Linear-Algebra-Done-Right · Unit 4 · Video 1 · Interactive Practice

Bases: Reduce a Spanning List, Extend an Independent List

IKey Formulas

Formula	Name	What it says
$v_1, \dots, v_n$ linearly independent and $\operatorname{span}(v_1, \dots, v_n) = V$	Basis of $V$	The standard basis of $\mathbf{F}^n$ , the list $(1, 2), (3, 5)$ of $\mathbf{F}^2$ , and $1, z, \dots, z^m$ of $\mathcal{P}_m(\mathbf{F})$ are bases
$v = a_1 v_1 + \cdots + a_n v_n$ for exactly one choice of $a_1, \dots, a_n \in \mathbf{F}$	Criterion for a basis	Basis $\iff$ every $v \in V$ has exactly one representation; for $(1, 2), (3, 5)$ : $a = -5x + 3y$ , $b = 2x - y$
Step $k$ : delete $v_k$ if $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$	Reduce a spanning list	What survives is a basis, so every finite-dimensional vector space has a basis
Reduce $u_1, \dots, u_m, w_1, \dots, w_n$ ; $\ W = \operatorname{span}(\text{kept } w\text{'s})$	Extend an independent list	No $u_k$ is deleted, and $V = U \oplus W$ for $U = \operatorname{span}(u_1, \dots, u_m)$

Key Insight: One deletion procedure does both jobs: run on a spanning list it cuts down to a basis, and run on $u_1, \dots, u_m$ followed by a spanning list it builds $u_1, \dots, u_m$ up to a basis.

IIVisualization 1 — Reducing a Spanning List

Step $k$ deletes $v_k$ exactly when $v_k \in \operatorname{span}(v_1, \dots, v_{k-1})$ ; the surviving list $B$ is a basis of the span.

Step k 0

IIIVisualization 2 — Extending an Independent List

Reducing $u_1, \dots, u_m, w_1, w_2, w_3$ keeps every $u_k$ and only the $w$ 's needed to reach all of $\mathbf{F}^3$ .

Independent list Spanning list Step k 0

IVVisualization 3 — Complements Are Not Unique

Every line $W$ through $0$ other than the $x$ -axis $U$ is a complement of $U$ : each $v$ splits uniquely as $u + w$ .

VQuiz Questions

Problem 1 · Run the Reduction Procedure

Given: the list $(0, 0)$ , $(2, -1)$ , $(-4, 2)$ , $(1, 3)$ , $(5, 1)$ spans $\mathbf{F}^2$ — find the list $B$ that the reduction procedure returns.

✅ Correct! The procedure deletes

(0, 0)

(-4, 2) = -2(2, -1)

and

(5, 1) = 2(2, -1) + (1, 3)

, leaving the basis

(2, -1), (1, 3)

❌ Step 1 deletes the zero vector. At step 1 the test is against

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

, and

(0, 0) \in \{0\}

; a list containing

0

is never linearly independent.

❌ Step 5 also deletes. By step 5 the kept vectors

(2, -1), (1, 3)

already reach

(5, 1) = 2(2, -1) + (1, 3)

, so

(5, 1)

lies in the span of the earlier vectors.

❌ The test looks backwards, not forwards.

(2, -1)

is tested only against

(0, 0)

and is kept; it is the later vector

(-4, 2) = -2(2, -1)

that lies in the span of the vectors before it.

Show solution

Write $v_1, \dots, v_5$ for the list and test each $v_k$ against $\operatorname{span}(v_1, \dots, v_{k-1})$ .

