v is a multiple of u” and you see that case: the plane has collapsed to a line, and two vectors on one line are dependent (the base case of the proof). Undoing the subtraction turns that dependence of $z_2, z_3$ into a dependence of $y_1, y_2, y_3$ .

ProofApostol 1.7

The idea: Induction on k. Write each of the k+1 vectors as a combination of x₁,…,x_k. Use y₁ to cancel the x₁-part from all the others (exactly like eliminating c₁ in the step-through). The k leftover vectors live in the span of only k−1 building blocks, so the induction hypothesis says they are dependent. Rewrite that dependence among the leftovers as a dependence among the original y's.

Claim: for every

k \ge 1

, any

k + 1

vectors in the span of

k

vectors are dependent.

Induction on

k

. The base case

k = 1

is checked first.CS 240 induction: prove it for the first value, then prove 'true for k−1 implies true for k'.

Let

y_1 = a x_1

and

y_2 = b x_1

Base case,

k = 1

. Everything in

L(\{x_1\})

is a multiple of

x_1

a = 0

1y_1 + 0y_2 = O

. If

a \ne 0

b y_1 - a y_2 = ab x_1 - ab x_1 = O

. Either way

\{y_1, y_2\}

is dependent.

a = 0

, then

y_1 = O

, and

1 \cdot y_1 + 0 \cdot y_2 = O

has a non-zero coefficient. If

a \ne 0

, the combination below has the coefficient

-a \ne 0

Assume the claim holds for

k - 1

. Let

y_1, \dots, y_{k+1}

be in

L(\{x_1, \dots, x_k\})

Inductive step. The hypothesis is the claim for

k - 1

building blocks; we prove it for

k

Write

y_i = a_{i1} x_1 + a_{i2} x_2 + \dots + a_{ik} x_k

for

i = 1, \dots, k + 1

. The number

a_{ij}

is the amount of

x_j

y_i

Each

y_i

is in the span, so it is SOME combination of the

x

's.In the picture (

k = 2

y_i

is an arrow in the blue plane, and

a_{i1}

a_{i2}

say how far it goes along

u

and along

v

If every

a_{i1} = 0

, then

y_1, \dots, y_k

lie in

L(\{x_2, \dots, x_k\})

. The hypothesis (the claim with

k - 1

in place of

k

) says any

k

vectors in the span of

k - 1

vectors are dependent. So

y_1, \dots, y_k

are dependent, and the whole set is too: add

0 \cdot y_{k+1}

Case 1: no

y_i

uses

x_1

at all. Then they live in a span of

k - 1

vectors, and we can use the hypothesis directly.

Otherwise, renumber so that

a_{11} \ne 0

Case 2: some

y_i

uses

x_1

. Renumber the

y

's so that it is

y_1

.Renumbering does not change whether a set is dependent.

For

i = 2, \dots, k + 1

, let

z_i = y_i - \dfrac{a_{i1}}{a_{11}}\, y_1

. The

x_1

-coefficient of

z_i

a_{i1} - \dfrac{a_{i1}}{a_{11}}a_{11} = 0

Subtract the right multiple of

y_1

from each other

y_i

to cancel its

x_1

-part. This is elimination: the same move as subtracting a multiple of equation 1.In the picture: each

z_i

has lost its

u

-part, so it lies along

v

alone. The plane has been cut down to the line case.

By the hypothesis,

z_2, \dots, z_{k+1}

are dependent:

\sum_{i=2}^{k+1} c_i z_i = O

with some

c_i \ne 0

So the

k

vectors

z_2, \dots, z_{k+1}

lie in the span of the

k - 1

vectors

x_2, \dots, x_k

. That is exactly the induction hypothesis, with

k - 1

in place of

k

\displaystyle O = \sum_{i=2}^{k+1} c_i z_i = \sum_{i=2}^{k+1} c_i \left( y_i - \frac{a_{i1}}{a_{11}}\, y_1 \right).

Substitute the definition of each

z_i

\displaystyle O = \sum_{i=2}^{k+1} c_i y_i - \left( \sum_{i=2}^{k+1} \frac{c_i a_{i1}}{a_{11}} \right) y_1.

Multiply out and collect the

y_1

terms into one coefficient.

\{y_1, \dots, y_{k+1}\}

is dependent. By induction, the claim holds for every

k

This is a combination of

y_1, \dots, y_{k+1}

equal to

O

, and some

c_i

with

i \ge 2

is non-zero, so not all coefficients are zero.

Same idea elsewhere· Where the idea came from

Each round of the proof spends one building block ( $x_1$ ) and one vector ( $y_1$ ). After $k$ rounds no building blocks are left, but one vector is still left over. A vector built from no building blocks can only be $O$ , and that forces a dependence. In matrix language (chapter 2.6): a system of equations with more unknowns than equations always has a non-zero solution.

