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

Independent Lists Are No Longer Than Spanning Lists

IKey Formulas

Statement	Name	Role
$u_1, \dots, u_m$ independent, $w_1, \dots, w_n$ spans $V$ $\;\Rightarrow\; m \leq n$	Length inequality	Holds in every finite-dimensional $V$
$v_1, \dots, v_m$ dependent $\Rightarrow$ some $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$ , and removing $v_k$ keeps the span	Linear dependence lemma	Chooses the $w$ removed at each step
$B = u_1, \dots, u_k, \text{ then the } n - k \text{ remaining } w\text{'s}$	List after step $k$	Spans $V$ and has length $n$
$V$ finite-dimensional, $U$ a subspace of $V$ $\Rightarrow$ $U$ finite-dimensional	Subspaces	Proved with the length inequality

Key Insight: The lemma can never land on a $u$ : the $u$ 's come first, and $u_k \in \operatorname{span}(u_1, \dots, u_{k-1})$ would contradict their independence. So every one of the $m$ steps removes a different $w$ , and there must be at least $m$ of them.

IIThe Exchange, One Vector at a Time

Each step inserts one $u$ and removes one $w$ found by the lemma, so $B$ keeps spanning with length $n$ .

Lists

IIIDecided by Length Alone

Comparing a list's length with $n$ can rule out independence or spanning in $\mathbf{R}^n$ , but never establishes either.

IVSubspaces Are Finite-Dimensional

Choosing each $u_k \in U$ outside $\operatorname{span}(u_1, \dots, u_{k-1})$ builds an independent list, so the construction stops within $n$ choices.

U ⊆ R³

VQuiz Questions

Problem 1 · Five Spanning Vectors

Given: a vector space $V$ is spanned by a list $w_1, w_2, w_3, w_4, w_5$ . Which statement must be true?

✅ Correct! An independent list in

V

has length at most

5

, so a list of

6

vectors cannot be independent.

❌ Not quite. The inequality only bounds independent lists from above; it never makes a list independent. The list

w_1, w_1, w_1, w_1, w_1

has length

5

and is dependent.

❌ The inequality points the other way. Independent lists are at most as long as spanning lists:

m \leq 5

, not

m \geq 5

. The one-vector list

w_1

is independent whenever

w_1 \neq 0

❌ Not quite. Appending

0

w_1, \dots, w_5

gives a spanning list of length

6

; the theorem bounds the length of a spanning list only from below.

Show solution

Let $u_1, \dots, u_m$ be any linearly independent list in $V$ . Since $w_1, \dots, w_5$ spans $V$ , the length inequality gives

m \leq 5.

A list of $6$ vectors would be an independent list of length $6 > 5$ if it were independent, so every list of $6$ vectors in $V$ is linearly dependent. The other statements fail: $w_1, \dots, w_1$ (five copies) is a dependent list of length $5$ ; a single nonzero vector is an independent list of length $1$ ; and $w_1, \dots, w_5, 0$ spans $V$ with length $6$ .

Problem 2 · What the Lengths Alone Decide

Given: in $\mathbf{R}^4$ , a list $A$ has $5$ vectors, a list $B$ has $3$ vectors and a list $C$ has $4$ vectors. Nothing else is known about them. Which conclusion follows from the length inequality alone?

✅ Correct!

e_1, \dots, e_4

both spans

\mathbf{R}^4

and is independent, so independent lists have length

\leq 4

and spanning lists have length

\geq 4

; a length of exactly

4

decides nothing.

❌ Not quite. Length comparisons give only negative conclusions. Five copies of

e_1

do not span

\mathbf{R}^4

, and

e_1, e_1, e_2

is not independent.

❌ Close, but length $4$ is the case the theorem says nothing about.

e_1, e_2, e_3, e_1 + e_2

has length

4

and is dependent; only a computation can decide for a list of length

4

Show solution

The standard list $e_1, e_2, e_3, e_4$ spans $\mathbf{R}^4$ and is linearly independent. Use it on each side of the inequality:

It spans $\mathbf{R}^4$ , so every independent list has length $\leq 4$ . $A$ has length $5$ , so $A$ is linearly dependent.
It is independent, so every spanning list has length $\geq 4$ . $B$ has length $3$ , so $B$ does not span $\mathbf{R}^4$ .

For $C$ both bounds are met with equality, and the theorem is silent: $e_1, e_2, e_3, e_4$ is independent and spans, while $e_1, e_2, e_3, e_1 + e_2$ is dependent and does not span.

Problem 3 · One Exchange Step

Given: in $\mathbf{R}^3$ , the list $w_1 = (1, 0, 0)$ , $w_2 = (0, 1, 0)$ , $w_3 = (0, 0, 1)$ , $w_4 = (1, 1, 1)$ spans, and $u_1 = (0, 1, 1)$ . Step 1 forms $u_1, w_1, w_2, w_3, w_4$ and removes the first vector that lies in the span of the vectors before it.