Step 1: $\operatorname{span}(\,) = \{0\}$ and $v_1 = (0, 0)$ , so $v_1$ is deleted.
Step 2: $\operatorname{span}(v_1) = \{0\}$ and $(2, -1) \neq 0$ , so $v_2$ is kept.
Step 3: $\operatorname{span}(v_1, v_2) = \{c(2, -1) : c \in \mathbf{F}\}$ and $(-4, 2) = -2(2, -1)$ , so $v_3$ is deleted.
Step 4: $(1, 3)$ is not a multiple of $(2, -1)$ (a multiple $c(2, -1)$ with first coordinate $1$ is $(1, -\tfrac{1}{2})$ ), so $v_4$ is kept.
Step 5: solving $a(2, -1) + b(1, 3) = (5, 1)$ gives $2a + b = 5$ , $-a + 3b = 1$ , so $a = 2$ , $b = 1$ ; thus $v_5 \in \operatorname{span}(v_1, \dots, v_4)$ is deleted.

So $B = (2, -1), (1, 3)$ . It spans $\mathbf{F}^2$ because the procedure never changes the span (each $(x, y) = \tfrac{3x - y}{7}(2, -1) + \tfrac{x + 2y}{7}(1, 3)$ ), and it is linearly independent because neither vector is a scalar multiple of the other. Hence $B$ is a basis of $\mathbf{F}^2$ .

Problem 2 · Three Vectors, No Multiples

Given: the list $(1, 0, 1)$ , $(0, 1, 1)$ , $(1, 1, 2)$ in $\mathbf{F}^3$ , in which no vector is a scalar multiple of another — which statement is true?

✅ Correct!

(1, 1, 2) = (1, 0, 1) + (0, 1, 1)

, so the list is dependent, and its span equals the span of a list of length

2

, too short to span

\mathbf{F}^3

❌ The scalar-multiple test only works for two vectors. With three vectors, one can be a combination of the other two:

(1, 1, 2) = (1, 0, 1) + (0, 1, 1)

❌ The list is linearly dependent.

1 \cdot (1, 0, 1) + 1 \cdot (0, 1, 1) - 1 \cdot (1, 1, 2) = 0

is a combination equal to

0

with nonzero scalars.

❌ This list does not span $\mathbf{F}^3$ . Deleting

(1, 1, 2)

leaves the span unchanged, so the span equals

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

; a list of length

2

cannot span

\mathbf{F}^3

, because the standard basis is a linearly independent list of length

3

Show solution

Dependent. Run the reduction test: $(1, 1, 2) \in \operatorname{span}((1, 0, 1), (0, 1, 1))$ since $(1, 1, 2) = (1, 0, 1) + (0, 1, 1)$ . Hence $(1, 0, 1) + (0, 1, 1) - (1, 1, 2) = 0$ , a combination equal to $0$ whose scalars are not all $0$ .

Not spanning. By the linear dependence lemma, removing $(1, 1, 2)$ does not change the span, so the span of the list equals $\operatorname{span}((1, 0, 1), (0, 1, 1))$ . The standard basis of $\mathbf{F}^3$ is linearly independent of length $3$ , so every list spanning $\mathbf{F}^3$ has length at least $3$ ; a list of length $2$ does not span $\mathbf{F}^3$ . For instance $(1, 0, 0)$ is missed: $a(1, 0, 1) + b(0, 1, 1) = (1, 0, 0)$ forces $a = 1$ , $b = 0$ and then third coordinate $1 \neq 0$ .

The test "no vector is a multiple of another" decides independence only for lists of length two.

Problem 3 · The Complement the Proof Builds

Given: $U = \{(x, y, z) \in \mathbf{F}^3 : x + y + z = 0\}$ with basis $u_1 = (1, -1, 0)$ , $u_2 = (1, 0, -1)$ , and the spanning list $w_1 = (1, 1, -2)$ , $w_2 = (0, 1, 0)$ , $w_3 = (1, 0, 0)$ of $\mathbf{F}^3$ — find the complement $W$ produced by reducing $u_1, u_2, w_1, w_2, w_3$ .

✅ Correct!

w_1 = -u_1 + 2u_2

and

w_3 = u_1 + w_2

are deleted; only

w_2

survives, so

W = \operatorname{span}((0, 1, 0))

❌ $w_1$ is in $U$ . It is not a multiple of

u_1

or of

u_2

, but its coordinates sum to

0

w_1 = -(1, -1, 0) + 2(1, 0, -1)

, so step 3 deletes it.

