Comparing Two Vectors with a Longer List
To compare two vectors, it is natural to ask whether one is just a scalar multiple of the other. If both are nonzero, this asks whether they point along the same line through the origin. For a pair of vectors, this comparison gives a direct test for independence. But a longer list can have a dependence involving three or more vectors even when no two vectors are scalar multiples. Checking pairs can miss that kind of relation.
Throughout, vectors are columns in \(\mathbb{R}^n\), and scalars are real. A list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is linearly independent if the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) forces every coefficient \(c_i\) to be zero. It is linearly dependent if there is a choice of coefficients, not all zero, for which that equation holds. Such a choice is a dependence relation.
Pairwise nonproportionality compares the vectors two at a time. Independence asks whether any nontrivial combination of the entire list can equal \(\mathbf{0}\). Those are different questions once a list has at least three vectors. First, however, we can settle exactly what the pairwise comparison tells us when there are only two nonzero vectors.
The Exact Test for a Pair
Proof. Suppose first that \(\mathbf{u}\) and \(\mathbf{v}\) are dependent. By the definition of dependence, there are real numbers \(a,b\), not both zero, such that \(a\mathbf{u}+b\mathbf{v}=\mathbf{0}\). If \(a\ne0\), rearranging and dividing by \(a\) gives \(\mathbf{u}=-(b/a)\mathbf{v}\), so the vectors are proportional. If \(a=0\), then \(b\ne0\), and the relation becomes \(b\mathbf{v}=\mathbf{0}\). Dividing by \(b\) gives \(\mathbf{v}=\mathbf{0}\), contradicting the hypothesis that \(\mathbf{v}\) is nonzero. Therefore dependence implies proportionality. Conversely, if the vectors are proportional, then either \(\mathbf{u}=c\mathbf{v}\) or \(\mathbf{v}=c\mathbf{u}\) for some scalar \(c\). In the first case, \(\mathbf{u}-c\mathbf{v}=\mathbf{0}\) is a nontrivial relation, because the coefficient on \(\mathbf{u}\) is \(1\). In the second case, \(\mathbf{v}-c\mathbf{u}=\mathbf{0}\) is nontrivial for the same reason. Thus proportionality implies dependence. By the definitions of independence and dependence, the two nonzero vectors are independent exactly when they are not proportional. \(\square\)
Worked Example: A Nonproportional Pair in \(\mathbb{R}^2\)
Test whether \(\mathbf{u}=(1,2)\) and \(\mathbf{v}=(2,-1)\) are independent. The vectors are columns, so test a general combination using real coefficients \(a\) and \(b\):
For this to be \(\mathbf{0}\), its coordinates must satisfy \(a+2b=0\) and \(2a-b=0\). The first equation gives \(a=-2b\). Substituting into the second gives \(2(-2b)-b=-5b=0\), so \(b=0\), and then \(a=0\). Only the trivial coefficients work, so the pair is independent.
Check: The first vector cannot be a multiple of the second: if \((1,2)=c(2,-1)\), then the first coordinate requires \(c=\tfrac12\), but the second would require \(2=-\tfrac12\). Nor can the second be a multiple of the first: \((2,-1)=c(1,2)\) would require \(c=2\) from the first coordinate and \(-1=4\) from the second. This agrees with the two-vector theorem.
The nonzero hypothesis in the theorem matters. For example, a pair containing \(\mathbf{0}\) is dependent: assigning coefficient \(1\) to the zero vector and \(0\) to the other vector gives a nontrivial combination equal to \(\mathbf{0}\). That pair is also proportional under the definition above. The zero vector therefore cannot serve as an exception to the idea that proportional pairs are dependent.
Why Pairwise Checks Miss a Relation
Worked Example: Three Vectors in \(\mathbb{R}^2\)
Consider \(\mathbf{v}_1=(1,0)\), \(\mathbf{v}_2=(0,1)\), and \(\mathbf{v}_3=(1,1)\). First check the pairs. Neither \(\mathbf{v}_1\) nor \(\mathbf{v}_2\) is a multiple of the other: a multiple of \((1,0)\) has second coordinate \(0\), while a multiple of \((0,1)\) has first coordinate \(0\). The vector \((1,1)\) is not a multiple of \((1,0)\), because its second coordinate is nonzero; it is not a multiple of \((0,1)\), because its first coordinate is nonzero. Thus the three vectors are pairwise nonproportional.
Now test the full list by combining all three vectors. Directly,
The coefficients \(1,1,-1\) are not all zero, so this is a dependence relation. The list is dependent even though each pair is nonproportional. Check: rearranging the relation gives \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\), and the coordinates on the right are \((1+0,0+1)=(1,1)\), exactly \(\mathbf{v}_3\).
The relation in this example involves all three vectors. No comparison of just two vectors can detect it: the issue is not that one vector is a multiple of another, but that one can be formed by adding the other two. A list may therefore have no redundant pair and still have redundancy across the full list.