Quick checkWhich statement does the

k + 1

theorem guarantee?

Worked example

Are

(2, 1)

(0, 3)

and

(5, -4)

independent in

\mathbb R^2

? If not, find a dependence.

Solution.

Count first. There are three vectors in $\mathbb R^2$ , which is spanned by two vectors, $(1, 0)$ and $(0, 1)$ . The $k + 1$ theorem with $k = 2$ says any 3 vectors there are dependent, so the answer is known before any algebra.
Set up the equation to find the dependence. $c_1(2, 1) + c_2(0, 3) + c_3(5, -4) = (0, 0)$ gives $2c_1 + 5c_3 = 0$ and $c_1 + 3c_2 - 4c_3 = 0$ . The theorem says a non-zero solution exists. Solving shows what it is.
Choose the free unknown. There are two equations and three unknowns, so $c_3$ can be chosen. Take $c_3 = 6$ . The first equation gives $c_1 = -\tfrac52 c_3$ , and $c_3 = 6$ clears the fraction.
Solve for the rest. $c_1 = -15$ . The second equation gives $3c_2 = -c_1 + 4c_3 = 15 + 24 = 39$ , so $c_2 = 13$ . Back-substitution, as in the step-through above.
Check by multiplying out. $-15(2, 1) + 13(0, 3) + 6(5, -4) = (-30 + 0 + 30,\ -15 + 39 - 24) = (0, 0)$ . The coefficients $-15, 13, 6$ are not all zero, so this is a genuine dependence.

Answer. Dependent. Equivalently $(5, -4) = \tfrac52 (2, 1) - \tfrac{13}{6}(0, 3)$ : the third vector is wasted.

No waste and nothing missing: basis and dimension

Where this came from· Where basis and dimension came from

Hermann Grassmann, a school teacher in Stettin, published his Ausdehnungslehre (theory of extension) in 1844. He wanted an algebra for geometric objects that works in any number $n$ of dimensions, not only the 2 and 3 of everyday pictures. For that he needed a way to say “how many independent directions”, which is the idea of independence and dimension used in this chapter. Few mathematicians read the book at the time, partly because it was so abstract. Giuseppe Peano restated its ideas with axioms in 1888, close to the ten axioms of 2.1. The $k + 1$ theorem is today called the Steinitz exchange lemma, after Ernst Steinitz, who proved it in the early 1910s.

Definition T

Basis

A finite set $B = \{e_1, \dots, e_n\}$ in $V$ is a basis of $V$ if it is linearly independent and spans $V$ .

In one lineAn independent set that spans the whole space: enough vectors to build everything, and none wasted.

Why it existsIt lets you describe every vector by a list of numbers, its coordinates, in exactly one way.

Spagnolie: “There is never a unique basis for a space!” For the polynomials $P_2$ of degree at most 2, both $\{1, t, t^2\}$ (the standard basis) and $\{1, t + 2, t^2 - t\}$ work. To see that the second spans, write $t = (t + 2) - 2 \cdot 1$ and $t^2 = (t^2 - t) + t$ . It is independent too: if $c_0 + c_1(t + 2) + c_2(t^2 - t) = 0$ , the $t^2$ coefficient gives $c_2 = 0$ , then the $t$ coefficient gives $c_1 = 0$ , then $c_0 = 0$ .

Quick checkThe set

\{(1, 0), (0, 1), (1, 1)\}

spans

\mathbb R^2

. Is it a basis of

\mathbb R^2

Theorem T Apostol Thm 1.6

All bases have the same size

If a vector space $V$ has a basis of $n$ elements, then every basis of $V$ has exactly $n$ elements.

In one lineIf $V$ has a basis with $n$ elements, every basis of $V$ has $n$ elements.

Why it is trueEach basis spans, and the other basis is independent. By the $k + 1$ theorem an independent set cannot outnumber a spanning set, so each basis is at most as big as the other.

ProofApostol 1.8

The idea: Compare two bases using the theorem above, in both directions.

Let

B

be a basis with

n

elements and

B'

a basis with

m

elements.

Two bases, sizes

n

and

m

Every element of

B'

lies in

V = L(B)

. If

m > n

, some

n + 1

elements of

B'

would be dependent, which is impossible since

B'

is independent. So

m \le n

B

spans

V

. If

m \ge n + 1

, then any

n + 1

elements of

B'

are

n + 1

vectors in the span of the

n

vectors of

B

, so by the

k + 1

theorem they are dependent. But a subset of an independent set is independent (proved in the problems below).

Likewise

n \le m

. So

m = n

The same argument with the roles of

B

and

B'

swapped.