❌ The procedure keeps the first $w$ that leaves the span.

w_2 = (0, 1, 0)

is tested before

w_3

and is kept; then

w_3 = (1, 0, 0) = u_1 + w_2

is deleted.

❌ $w_3$ is already reached. Once

w_2

is kept,

w_3 = (1, 0, 0) = (1, -1, 0) + (0, 1, 0) = u_1 + w_2

lies in the span of the earlier vectors, so step 5 deletes it.

Show solution

Steps 1 and 2 keep $u_1, u_2$ , since $u_1, u_2$ is linearly independent.

Step 3. $a u_1 + b u_2 = (a + b, -a, -b)$ . Matching $(1, 1, -2)$ gives $a = -1$ , $b = 2$ , and indeed $a + b = 1$ . So $w_1 = -u_1 + 2u_2$ is deleted.

Step 4. $u_1, u_2, w_1$ all lie in the subspace $U$ , so their span is contained in $U$ . The coordinates of $w_2 = (0, 1, 0)$ sum to $1 \neq 0$ , so $w_2 \notin U$ and $w_2$ is kept.

Step 5. $u_1 + w_2 = (1, -1, 0) + (0, 1, 0) = (1, 0, 0) = w_3$ , so $w_3$ is deleted.

The basis of $\mathbf{F}^3$ is $u_1, u_2, w_2$ , and the proof sets $W = \operatorname{span}(w_2) = \operatorname{span}((0, 1, 0))$ , giving $\mathbf{F}^3 = U \oplus W$ .

Problem 4 · Two Representations

Given: $v_1, v_2, v_3$ spans $V$ , and some $v \in V$ satisfies $v = a_1 v_1 + a_2 v_2 + a_3 v_3 = c_1 v_1 + c_2 v_2 + c_3 v_3$ with $(a_1, a_2, a_3) \neq (c_1, c_2, c_3)$ — which conclusion is correct?

✅ Correct! Subtracting gives a combination equal to

0

with scalars

a_k - c_k

not all

0

; adding it to any representation of any

u

produces a second one.

❌ The extra freedom transfers to every vector. If

u = d_1 v_1 + d_2 v_2 + d_3 v_3

, then also

u = (d_1 + a_1 - c_1) v_1 + (d_2 + a_2 - c_2) v_2 + (d_3 + a_3 - c_3) v_3

, a different list of scalars.

❌ The criterion rules this out. A list is a basis exactly when every vector has only one representation;

v

has two, so the list is not a basis.

❌ Spanning is about existence. The list spans

V

by hypothesis; two representations say that uniqueness, hence linear independence, fails.

Show solution

Subtracting the two representations of $v$ gives

0 = (a_1 - c_1) v_1 + (a_2 - c_2) v_2 + (a_3 - c_3) v_3,

and some $a_k - c_k \neq 0$ , so $v_1, v_2, v_3$ is linearly dependent (and not a basis, by the criterion).

Now take any $u \in V$ . The list spans $V$ , so $u = d_1 v_1 + d_2 v_2 + d_3 v_3$ for some scalars. Adding the displayed combination (which is $0$ ) gives

u = (d_1 + a_1 - c_1) v_1 + (d_2 + a_2 - c_2) v_2 + (d_3 + a_3 - c_3) v_3,

whose scalars differ from $d_1, d_2, d_3$ in slot $k$ . So every vector of $V$ has at least two representations. Example in $\mathbf{F}^2$ with $(1, 0), (0, 1), (1, 1)$ : $(1, 1) = 1 \cdot (1, 1) = 1 \cdot (1, 0) + 1 \cdot (0, 1)$ , and then every $(x, y) = x(1, 0) + y(0, 1) = (x - 1)(1, 0) + (y - 1)(0, 1) + 1 \cdot (1, 1)$ .

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