Linear algebraConcept reference

Linear independence

A vector list is linearly independent when its zero-combination equation has only the all-zero coefficient solution.

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Overview
For u=(1,0) and v=(0,1), the equation au+bv=0 becomes (a,b)=(0,0)
A list containing the zero vector is dependent
For u=(1,0), v=(0,1) and w=(1,1), no two vectors are multiples
Key takeaway
Sources & further reading
Concept connections

01For u=(1,0) and v=(0,1), the equation au+bv=0 becomes (a,b)=(0,0)#

For $u=(1,0)$ and $v=(0,1)$ , the equation $au+bv=0$ becomes $(a,b)=(0,0)$ . Every coefficient must vanish.

This list is linearly independent: only the all-zero coefficient choice produces the zero vector.

Solve for real $a,b$ .

au+bv=0

All-zero coefficients are the trivial solution. A free coefficient permits a nonzero choice: dependence. No free coefficient means independence.

Figure 1For u=(1,2) and v=(2,4), the arrow −2u ends at (−2,−4). Adding v returns to the origin. Coefficients (−2,1) are not both zero, so this is a nontrivial cancellation and the two-vector list is dependent.

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Here $u=(1,2)$ and $v=(2,4)$ . Choosing coefficients $(-2,1)$ cancels both coordinates, so the list is dependent.

Check your reasoning

$u=(1,3)$ , $v=(2,6)$ . Which $(a,b)$ proves dependence via $au+bv=0$ ?

A $(0,0)$
B $(-2,1)$
C $(2,1)$

Show answer and explanation

$(-2,1)$

$-2u+v=(-2,-6)+(2,6)=0$ .

02A list containing the zero vector is dependent#

A list containing the zero vector is dependent. For $u=(2,1)$ and $v=(0,0)$ , the coefficients $(0,1)$ give $0u+1v=0$ .

A dependence relation needs at least one nonzero coefficient. It need not use every vector.

Check your reasoning

Channels: A adds $(1,0)$ , B adds $(0,2)$ per real weight. Nontrivial cancellation?

ANo: only (0,0).
BYes: weights (1,-1).
CYes: weights (2,-1).

Show answer and explanation

No: only (0,0).

$(a,2b)=0$ forces $a=b=0$ .

03For u=(1,0), v=(0,1) and w=(1,1), no two vectors are multiples#

For $u=(1,0)$ , $v=(0,1)$ and $w=(1,1)$ , no two vectors are multiples. Yet $u+v-w=0$ .

Checking each pair is insufficient for a larger list. Test the whole coefficient equation.

Check your reasoning

For $u=(1,0),v=(0,1),w=(1,1)$ , refute Sam’s claim of independence.

A $u$ and $v$ are different.
B $0u+0v+0w=0$ .
C $u+v-w=0$ .

Show answer and explanation

$u+v-w=0$ .

The relation uses nonzero coefficients.

Key takeaway

Only the all-zero coefficient solution means independence.

Test independence using the whole zero-combination equation.

Sources & further reading

[1]
Dan Margalit and Joseph Rabinoff, Interactive Linear Algebra (June 3, 2019), §2.5.1 ↗Georgia Tech · Interactive Linear Algebra · Book

01For u=(1,0) and v=(0,1), the equation au+bv=0 becomes (a,b)=(0,0)#

02A list containing the zero vector is dependent#

03For u=(1,0), v=(0,1) and w=(1,1), no two vectors are multiples#

Only the all-zero coefficient solution means independence.

Sources & further reading

Reference this concept