Definition T

Dimension

If $V$ has a basis with $n$ elements, $V$ has dimension $n$ , written $\dim V = n$ . The space $\{O\}$ has dimension 0. Its basis is the empty set: the empty set is independent (there is no combination to check), and by convention its span is $\{O\}$ . If $V$ has no finite basis, then $V$ is infinite-dimensional.

In one lineThe number of vectors in any basis. It is the number of independent directions, the number of coordinates you need.

Why it existsThe theorem above makes it well defined: it does not depend on which basis you pick.

Examples from Lecture 3:

$\dim \mathbb R^3 = 3$ , with standard basis $(1,0,0), (0,1,0), (0,0,1)$ . In general $\dim \mathbb R^n = n$ .
$\dim P_4 = 5$ , with basis $1, t, t^2, t^3, t^4$ . Watch the count: degree at most 4 means five coefficients.
The line $\{(x, y) : x + y = 0\}$ has basis $\{(1, -1)\}$ , so its dimension is 1.
All polynomials, with no limit on the degree, form an infinite-dimensional space. If some $k$ polynomials spanned them, any $k + 1$ polynomials would be dependent by the $k + 1$ theorem. But $1, t, \dots, t^k$ are $k + 1$ independent polynomials.

Worked example

Find a basis and the dimension of

W = \{(x, y, z, w) \in \mathbb R^4 : x + y = 0 \text{ and } z = 2w\}

Solution.

Solve the conditions for some coordinates. The conditions give $y = -x$ and $z = 2w$ . That leaves $x$ and $w$ free to choose. Each independent condition fixes one coordinate in terms of the others. The coordinates still free tell you the dimension.
Write a general element of $W$ . $(x, -x, 2w, w) = x(1, -1, 0, 0) + w(0, 0, 2, 1)$ . Pull out the free numbers $x$ and $w$ as coefficients.
Spanning. Every element of $W$ is a combination of $(1, -1, 0, 0)$ and $(0, 0, 2, 1)$ . Both vectors are in $W$ : $1 + (-1) = 0$ , and $0 + 0 = 0$ with $2 = 2 \cdot 1$ . The two vectors must lie in $W$ , or their span could be bigger than $W$ .
Independence. If $a(1, -1, 0, 0) + b(0, 0, 2, 1) = O$ , the first entry gives $a = 0$ and the fourth gives $b = 0$ . Compare entries, as in the independence checks above.
Count. The basis has 2 vectors. Dimension is the size of any basis.

Answer. A basis is $\{(1, -1, 0, 0),\ (0, 0, 2, 1)\}$ , and $\dim W = 2$ . Two independent conditions removed two dimensions from $\dim \mathbb R^4 = 4$ . A repeated condition removes nothing: $x + y = 0$ together with $2x + 2y = 0$ is only one condition.

Your answerWhat is the dimension of

\{(x_1, \dots, x_5) \in \mathbb R^5 : x_1 + x_2 + x_3 = 0 \text{ and } x_4 = x_5\}

Your answerWhat is the dimension of the space of symmetric

3 \times 3

matrices (those with

A_{ij} = A_{ji}

Theorem T Apostol Thm 1.7

Extending to a basis

Let $\dim V = n$ . Then:

(a) every independent set in $V$ is part of some basis of $V$ ;

(b) every set of $n$ independent vectors in $V$ is a basis.

In one lineIn an $n$ -dimensional space, any independent set can be extended to a basis, and any $n$ independent vectors already form a basis.

Why it is trueKeep adding a vector outside the current span; this keeps independence and must stop after $n$ steps.

ProofApostol 1.8

The idea: First a lemma: adding a vector from OUTSIDE the span keeps a set independent. Then add such vectors one at a time. The process must stop, because there can never be more than n independent vectors.

Lemma. If

S = \{x_1, \dots, x_k\}

is independent and

y \notin L(S)

, then

S \cup \{y\}

(the set

S

with

y

added) is independent.

We prove this lemma first, then use it.

Suppose

c\,y + \sum c_i x_i = O

To show a set is independent, take any combination of it equal to

O

and show every coefficient is 0.

Then

c = 0

, since otherwise

y = -\frac1c \sum c_i x_i \in L(S)

c \ne 0

, solve for

y

. That makes

y

a combination of the

x_i

, so

y \in L(S)

, contradicting the choice of

y

\sum c_i x_i = O

, which forces every

c_i = 0

. Hence

S \cup \{y\}

is independent.

With

c = 0

, what is left is a combination of the independent

x_i

equal to zero.

Starting from an independent

S

: while

L(S) \ne V

, add a vector outside

L(S)

. By the lemma,

S

stays independent.

Part (a). If

S

already spans, it is a basis. If not, some

y

is outside its span; add it.

An independent set in

V

has at most

n

elements, so the process stops after at most

n

additions. It stops only when

L(S) = V

. Then

S

is independent and spans: a basis containing the original set.

