Recipe

Test whether a set of vectors is linearly independent

Use when: you're given a specific list of vectors \mathbf v_1,\dots,\mathbf v_k (in any vector space, not necessarily \mathbb{R}^n written as columns) and asked whether they're linearly independent

Goal: either a proof that only c_1=\cdots=c_k=0 solves the dependence equation, or an explicit nontrivial dependence relation showing which vector is redundant

Taught in Linear independence · a later section extends this recipe with more cases

1. 1

   Set c_1\mathbf v_1 + c_2\mathbf v_2 + \cdots + c_k\mathbf v_k = \mathbf 0, one unknown scalar coefficient per vector.

   Example

   For \mathbf v_1=(1,2,1), \mathbf v_2=(2,1,-1), \mathbf v_3=(0,1,1): c_1(1,2,1)+c_2(2,1,-1)+c_3(0,1,1)=(0,0,0)

   ⚡ Shortcut

   For exactly two vectors, skip the whole system: \mathbf v_1,\mathbf v_2 are dependent exactly when one is a scalar multiple of the other, which you can check by comparing the ratio of corresponding entries. A single vector needs even less — it's independent unless it's \mathbf 0 itself. Neither shortcut extends to three or more vectors, where "one is a multiple of another" stops being the only way to be dependent.

   Why: independence is defined as exactly this equation having only the trivial solution — there's no other starting point for the test.

2. 2

   Each coordinate of the vector equation gives one ordinary linear equation in c_1,\dots,c_k — write them out as a stacked system.

   Example

   c_1+2c_2=0, \qquad 2c_1+c_2+c_3=0, \qquad c_1-c_2+c_3=0

   Why: a single vector equation among vectors with n coordinates is really n ordinary scalar equations holding simultaneously, one per coordinate — exactly what "two vectors are equal" always means.

3. 3

   Solve one equation for one variable, substitute into the rest, and repeat until either every c_i is forced to 0 or a variable survives with no constraint left to pin it down.

   Example

   Equation (1) gives c_1=-2c_2. Substituting into (3): -2c_2-c_2+c_3=0 \Rightarrow c_3=3c_2. Substituting both into (2): 2(-2c_2)+c_2+3c_2 = -4c_2+c_2+3c_2 = 0 — this holds for every value of c_2, so c_2 is never pinned down to a single number.

   • the substitution forces every c_i=0, with no variable left free — stop; the vectors are linearly independent.
   • at least one c_i remains free after substitution — continue to Step 4.

   Why: this is the trivial-vs-nontrivial-solution question from a homogeneous system, applied to the unknown coefficients c_i instead of the vectors' own coordinates.

4. 4

   Assign the free variable a convenient nonzero value (usually 1), compute the remaining c_i from it, and write out the resulting equation — that IS the dependence relation.

   Example

   Set c_2=1: then c_1=-2(1)=-2 and c_3=3(1)=3, none of them zero. Check it balances: -2(1,2,1)+1(2,1,-1)+3(0,1,1) = (-2+2+0,\ -4+1+3,\ -2-1+3) = (0,0,0) \boxed{-2\mathbf v_1+\mathbf v_2+3\mathbf v_3=\mathbf 0} Rearranged, \mathbf v_2 = 2\mathbf v_1-3\mathbf v_3 — \mathbf v_2 was reachable from the other two all along, so the set is linearly dependent.

   Why: a dependence relation is nothing more than the vector equation with its nontrivial coefficients plugged back in — concrete proof the "free lunch" the definition describes actually exists here.