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

Linear Combinations, Span, and Finite-Dimensional Spaces

IKey Formulas

Formula	Name	What it says
$a_1 v_1 + \cdots + a_m v_m$ , $\ a_1, \dots, a_m \in \mathbf{F}$	Linear combination	A sum of scalar multiples of the vectors in the list $v_1, \dots, v_m$
$\operatorname{span}(v_1, \dots, v_m) = \{a_1 v_1 + \cdots + a_m v_m : a_k \in \mathbf{F}\}$ , $\ \operatorname{span}(\,) = \{0\}$	Span	The set of all linear combinations; the empty list spans $\{0\}$
$\operatorname{span}(v_1, \dots, v_m) = V$ for some list	Finite-dimensional	Finitely many vectors produce every vector of $V$ ; $\mathbf{F}^n$ is spanned by the $n$ standard vectors
$\mathcal{P}_m(\mathbf{F}) = \operatorname{span}(1, z, \dots, z^m)$	Polynomials of degree at most $m$	A list of length $m + 1$ spans it, but no list spans $\mathcal{P}(\mathbf{F})$

Key Insight: $\operatorname{span}(v_1, \dots, v_m)$ is the smallest subspace of $V$ containing $v_1, \dots, v_m$ : it is a subspace, and every subspace $U$ with each $v_k \in U$ satisfies $\operatorname{span}(v_1, \dots, v_m) \subseteq U$ .

IIVisualization 1 — Linear Combinations in $\mathbf{R}^2$

Combinations of $v_1 = (1, 1)$ and $v_2 = (1, -1)$ reach every point of $\mathbf{R}^2$ ; a pair on one line reaches only that line.

IIIVisualization 2 — Is a Vector in the Span?

A vector of $\mathbf{R}^3$ lies in $\operatorname{span}((2, 1, -3), (1, -2, 4))$ only if one pair $a_1, a_2$ solves all three coordinate equations.

x₁ 17 x₂ −4 x₃ 5

💡 The first two coordinates already determine $a_1$ and $a_2$ , so the third is forced: this span is the plane $2x_1 + 11x_2 + 5x_3 = 0$ , not all of $\mathbf{R}^3$ .

IVVisualization 3 — Why $\mathcal{P}(\mathbf{F})$ Is Infinite-Dimensional

Every list in $\mathcal{P}(\mathbf{F})$ has a highest degree $m$ , so its span sits inside $\mathcal{P}_m(\mathbf{F})$ and misses $z^{m+1}$ .

p₁ p₂ p₃

💡 Since every list misses some $z^{m+1}$ , no list spans $\mathcal{P}(\mathbf{F})$ , while each $\mathcal{P}_m(\mathbf{F})$ is spanned by $1, z, \dots, z^m$ .

VQuiz Questions

Problem 1 · Finding the Scalars

Given: $(5, 5, -13) = a_1 (2, 1, -3) + a_2 (1, -2, 4)$ in $\mathbf{R}^3$ — find $a_1$ and $a_2$ .

✅ Correct!

3(2, 1, -3) - (1, -2, 4) = (6 - 1, 3 + 2, -9 - 4) = (5, 5, -13)

❌ The scalars are swapped.

-(2, 1, -3) + 3(1, -2, 4) = (1, -7, 15)

;

a_1

multiplies the first vector

(2, 1, -3)

❌ Check the sign of $a_2$ .

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

, which misses already in the first coordinate.

❌ Not quite. Compare coordinates:

5 = 2a_1 + a_2

and

5 = a_1 - 2a_2

; solve these two, then check the third.

Show solution

Comparing coordinates gives three equations:

5 = 2a_1 + a_2, \qquad 5 = a_1 - 2a_2, \qquad -13 = -3a_1 + 4a_2.

The first gives $a_2 = 5 - 2a_1$ . Substituting into the second:

5 = a_1 - 2(5 - 2a_1) = 5a_1 - 10 \implies a_1 = 3, \quad a_2 = 5 - 6 = -1.

Check the third equation: $-3(3) + 4(-1) = -9 - 4 = -13$ ✓. So $a_1 = 3$ and $a_2 = -1$ , and $(5, 5, -13) \in \operatorname{span}((2, 1, -3), (1, -2, 4))$ .

Problem 2 · Don't Stop at Two Equations

Given: the list $(2, 1, -3), (1, -2, 4)$ in $\mathbf{R}^3$ — is $(1, 3, -5)$ in its span?

✅ Correct! The only scalars that fit the first two equations give

-3(1) + 4(-1) = -7

, so the third equation fails and no linear combination equals

(1, 3, -5)

❌ Those scalars only fit two coordinates.

(2, 1, -3) - (1, -2, 4) = (1, 3, -7)

, not

(1, 3, -5)

: the third equation was never checked.

❌ That combination is $(-1, -3, 7)$ . It already misses the first coordinate.

❌ These two vectors do not span $\mathbf{R}^3$ . Their span here is the plane

2x_1 + 11x_2 + 5x_3 = 0

, and

2 + 33 - 25 = 10 \neq 0

❌ Not quite. Solve the first two coordinate equations, then test the third.