By the

k + 1

theorem, with the

n

basis vectors as the spanning set, any

n + 1

vectors of

V

are dependent.

S

has

n

independent vectors, the basis from (a) contains

S

and has exactly

n

elements, so it is

S

itself.

Part (b). Extend the

n

independent vectors to a basis by part (a). Every basis has

n

elements, so nothing was added.

Proof workshopGive the reason for each line of the lemma: adding a vector outside the span keeps independence

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $S = \{x_1, \dots, x_k\}$ be independent and $y \notin L(S)$ . Suppose $c\,y + \sum c_i x_i = O$ .
If $c \ne 0$ , then $y = -\frac1c \sum c_i x_i$ .
Then $y \in L(S)$ , which contradicts the choice of $y$ . So $c = 0$ .
Now $\sum c_i x_i = O$ .
Every $c_i = 0$ , so $S \cup \{y\}$ is independent.

Quick checkYou have 3 independent vectors in

\mathbb R^4

. Which statement is guaranteed?

Worked example

Extend the independent set

\{(1, 1, 0),\ (0, 1, 1)\}

to a basis of

\mathbb R^3

Solution.

Count what is missing. $\dim \mathbb R^3 = 3$ and we have 2 vectors, so we need to add exactly 1. Every basis of $\mathbb R^3$ has 3 vectors.
Describe the span we already have. $a(1, 1, 0) + b(0, 1, 1) = (a,\ a + b,\ b)$ . To add a vector OUTSIDE the span, we must know what the span is.
Try a candidate. Is $(1, 0, 0)$ of the form $(a, a + b, b)$ ? That needs $a = 1$ , $b = 0$ and $a + b = 0$ , but $a + b = 1$ . Not possible, so $(1, 0, 0)$ is outside the span. Standard basis vectors are the quickest candidates to test. If one is inside the span, try the next.
Add it. The set $\{(1,1,0), (0,1,1), (1,0,0)\}$ is independent by the lemma. Adding a vector outside the span keeps independence.
Finish. Three independent vectors in the 3-dimensional $\mathbb R^3$ form a basis. Part (b) of ‘Extending to a basis’.

Answer. $\{(1,1,0), (0,1,1), (1,0,0)\}$ is a basis. Many other choices work: any vector off the plane $\{(a, a+b, b)\}$ would do.

Coordinates

Theorem T Apostol 1.9

Unique coordinates

Let $\{e_1, \dots, e_n\}$ be a basis of $V$ . Every $x \in V$ can be written as $x = \sum_{i=1}^n c_i e_i$ for exactly one choice of scalars $c_1, \dots, c_n$ .

In one lineIf $e_1, \dots, e_n$ is a basis, each vector is $\sum c_i e_i$ for exactly one list $(c_1, \dots, c_n)$ .

Why it is trueSpanning gives at least one way; independence gives at most one, since two ways would subtract to a non-zero combination equal to zero.

Why· The picture behind the proof

In the Basis2D picture above, the dashed route to the point $x$ is “go $c_1$ steps along $b_1$ , then $c_2$ steps along $b_2$ ”. Each grid point has exactly one such route. Now drag $b_1$ and $b_2$ onto one line. The grid collapses, most points cannot be reached at all, and the points on the line can be reached in many ways. That is the proof in a picture. Two different routes to the same point would give a closed loop that moves along $b_1$ and $b_2$ and ends where it started. Such a loop with non-zero steps is a dependence of $b_1$ and $b_2$ , and it exists only when they lie on one line.

ProofApostol 1.9

The idea: Existence comes from spanning. For uniqueness, take two representations and subtract them.

Some representation

x = \sum c_i e_i

exists.

The basis spans

V

.In the picture: every point of the plane has a route along

b_1

then

b_2

Suppose also

x = \sum d_i e_i

Suppose there is a second one.

O = x - x = \sum (c_i - d_i)\, e_i

Subtract the two expressions for the same vector

x

, and collect terms with axiom 9.In the picture: this is the closed loop from the box above, with steps

c_i - d_i

along the

b_i

c_i - d_i = 0

for every

i

, so the two representations are the same.

The basis is independent, so every coefficient of a combination equal to

O

is zero.In the picture: independent arrows give no closed loop with non-zero steps.

Proof workshopRebuild the proof that coordinates are unique

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Quick checkWhy can a vector have only ONE list of coordinates in a basis?

Definition T

Coordinates

A set has no order, but a list of coordinates does, so we fix which basis vector is first, second, and so on. That choice is an ordered basis. Given an ordered basis $(e_1, \dots, e_n)$ of $V$ , the numbers $c_1, \dots, c_n$ with $x = \sum c_i e_i$ are the coordinates (Apostol: components) of $x$ in that basis.

