What does it mean for vectors to be linearly independent?

Use the zero-combination test. For vectors \(v_1,\ldots,v_n\), set up

\(c_1v_1+\cdots+c_nv_n=0\),

where each \(c_i\) is an unknown number and \(0\) is the zero vector. If the equation forces every coefficient to be zero, the vectors are linearly independent. If you can find even one solution with a nonzero coefficient, they are dependent. Independence means the all-zero coefficients are the only ones that combine the vectors to make zero. [S1]

The phrase “repeat one another” is a helpful first picture, but it is not the full test for larger sets. With two vectors, dependence means one is a scalar multiple of the other. For three or more, a vector might be made from a combination of several others, even if it is not just a copy or rescaling of any one of them.

Take \(v_1=(1,0)\) and \(v_2=(0,1)\). Ask whether some coefficients \(a,b\), not both zero, can satisfy

\(a(1,0)+b(0,1)=(0,0)\).

Adding the components gives \((a,b)=(0,0)\), so \(a=0\) and \(b=0\) are the only possibility. The pair is linearly independent.

Geometrically, the first vector points along the horizontal coordinate direction and the second along the vertical one. Neither is a scalar multiple of the other: changing the scale of one cannot turn it into the other. This matches the two-vector picture that independent vectors do not lie on the same line. [S1] The zero-combination check is the reliable way to state the result, rather than relying only on how a sketch looks.

Compare \(u_1=(1,0)\) and \(u_2=(2,0)\). Since \(u_2=2u_1\), one vector can be built from the other. The zero-combination equation also shows dependence directly:

\(2u_1-u_2=(0,0)\).

Those coefficients, 2 and −1, are not both zero, so the pair is dependent under the test. [S1] Both vectors lie along the same line; the second gives a longer vector in the same direction, not a new direction. That is the sense in which it is redundant: either vector can supply the direction represented by the pair. It does not mean the vectors have identical coordinates.

A useful write-up is: “The set is dependent because \(u_2=2u_1\). Equivalently, \(2u_1-u_2=0\) is a nontrivial zero combination, so the pair supplies only one line direction.”

For any set, write one coefficient per vector and set their weighted sum equal to the zero vector. Then solve the resulting component equations. If the only solution has every coefficient zero, the set is independent; if a nonzero coefficient choice works, it is dependent. For three vectors, check combinations of all three rather than looking only for a pair of scalar multiples: one vector could be made by combining the other two.

For instance, if a set contains \(p=(1,0)\), \(q=(0,1)\), and \(r=(1,1)\), then \(p+q-r=(0,0)\). That is a nontrivial zero combination, so this set is dependent, even though that relation involves all three vectors. The example illustrates the same test; the two-vector shortcut alone would not find that particular relation.

Reusable write-up template:

1. “Let the coefficients be ___. Set their linear combination equal to the zero vector.”
2. “Solving the component equations gives ___.”
3. “Because the only solution is all zero / because this solution is nontrivial, the set is independent / dependent.”
4. “The dependence means ___ can be formed from ___,” if you have identified such a relation.

The examples use two-dimensional coordinates. The same coefficient test applies to vectors with more components; the number of component equations changes.