Which vector is removed?

Which equation shows that it lies in the span of the vectors before it?

✅ Correct!

w_3 = u_1 - w_2 \in \operatorname{span}(u_1, w_1, w_2)

, and removing it leaves

u_1, w_1, w_2, w_4

, which still spans

\mathbf{R}^3

and has length

4

❌ Not quite. Test the vectors in order: is

w_1

a multiple of

u_1

? Is

w_2 = a u_1 + b w_1

possible? The third coordinate forces

a = 0

, and then the second coordinate gives

0 = 1

❌ It cannot be $u_1$ . Nothing comes before

u_1

, and

\operatorname{span}(\,) = \{0\}

, while

u_1 \neq 0

❌ Close.

w_4

also lies in the span of the vectors before it, but an earlier vector already does, and the step removes the first one.

❌ Check the coordinates. Compute the right-hand side: the result must be

(0, 0, 1)

❌ True equation, wrong vectors.

w_4 - w_1 - w_2 = (0, 0, 1)

, but

w_4

comes after

w_3

; the lemma needs

w_3

in the span of

u_1, w_1, w_2

Show solution

Check the vectors of $u_1, w_1, w_2, w_3, w_4$ in order.

$u_1 \neq 0$ , so $u_1 \notin \operatorname{span}(\,) = \{0\}$ .
$w_1 = (1, 0, 0)$ is not a multiple of $(0, 1, 1)$ .
$w_2 = a(0, 1, 1) + b(1, 0, 0) = (b, a, a)$ would need $a = 1$ and $a = 0$ : impossible.
$w_3$ : since $u_1 - w_2 = (0, 1, 1) - (0, 1, 0) = (0, 0, 1)$ , we get $w_3 = u_1 - w_2 \in \operatorname{span}(u_1, w_1, w_2)$ .

So $w_3$ is removed. By the second part of the lemma the span is unchanged:

\operatorname{span}(u_1, w_1, w_2, w_4) = \operatorname{span}(u_1, w_1, w_2, w_3, w_4) = \mathbf{R}^3,

and the new list has length $4 = n$ . The identity $w_3 = w_4 - w_1 - w_2$ is true but uses $w_4$ , which comes later in the list.

Problem 4 · A Subspace of Polynomials

Given: $U$ is a subspace of $\mathcal{P}_4(\mathbf{R})$ , the polynomials of degree at most $4$ , which is spanned by $1, x, x^2, x^3, x^4$ . The construction from the video chooses a nonzero $u_1 \in U$ , then each $u_k \in U$ with $u_k \notin \operatorname{span}(u_1, \dots, u_{k-1})$ , until the chosen vectors span $U$ .

At most how many vectors can it choose?

Why is $u_1, \dots, u_k$ linearly independent at every stage?

✅ Correct! The chosen list is independent by the lemma, it lies in

\mathcal{P}_4(\mathbf{R})

, which a list of length

5

spans, so at most

5

vectors are chosen and

U

is finite-dimensional.

❌ Not quite. The chosen vectors form an independent list in a space spanned by

1, x, x^2, x^3, x^4

❌ Too small. When

U = \mathcal{P}_4(\mathbf{R})

the construction can choose

1, x, x^2, x^3, x^4

: five vectors.

❌ That counts steps, not choices. The construction may run up to a sixth step, but that step only checks that the five chosen vectors span

U

; six chosen vectors would be an independent list of length

6

in a space spanned by

5

❌ That is what the construction rules out. The chosen vectors are independent and lie in

\mathcal{P}_4(\mathbf{R})

, so the length inequality caps them at

5

, and the process must stop.

❌ Not quite. Lying in a subspace says nothing about independence:

u, 2u

both lie in any subspace containing

u

❌ That is circular. The finite-dimensionality of

U

is the conclusion of this argument, not an ingredient.

❌ The logic runs the other way. Independence comes first, from the lemma; the bound of

5

is then a consequence of the length inequality.

Show solution

Independence. By construction $u_1 \neq 0$ , so $u_1 \notin \operatorname{span}(\,) = \{0\}$ , and each later $u_j \notin \operatorname{span}(u_1, \dots, u_{j-1})$ . If $u_1, \dots, u_k$ were dependent, the linear dependence lemma would give some $u_j \in \operatorname{span}(u_1, \dots, u_{j-1})$ , which never happens. So the list is independent.

The bound. The list lies in $\mathcal{P}_4(\mathbf{R})$ , which is spanned by the list $1, x, x^2, x^3, x^4$ of length $5$ . By the length inequality,

k \leq 5.

So the construction stops after at most $5$ choices, at a step where $U = \operatorname{span}(u_1, \dots, u_k)$ , and $U$ is finite-dimensional. The bound $5$ is reached when $U = \mathcal{P}_4(\mathbf{R})$ , for instance with $u_1, \dots, u_5 = 1, x, x^2, x^3, x^4$ .

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