In one lineThe unique numbers $(c_1, \dots, c_n)$ with $x = \sum c_i e_i$ for a chosen, ordered basis.

Why it existsThey turn any $n$ -dimensional space into $\mathbb R^n$ , so everything can be computed with lists of numbers.

Coordinates are why $\mathbb R^n$ matters so much. Once you choose a basis, a polynomial in $P_2$ becomes a list of 3 numbers, and a 2 by 2 matrix becomes a list of 4. Any $n$ -dimensional space then behaves exactly like $\mathbb R^n$ . Chapter 2.4 makes this precise: it calls such spaces isomorphic.

Worked example

Find the coordinates of

x = (7, 4)

in the ordered basis

b_1 = (1, 2)

b_2 = (3, 1)

\mathbb R^2

. Then find the coordinates of

p = 4 + 3t + 2t^2

in the ordered basis

(1,\ t + 2,\ t^2 - t)

P_2

Solution, part 1 (vectors in $\mathbb R^2$ ).

Check that it is a basis. $b_2$ is not a multiple of $b_1$ , so the two are independent. Two independent vectors in the 2-dimensional $\mathbb R^2$ form a basis, by part (b) of ‘Extending to a basis’. Without a basis, coordinates may not exist or may not be unique.
Write the defining equation. $c_1(1, 2) + c_2(3, 1) = (7, 4)$ , so $c_1 + 3c_2 = 7$ and $2c_1 + c_2 = 4$ . The coordinates are the numbers that build $x$ from the basis.
Solve. The first equation gives $c_1 = 7 - 3c_2$ . Put that into the second: $14 - 6c_2 + c_2 = 4$ , so $c_2 = 2$ and $c_1 = 1$ . Substitution, as in the independence checks.
Check. $1(1, 2) + 2(3, 1) = (1 + 6,\ 2 + 2) = (7, 4)$ . Multiplying out confirms the numbers.

Answer. The coordinates are $(1, 2)$ . In the standard basis the same vector has coordinates $(7, 4)$ . The list of numbers changes with the basis, and the vector does not.

Solution, part 2 (polynomials).

Write the defining equation. $p = c_0 \cdot 1 + c_1 (t + 2) + c_2 (t^2 - t)$ . Same idea as part 1, with polynomials as the vectors.
Expand and collect powers of $t$ . The right side is $c_2 t^2 + (c_1 - c_2)\,t + (c_0 + 2c_1)$ . Two polynomials are equal exactly when their coefficients match, because $1, t, t^2$ are independent.
Match coefficients, from the top. $t^2$ gives $c_2 = 2$ . Then $t$ gives $c_1 - 2 = 3$ , so $c_1 = 5$ . Then the constant term gives $c_0 + 10 = 4$ , so $c_0 = -6$ . The top equation has one unknown, and each next equation has one new unknown.
Check. $-6 + 5(t + 2) + 2(t^2 - t) = -6 + 5t + 10 + 2t^2 - 2t = 4 + 3t + 2t^2$ . It returns $p$ .

Answer. The coordinates are $(-6, 5, 2)$ . In the standard basis $1, t, t^2$ they are $(4, 3, 2)$ .

Your answerWrite

(10, 5) = c_1 (1, 2) + c_2 (3, 1)

in the basis

b_1 = (1, 2)

b_2 = (3, 1)

. What is

c_2

Examples and non-examples

Polynomials $1, t, \dots, t^n$ are independent. This is Spagnolie’s Lecture 3 proof. Suppose $F(t) = c_0 + c_1 t + \dots + c_n t^n$ is the zero function. Since $F$ is zero for every $t$ , so are all its derivatives. Then $F(0) = c_0 = 0$ . Differentiate: $F'(0) = c_1 = 0$ . Differentiating $j$ times and setting $t = 0$ gives $j!\,c_j = 0$ . So every $c_j = 0$ .

$\cos^2 t$ , $\sin^2 t$ and $1$ are dependent, because $1 \cdot \cos^2 t + 1 \cdot \sin^2 t - 1 \cdot 1 = 0$ for every $t$ . A dependence among functions must hold at every $t$ , and this one does.

$e^{at}$ and $t\,e^{at}$ are independent (Apostol 1.10 #23(b)). If $c_0 e^{at} + c_1 t e^{at} = 0$ for all $t$ , divide by $e^{at}$ , which is never 0: $c_0 + c_1 t = 0$ for all $t$ . A polynomial that is zero everywhere has zero coefficients, so $c_0 = c_1 = 0$ .

Worked example

Show that

\sin t

and

\cos t

are independent. Then show that

1

t

and

e^{2t}

are independent.

For functions, " $F = 0$ " means $F(t) = 0$ for EVERY $t$ . That lets us pick values of $t$ , or differentiate, to get equations for the coefficients.

Solution, part 1 (evaluate at chosen points).

