Linear Algebra 23 | Linear Independence (Examples) [dark version]

Q: Why does including the zero vector automatically destroy linear independence?

If a family contains the zero vector $0$, then $\lambda\cdot 0=0$ holds for any scalar $\lambda$. That means there exists a non-trivial linear combination (some coefficient not equal to zero) that still produces the zero vector, which violates the definition of linear independence.

Q: What guarantees that three vectors in $\mathbb{R}^2$ are linearly dependent?

In $\mathbb{R}^2$, three vectors form a family larger than the dimension. The transcript illustrates this concretely with vectors like $(1,1)$, $(1,0)$, and $(0,1)$: $(1,1) - (1,0) - (0,1) = (0,0)$. The existence of such a non-trivial combination proves dependence.

Q: How do canonical unit vectors $e_1,\dots,e_N$ demonstrate linear independence?

Take $\lambda_1 e_1 + \cdots + \lambda_N e_N$. Since each $e_j$ has a 1 in the $j$-th position and zeros elsewhere, the sum equals $(\lambda_1,\lambda_2,\dots,\lambda_N)$. If this equals the zero vector, then every component must be zero, forcing $\lambda_1=\cdots=\lambda_N=0$. That matches the independence definition.

Q: What is the span-based characterization of linear dependence?

A family of $K$ vectors is linearly dependent if and only if some vector $v_L$ can be removed without changing the span of the remaining vectors. Dependence means at least one vector is unnecessary for generating the same subspace; independence means no vector can be omitted while preserving the span.

TL;DR

A family of vectors is linearly independent exactly when the only solution to a linear combination equaling the zero vector uses all zero coefficients.

Briefing Cornell Notes

Briefing

Linear independence hinges on one test: a family of vectors is linearly independent exactly when the only way to combine them to get the zero vector uses all zero coefficients. That definition immediately yields a sharp rule of thumb—if the family includes the zero vector, independence is impossible. For a single vector in $R^{N}$ , the family ${v}$ is linearly independent as long as $v \neq = 0$ ; the equation $λ v = 0$ forces $λ = 0$ . But if $v = 0$ , then $λ \cdot 0 = 0$ holds for any $λ$ , producing a non-trivial linear combination and proving linear dependence.

The examples then scale this idea up. In $R^{2}$ , three vectors with two components cannot all be independent. With vectors like $(1, 0)$ , $(0, 1)$ , and $(1, 1)$ , there is a non-trivial combination that lands on the zero vector: $(1, 1) - (1, 0) - (0, 1) = (0, 0)$ . The takeaway is structural rather than numerical: in $R^{2}$ , any set larger than the dimension must be linearly dependent.

A contrasting “canonical” example comes from the standard basis vectors $e_{1}, e_{2}, \dots, e_{N}$ in $R^{N}$ . Consider an arbitrary linear combination $λ_{1} e_{1} + \dots + λ_{N} e_{N}$ . Because each $e_{j}$ has a single 1 in the $j$ -th position and zeros elsewhere, the sum becomes the vector $(λ_{1}, λ_{2}, \dots, λ_{N})$ . Setting this equal to the zero vector forces every component $λ_{j}$ to be zero, matching the definition of linear independence. This standard basis is therefore a foundational example of a linearly independent family.

The transcript also highlights a common way to generate dependence: if a linearly independent family is extended by adding another vector $v$ to the end, the enlarged family becomes linearly dependent. Intuitively, once the original family already spans its “maximal efficient” structure, adding a new vector creates redundancy—there will exist coefficients that cancel out to produce the zero vector.

Finally, linear dependence is characterized in a more geometric way. A family of $K$ vectors in $R^{N}$ is linearly dependent if and only if at least one vector in the family can be removed without changing the span of the remaining vectors. In other words, dependence means some vector is unnecessary for generating the same subspace; independence means no vector can be omitted while preserving the span. This equivalence explains why linear independence is so tightly linked to the efficiency of describing subspaces—an idea that becomes central when constructing bases.

Cornell Notes

Linear independence is defined by a zero-combination test: a family of vectors is linearly independent if the only way to form the zero vector is to use all zero coefficients. The presence of the zero vector in a family guarantees dependence, since any coefficient works in $λ \cdot 0 = 0$ . In $R^{2}$ , any three vectors are automatically dependent because a non-trivial combination can be found that sums to $(0, 0)$ . The standard basis vectors $e_{1}, \dots, e_{N}$ form a key independent example: $λ_{1} e_{1} + \dots + λ_{N} e_{N} = (λ_{1}, \dots, λ_{N})$ , so equality to the zero vector forces every $λ_{j} = 0$ . Dependence can also be recognized by span: a family is dependent exactly when some vector can be omitted without changing the span.