Worked Example: A Dependent Triple in \(\mathbb{R}^3\)
Consider \(\mathbf{w}_1=(1,0,1)\), \(\mathbf{w}_2=(0,1,1)\), and \(\mathbf{w}_3=(1,1,2)\). Check that no pair is proportional. The first two cannot be multiples because \(\mathbf{w}_1\) has first coordinate \(1\) and \(\mathbf{w}_2\) has first coordinate \(0\), while \(\mathbf{w}_2\) also has second coordinate \(1\) and \(\mathbf{w}_1\) has second coordinate \(0\). The third vector is not a multiple of \(\mathbf{w}_1\), since the second coordinate of \(\mathbf{w}_1\) is \(0\) but that of \(\mathbf{w}_3\) is \(1\). It is not a multiple of \(\mathbf{w}_2\), since the first coordinate of \(\mathbf{w}_2\) is \(0\) but that of \(\mathbf{w}_3\) is \(1\). Thus the triple is pairwise nonproportional.
Nevertheless, adding the first two vectors gives the third:
This is a nontrivial dependence relation, so the list is dependent. Check: the coordinate sum \(\mathbf{w}_1+\mathbf{w}_2=(1+0,0+1,1+1)=(1,1,2)\) is \(\mathbf{w}_3\). All three vectors lie in the plane described by \(z=x+y\), but none is a scalar multiple of either of the others.
Pairwise nonproportionality does not prove dependence either. It can occur in an independent list as well. This matters because the phrase “not enough to prove independence” should not be misread as “evidence that the list must be dependent.”
Worked Example: A Pairwise Nonproportional Independent Triple in \(\mathbb{R}^3\)
Let \(\mathbf{e}_1=(1,0,0)\), \(\mathbf{e}_2=(0,1,0)\), and \(\mathbf{e}_3=(0,0,1)\). No two are proportional: each has its \(1\) in a different coordinate, and multiplying one by a scalar cannot move that nonzero coordinate to another position.
To test the whole list, take arbitrary real coefficients \(a,b,c\):
If this equals \(\mathbf{0}\), then its first, second, and third coordinates give \(a=0\), \(b=0\), and \(c=0\), respectively. The list is independent by definition. Check: substituting those coefficients gives \(0\mathbf{e}_1+0\mathbf{e}_2+0\mathbf{e}_3=\mathbf{0}\), and the coordinate equations show there is no other choice.
What Pairwise Nonproportionality Does Guarantee
Proof. Suppose a list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is independent, and choose two distinct entries \(\mathbf{v}_i,\mathbf{v}_j\). If they were proportional, then either \(\mathbf{v}_i=c\mathbf{v}_j\) or \(\mathbf{v}_j=c\mathbf{v}_i\) for some real \(c\). In the first case, \(\mathbf{v}_i-c\mathbf{v}_j=\mathbf{0}\) gives a relation for the full list by assigning coefficient \(1\) to \(\mathbf{v}_i\), coefficient \(-c\) to \(\mathbf{v}_j\), and \(0\) to every other entry. This relation is nontrivial because the coefficient on \(\mathbf{v}_i\) is \(1\). In the second case, use coefficient \(1\) on \(\mathbf{v}_j\), coefficient \(-c\) on \(\mathbf{v}_i\), and \(0\) elsewhere; again the relation is nontrivial. Either case contradicts independence. Therefore every pair of distinct vectors must be nonproportional. \(\square\)
This result gives a necessary condition, not a complete test for a list of three or more vectors: independence forces pairwise nonproportionality, but pairwise nonproportionality alone does not force independence. The two triples above demonstrate both possibilities in \(\mathbb{R}^3\): one is dependent and one is independent, even though each has no proportional pair.
Common Mistake: Checking Only the Pairs
A tempting but incorrect argument is: “The three vectors \((1,0)\), \((0,1)\), and \((1,1)\) are not multiples of one another, so they are independent.” The pairwise checks are correct, but the conclusion does not follow. The full-list calculation gives
and the coefficients \(1,1,-1\) are not all zero. Therefore the list is dependent. The correction is to distinguish what the pairwise check establishes from what the definition asks: pairwise nonproportionality rules out a dependence using just a pair, while independence requires ruling out every nontrivial combination of the entire list.
When you are given two nonzero vectors, the proportionality test is decisive. For three or more vectors, treat pairwise comparisons as an initial check only. To establish independence, use the definition and show that a general combination equalling \(\mathbf{0}\) forces every coefficient to vanish. To establish dependence, provide a single nontrivial relation and substitute it back to verify the result.
Check Your Understanding
For each question, distinguish a check on individual pairs from a test of the full list.
- Are \((1,3)\) and \((2,-1)\) proportional? Use the two-vector theorem to decide whether they are independent.
- Show that \((1,0)\), \((0,1)\), and \((2,1)\) are pairwise nonproportional, then find a dependence relation or prove that none exists.
- For \((1,0,1)\), \((0,1,0)\), and \((1,1,1)\), check every pair and test the full list for independence.
- Give a one-sentence explanation of why pairwise nonproportionality is necessary but not sufficient for independence.
- Why does the pair test require nonzero vectors in the stated two-vector theorem?