Set up. Suppose $c_1 \sin t + c_2 \cos t = 0$ for every $t$ . Independence means this forces $c_1 = c_2 = 0$ .
Put $t = 0$ . $c_1 \cdot 0 + c_2 \cdot 1 = 0$ , so $c_2 = 0$ . $\sin 0 = 0$ and $\cos 0 = 1$ , so one point kills one unknown.
Put $t = \pi/2$ . $c_1 \cdot 1 + c_2 \cdot 0 = 0$ , so $c_1 = 0$ . $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$ . The equation holds at every $t$ , so it holds at this one.

Solution, part 2 (differentiate).

Set up. Suppose $F(t) = c_0 + c_1 t + c_2 e^{2t} = 0$ for every $t$ . Same idea.
Differentiate twice. $F'(t) = c_1 + 2c_2 e^{2t} = 0$ and $F''(t) = 4c_2 e^{2t} = 0$ . The zero function has zero derivatives. Each derivative removes a term: the constant first, then $t$ .
Solve from the bottom. $4c_2 e^{2t} = 0$ and $e^{2t} \ne 0$ , so $c_2 = 0$ . Then $F'$ gives $c_1 = 0$ , and $F$ gives $c_0 = 0$ . Each equation has only one new unknown, so we can climb back up.

Answer. Both sets are independent. Evaluating at points works when a few values of $t$ give a solvable system. Differentiating works when each derivative removes a term.

Worked example MATH 375 HW2, Apostol 1.10 #14

Let

S = \{f \in P_n : f(0) + f'(0) = 0\}

. Find a basis of

S

and its dimension.

Solution.

Write a general $f$ . $f = c_0 + c_1 t + \dots + c_n t^n$ , so $f(0) = c_0$ and $f'(0) = c_1$ . The condition is about $f$ and $f'$ at 0, and only the constant and linear coefficients show up there.
Translate the condition. $c_0 + c_1 = 0$ , so $c_1 = -c_0$ . The condition is one linear equation on the $n + 1$ coefficients.
Substitute and factor. $f = c_0 (1 - t) + c_2 t^2 + \dots + c_n t^n$ . So $\{1 - t, t^2, \dots, t^n\}$ spans $S$ . Every element of $S$ has this form, and each of these polynomials satisfies the condition itself.
Check independence. If $a(1 - t) + b_2 t^2 + \dots + b_n t^n = 0$ , the constant term gives $a = 0$ , and then each $b_j = 0$ . Only $1 - t$ has a constant term, and $t^2, \dots, t^n$ are independent.
Count. The basis has $1 + (n - 1) = n$ vectors. Dimension is the size of any basis.

Answer. $\dim S = n$ . One linear condition on $n + 1$ coefficients removed exactly one dimension.

Where it is used

Used inside maths· Inside maths: solution spaces of differential equations

The solutions of $y'' + y = 0$ form a vector space of dimension 2, with basis $\{\cos t, \sin t\}$ . Chapter 7.2 proves that an order- $n$ linear homogeneous equation always has an $n$ -dimensional solution space. So you never need more than $n$ “basic” solutions, and you need all $n$ of them. That is why MATH 376 looked for exactly two solutions of a second-order equation.

Used in the real world· Signals: JPEG is a change of basis

A JPEG cuts an image into 8 by 8 blocks. Each block is a vector in $\mathbb R^{64}$ : 64 brightness values. In the standard basis, its coordinates are those 64 pixel values. JPEG switches to a different basis of $\mathbb R^{64}$ , made of 64 cosine patterns running from smooth to finely striped (the discrete cosine transform). In the new basis, most coordinates of a typical photo block are tiny, because photos are mostly smooth. JPEG stores the few large coordinates and throws the tiny ones away. The change of basis itself saves nothing. It is what makes the saving possible: after it, rounding off the tiny coordinates and storing the few large ones compactly is what shrinks a 20 MB photo to about 2 MB.

Used in the real world· AI: dimension of a representation

A language model such as BERT-base turns each token into a vector in $\mathbb R^{768}$ . “768-dimensional” is this chapter’s dimension: 768 coordinates in the standard basis. LoRA, a popular way to fine-tune such models, learns a change $\Delta W$ to each weight matrix that has low rank (for example rank 8). For a 768 by 768 matrix that means about 12,000 trainable numbers instead of about 590,000, roughly 48 times fewer. Chapter 3.4 explains rank and why low rank works.

Traps

Watch out· Independence belongs to the whole set

Independence is a property of a whole set, not of the vectors one at a time. Vectors that look fine individually can still be dependent together. $(1,2,3)$ , $(4,5,6)$ and $(7,8,9)$ are each non-zero, and no two lie on one line, yet the three are dependent, as the step-through above showed.