Show solution

Suppose $(1, 3, -5) = a_1 (2, 1, -3) + a_2 (1, -2, 4)$ . Comparing coordinates:

1 = 2a_1 + a_2, \qquad 3 = a_1 - 2a_2, \qquad -5 = -3a_1 + 4a_2.

The first gives $a_2 = 1 - 2a_1$ ; then $3 = a_1 - 2 + 4a_1$ , so $5a_1 = 5$ , $a_1 = 1$ and $a_2 = -1$ . These are the only values satisfying the first two equations.

-3a_1 + 4a_2 = -3 - 4 = -7 \neq -5.

The system has no solution, so $(1, 3, -5) \notin \operatorname{span}((2, 1, -3), (1, -2, 4))$ .

Problem 3 · Degree After Simplifying

Given: $p(z) = (z + 1)^2 - z(z + 2)$ and $q(z) = (z + 1)^2 - z(z + 2) - 1$ for all $z \in \mathbf{F}$ — find $\deg p$ and $\deg q$ .

What is $\deg p$ ?

What is $\deg q$ ?

✅ Correct!

p

is the constant polynomial

1

, of degree

0

, and

q

is the zero polynomial, of degree

-\infty

❌ Expand first. The

z^2

terms cancel:

(z^2 + 2z + 1) - (z^2 + 2z) = 1

. Degree is read from the polynomial, whose coefficients are unique, not from how it is written.

❌ $p$ is not the zero polynomial.

p(z) = 1

for all

z

, a nonzero constant:

a_0 = 1 \neq 0

, so

\deg p = 0

❌ Degree $0$ needs $a_0 \neq 0$ . Here

q(z) = 0

for every

z

, so no coefficient is nonzero.

❌ The zero polynomial does have a degree. By convention it is

-\infty

, which is less than every integer, so

0 \in \mathcal{P}_m(\mathbf{F})

for every

m

❌ Check $\deg p$ . Expand

(z + 1)^2

and

z(z + 2)

and subtract.

❌ Check $\deg q$ .

q = p - 1

; what polynomial is that?

Show solution

Expand for every $z \in \mathbf{F}$ :

p(z) = (z^2 + 2z + 1) - (z^2 + 2z) = 1.

So $p(z) = a_0$ with $a_0 = 1 \neq 0$ , and $\deg p = 0$ . Since coefficients of a polynomial are uniquely determined, the $z^2$ in the way $p$ was written does not count.

q(z) = p(z) - 1 = 0 \quad \text{for all } z \in \mathbf{F}.

$q$ is the zero polynomial, so $\deg q = -\infty$ .

Problem 4 · What a Subspace Must Contain

Given: $U$ is a subspace of $\mathcal{P}(\mathbf{F})$ with $z^2 \in U$ and $1 + z \in U$ — which polynomial must lie in $U$ ?

✅ Correct!

3 + 3z - 2z^2 = 3(1 + z) + (-2)z^2 \in \operatorname{span}(z^2, 1 + z)

, and the span is contained in every subspace containing

z^2

and

1 + z

❌ Not a linear combination.

a(z^2) + b(1 + z) = b + bz + az^2

always has equal constant and

z

coefficients, so it cannot equal this polynomial;

U

might be exactly

\operatorname{span}(z^2, 1 + z)

, which excludes it.

❌ Too much.

\operatorname{span}(z^2, 1 + z)

is itself a subspace containing both polynomials, and it misses

z \in \mathcal{P}_2(\mathbf{F})

, so

U

need not contain all of

\mathcal{P}_2(\mathbf{F})

❌ Not quite.

U

must contain

\operatorname{span}(z^2, 1 + z)

and nothing more is guaranteed.

Show solution

Since $\operatorname{span}(z^2, 1 + z)$ is the smallest subspace containing $z^2$ and $1 + z$ , every such $U$ satisfies $\operatorname{span}(z^2, 1 + z) \subseteq U$ . Conversely $U = \operatorname{span}(z^2, 1 + z)$ is one such subspace, so the polynomials that must lie in $U$ are exactly the linear combinations

a z^2 + b(1 + z) = b + bz + a z^2, \qquad a, b \in \mathbf{F}.

Their constant and $z$ coefficients are equal. For $3 + 3z - 2z^2$ take $b = 3$ , $a = -2$ ✓. For $1 + z^2$ the $z$ coefficient would be $b = 1$ but it is $0$ ; for $z$ the constant would be $b = 1$ but it is $0$ . Neither is in the span; in particular $z \in \mathcal{P}_2(\mathbf{F})$ is not, so not all of $\mathcal{P}_2(\mathbf{F})$ is forced. The answer is $3 + 3z - 2z^2$ .

Solved: 0 / 4

Linear Combinations, Span, and Finite-Dimensional Spaces

IKey Formulas

IIVisualization 1 — Linear Combinations in R2\mathbf{R}^2R2

IIIVisualization 2 — Is a Vector in the Span?

IVVisualization 3 — Why P(F)\mathcal{P}(\mathbf{F})P(F) Is Infinite-Dimensional

VQuiz Questions

IIVisualization 1 — Linear Combinations in $\mathbf{R}^2$

IVVisualization 3 — Why $\mathcal{P}(\mathbf{F})$ Is Infinite-Dimensional