Why does including the zero vector automatically destroy linear independence?

If a family contains the zero vector

0

, then

λ \cdot 0 = 0

holds for any scalar

λ

. That means there exists a non-trivial linear combination (some coefficient not equal to zero) that still produces the zero vector, which violates the definition of linear independence.

What guarantees that three vectors in $R^{2}$ are linearly dependent?

R^{2}

, three vectors form a family larger than the dimension. The transcript illustrates this concretely with vectors like

(1, 1)

(1, 0)

, and

(0, 1)

(1, 1) - (1, 0) - (0, 1) = (0, 0)

. The existence of such a non-trivial combination proves dependence.

How do canonical unit vectors $e_{1}, \dots, e_{N}$ demonstrate linear independence?

Take

λ_{1} e_{1} + \dots + λ_{N} e_{N}

. Since each

e_{j}

has a 1 in the

j

-th position and zeros elsewhere, the sum equals

(λ_{1}, λ_{2}, \dots, λ_{N})

. If this equals the zero vector, then every component must be zero, forcing

λ_{1} = \dots = λ_{N} = 0

. That matches the independence definition.

Why does adding an extra vector $v$ to the end of a family lead to linear dependence?

Once the family already has the structure of a maximal independent set (as in the canonical example), appending another vector introduces redundancy. The transcript phrases this as the ability to choose coefficients so a linear combination of the original unit vectors produces

v

, meaning the enlarged family contains a non-trivial combination that yields the zero vector—so it becomes linearly dependent.

What is the span-based characterization of linear dependence?

A family of

K

vectors is linearly dependent if and only if some vector

v_{L}

can be removed without changing the span of the remaining vectors. Dependence means at least one vector is unnecessary for generating the same subspace; independence means no vector can be omitted while preserving the span.

Review Questions

Given a family that includes the zero vector, what specific coefficient-based argument shows it must be linearly dependent?
In $R^{2}$ , construct (or reason about) a non-trivial linear combination of three vectors that equals the zero vector.
Explain the equivalence: how does “can omit a vector without changing the span” translate into linear dependence?

Key Points

1
A family of vectors is linearly independent exactly when the only solution to a linear combination equaling the zero vector uses all zero coefficients.
2
Any family containing the zero vector is automatically linearly dependent because $λ \cdot 0 = 0$ for any $λ$ .
3
In $R^{2}$ , any set of three vectors is linearly dependent; a non-trivial combination can be found that sums to $(0, 0)$ .
4
The standard basis vectors $e_{1}, \dots, e_{N}$ are linearly independent because $λ_{1} e_{1} + \dots + λ_{N} e_{N} = (λ_{1}, \dots, λ_{N})$ forces all $λ_{j} = 0$ when the result is zero.
5
Adding a vector $v$ to the end of a canonical independent family creates linear dependence by introducing redundancy.
6
Linear dependence is equivalent to the ability to remove at least one vector while keeping the same span; independence means no such omission is possible.

Highlights

A single nonzero vector {v} in RN is linearly independent because λv=0 forces λ=0.

In R2, three vectors must be dependent; an explicit example uses (1,1)−(1,0)−(0,1)=(0,0).

The standard basis e1​,…,eN​ is the canonical independent family: ∑λj​ej​ produces (λ1​,…,λN​).

A family is dependent exactly when some vector can be omitted without changing the span of the rest—dependence equals redundancy in spanning power.

Linear Algebra 23 | Linear Independence (Examples) [dark version]

Briefing

Cornell Notes

Why does including the zero vector automatically destroy linear independence?

What guarantees that three vectors in $R^{2}$ are linearly dependent?

How do canonical unit vectors $e_{1}, \dots, e_{N}$ demonstrate linear independence?

Why does adding an extra vector $v$ to the end of a family lead to linear dependence?

What is the span-based characterization of linear dependence?

Review Questions

Key Points

Highlights

Topics

Get summaries like this for any content

Briefing

Cornell Notes

Why does including the zero vector automatically destroy linear independence?

What guarantees that three vectors in R2 are linearly dependent?

How do canonical unit vectors e1​,…,eN​ demonstrate linear independence?

Why does adding an extra vector v to the end of a family lead to linear dependence?

What is the span-based characterization of linear dependence?

Review Questions

Key Points

Highlights

Topics

Get summaries like this for any content

What guarantees that three vectors in $R^{2}$ are linearly dependent?

How do canonical unit vectors $e_{1}, \dots, e_{N}$ demonstrate linear independence?

Why does adding an extra vector $v$ to the end of a family lead to linear dependence?