Watch out· Count the coefficients

$\dim P_n = n + 1$ . The polynomials of degree at most 2 need three numbers: $c_0$ , $c_1$ and $c_2$ .

Watch out· Spanning and independence are separate checks

A basis needs both. $\{(1,0), (0,1), (1,1)\}$ spans $\mathbb R^2$ but is dependent. $\{(1,0)\}$ is independent but does not span. In an $n$ -dimensional space, ‘Extending to a basis’ (part b) saves half the work: $n$ independent vectors automatically span.

Watch out· A typo in the course notes

Lecture 3 lists $\{1, t, t^3\}$ as the standard basis of $P_2$ . It should be $\{1, t, t^2\}$ : $t^3$ is not even in $P_2$ .

Problems

checkSpagnolie Lecture 3

Is $\{(1, 0), (1, 1), (-1, 3)\}$ independent in $\mathbb R^2$ ? Answer without solving any equations.

Solution

No. Three vectors in a space spanned by two vectors are dependent (the $k + 1$ theorem). Concretely, $(-1, 3) = -4(1, 0) + 3(1, 1)$ .

check

What is the dimension of (a) $\mathbb R^{2 \times 3}$ , (b) $P_3$ , (c) the plane $\{(x, y, z) : x + y + z = 0\}$ ?

Solution

(a) 6: one basis matrix for each entry. (b) 4: $1, t, t^2, t^3$ . (c) 2. The condition gives $x = -y - z$ , so every element is $(-y - z, y, z) = y(-1, 1, 0) + z(-1, 0, 1)$ . The two vectors $(-1, 1, 0)$ and $(-1, 0, 1)$ span the plane and are independent, so they form a basis.

check

Does a set containing the zero vector ever count as independent?

Solution

Never. $1 \cdot O + 0 \cdot (\text{the rest}) = O$ is a combination equal to zero with a non-zero coefficient.

useApostol 1.10 #9

Show $S = \{(x, y, z) : y = 2x,\ z = 3x\}$ is a subspace of $\mathbb R^3$ and find its dimension.

Solution

Every element is $(x, 2x, 3x) = x(1, 2, 3)$ , so $S = L(\{(1, 2, 3)\})$ , which is a subspace because a span is a subspace. The set $\{(1, 2, 3)\}$ is independent (one non-zero vector) and spans $S$ , so $\dim S = 1$ . Geometrically, $S$ is a line through the origin.

useApostol 1.10 #17

Find a basis and the dimension of the even polynomials in $P_n$ (those with $f(-t) = f(t)$ ).

Solution

$f(-t) = f(t)$ means $c_j(-1)^j = c_j$ for every $j$ , which forces $c_j = 0$ for odd $j$ . So a basis is $\{1, t^2, t^4, \dots\}$ , up to the largest even power that is at most $n$ . Its dimension is $n/2 + 1$ if $n$ is even, and $(n + 1)/2$ if $n$ is odd.

useApostol 1.10 #23(a)

For $a \ne b$ , both non-zero, show that $1, e^{at}, e^{bt}$ are independent.

Solution

Suppose $F(t) = c_0 + c_1 e^{at} + c_2 e^{bt} = 0$ for all $t$ . Then $F'(t) = a c_1 e^{at} + b c_2 e^{bt} = 0$ and $F''(t) = a^2 c_1 e^{at} + b^2 c_2 e^{bt} = 0$ . At $t = 0$ : $a c_1 + b c_2 = 0$ and $a^2 c_1 + b^2 c_2 = 0$ . Multiply the first by $a$ and subtract from the second: $(b^2 - ab)c_2 = b(b - a)c_2 = 0$ . Since $b \ne 0$ and $b \ne a$ , $c_2 = 0$ . Then $a c_1 = 0$ with $a \ne 0$ , so $c_1 = 0$ . Finally $F(0) = c_0 = 0$ .

use

Find the coordinates of $x = (3, 1)$ in the basis $b_1 = (1, 1)$ , $b_2 = (1, -1)$ . Check them in the Basis2D visualizer by dragging.

Solution

Solve $c_1(1, 1) + c_2(1, -1) = (3, 1)$ : $c_1 + c_2 = 3$ and $c_1 - c_2 = 1$ . Adding gives $c_1 = 2$ , then $c_2 = 1$ . Coordinates $(2, 1)$ . Check: $2(1,1) + 1(1,-1) = (3, 1)$ .

prove

Prove that any subset of an independent set is independent.

Solution

Let $S$ be independent and $T \subseteq S$ . Suppose a combination of elements of $T$ equals $O$ . Give every element of $S$ that is not in $T$ the coefficient 0. That gives a combination of elements of $S$ equal to $O$ , so every coefficient is zero by the independence of $S$ , including those on $T$ . So $T$ is independent.

proveApostol 1.10, adapted

Let $\dim V = n$ . Prove that any $n$ vectors that span $V$ form a basis.

Solution

Suppose $S = \{x_1, \dots, x_n\}$ spans $V$ but is dependent. Then some $x_j$ is a combination of the others, so removing it leaves a set of $n - 1$ vectors that still spans $V$ : anything built using $x_j$ can be rebuilt with $x_j$ replaced by its expression. But $V$ has a basis of $n$ independent vectors, and by the $k + 1$ theorem, $n$ vectors in the span of $n - 1$ vectors are dependent. That is a contradiction, so $S$ is independent, and it is a basis.

prove

Prove: if $W$ is a subspace of $V$ and $\dim V = n$ , then $\dim W \le n$ , with equality only if $W = V$ .

Solution

By the $k + 1$ theorem, any independent set in $W$ , which is also in $V$ , has at most $n$ elements. So build a basis of $W$ as in ‘Extending to a basis’ (part a), starting from the empty set and adding vectors of $W$ outside the current span. This stops after at most $n$ steps, so $W$ has a basis and $\dim W \le n$ . If $\dim W = n$ , a basis of $W$ is $n$ independent vectors in $V$ , so it is a basis of $V$ by ‘Extending to a basis’ (part b). Then $W = L(\text{basis}) = V$ .

prove

Prove that the span of $S$ is the smallest subspace containing $S$ : if $W$ is a subspace with $S \subseteq W$ , then $L(S) \subseteq W$ .

Solution

Take any element $\sum c_i x_i$ of $L(S)$ . Each $x_i$ is in $S$ , hence in $W$ . By closure under scaling, each $c_i x_i \in W$ . By closure under addition, applied $k - 1$ times, the sum is in $W$ . So $L(S) \subseteq W$ .

stretchAxler 2C

Let $U$ and $W$ be subspaces of a finite-dimensional space. Let $U + W$ be the set of all sums $u + w$ with $u \in U$ and $w \in W$ . Show $U + W$ is a subspace, and prove

\dim(U + W) = \dim U + \dim W - \dim(U \cap W).

Solution

$U + W$ contains $O + O$ , and $(u + w) + (u' + w') = (u + u') + (w + w')$ and $a(u + w) = au + aw$ , so it is a subspace.

For the formula, take a basis $v_1, \dots, v_m$ of $U \cap W$ . Extend it to a basis $v_1, \dots, v_m, u_1, \dots, u_j$ of $U$ , and separately to a basis $v_1, \dots, v_m, w_1, \dots, w_k$ of $W$ (‘Extending to a basis’ (part a)). The combined list $v$ ‘s, $u$ ‘s, $w$ ‘s spans $U + W$ . Suppose $\sum a_i v_i + \sum b_i u_i + \sum c_i w_i = O$ . Then $\sum c_i w_i = -\sum a_i v_i - \sum b_i u_i$ lies in $U$ , and also in $W$ , so it is in $U \cap W$ . So $\sum c_i w_i = \sum d_i v_i$ for some $d_i$ , that is, $\sum c_i w_i - \sum d_i v_i = O$ . The $v$ ‘s and $w$ ‘s together are a basis of $W$ , hence independent, so all $c_i = 0$ (and all $d_i = 0$ ). Then $\sum a_i v_i + \sum b_i u_i = O$ forces all $a_i = b_i = 0$ . So the list is a basis, and

\dim(U + W) = m + j + k = (m + j) + (m + k) - m.

This is the vector-space version of inclusion–exclusion, $|A \cup B| = |A| + |B| - |A \cap B|$ , from CS 240.

Summary

The span of some vectors is everything they can build; it is always a subspace, the smallest one containing them.
A set is dependent if some combination with coefficients not all zero gives $O$ . Equivalently, one vector is a combination of the others: it is wasted.
To test independence, set $\sum c_i x_i = O$ and solve. Independent means the only solution is all $c_i = 0$ .
Key theorem: $k + 1$ vectors in the span of $k$ vectors are dependent. The proof is elimination plus induction.
A basis is independent and spanning. All bases of a space have the same size, the dimension.
In an $n$ -dimensional space, $n$ independent vectors form a basis, $n$ spanning vectors form a basis, and any independent set extends to a basis.
A basis gives every vector unique coordinates, so an $n$ -dimensional space can be computed in as if it were $\mathbb R^n$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

2.2 Span, independence, basis, dimension

The picture

Definitions and theorems

Building vectors: span

Linear combination

Span

The span is a subspace

Wasted vectors: dependence

Linearly dependent

Linearly independent

The key theorem: too many vectors are always dependent

More vectors than a spanning set means dependent

No waste and nothing missing: basis and dimension

Basis

All bases have the same size

Dimension

Extending to a basis

Coordinates

Unique coordinates

Coordinates

Examples and non-examples

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere