Counterexamples to Independence and Spanning Claims

Two Different Questions About a List

Independence and spanning ask different questions. Independence asks whether the list contains redundancy: can a nonzero choice of coefficients combine its entries to give the zero vector? Spanning asks whether every vector in the space can be expressed using the list. A list can be dependent and still span, or independent and still fail to span. Counterexamples using only a few coordinate vectors make these distinctions concrete.

Here, all scalars are real, and the spaces considered are \(\mathbb{R}^2\) or \(\mathbb{R}^3\). We use lists of vectors: unlike a set, a list can contain the same vector more than once. The standard coordinate vectors are \(\mathbf{e}_1=(1,0)\), \(\mathbf{e}_2=(0,1)\) in \(\mathbb{R}^2\); in \(\mathbb{R}^3\), they are \(\mathbf{e}_1=(1,0,0)\), \(\mathbf{e}_2=(0,1,0)\), and \(\mathbf{e}_3=(0,0,1)\).

To refute a universal claim, one valid example where the claim fails is enough. To verify a counterexample about spanning, it is useful to show either how an arbitrary vector can be written as a combination or to name a vector that cannot be. To verify dependence, exhibit a nontrivial zero-combination equation; to verify independence, start with an arbitrary zero-combination equation and show its coefficients must all vanish.

A Repeated Entry Can Leave Spanning Intact

The same reasoning applies to any repeated entry, not only a coordinate vector.

Proof. Let the repeated entries be \(\mathbf{v}_i=\mathbf{v}_j\) with \(i\ne j\). Choose coefficient \(1\) in position \(i\), coefficient \(-1\) in position \(j\), and coefficient \(0\) in every other position. These coefficients are not all zero, and their combination is \(\mathbf{v}_i-\mathbf{v}_j=\mathbf{0}\). This is a nontrivial dependence relation, so the list is dependent. \(\square\)

When using this proposition, keep the list and its entries in view. A set does not record repeated elements, but a list does: the dependence relation uses the two distinct positions. Also, the proposition gives no conclusion that the list fails to span. The worked example shows that a repeated entry can be redundant while the other entries still span the space.

An Independent List Can Miss Vectors

There is no conflict between these conclusions. Independence restricts which combinations can equal zero; spanning asks whether the combinations reach every target vector. The first condition can hold while the second fails, as it does here.

Repeated and Zero Vectors: What They Do and Do Not Prove

A zero vector also forces dependence, but does not by itself determine whether the list spans. The reason is the same distinction: the zero entry creates redundancy, while the other entries determine which vectors can be reached.

Proof. Suppose the zero vector occurs in position \(k\). Choose coefficient \(1\) in position \(k\) and \(0\) in every other position. This is a nontrivial choice of coefficients, but the resulting combination is \(1\mathbf{0}=\mathbf{0}\). By the definition of dependence, the list is dependent. \(\square\)

For contrast, \(\mathbf{0},\mathbf{e}_1\) in \(\mathbb{R}^2\) is also dependent, but its combinations have the form \(c\mathbf{e}_1=(c,0)\). It does not span, since \(\mathbf{e}_2=(0,1)\) is not of that form. The zero entry establishes dependence in both lists; the other entries distinguish their spanning behavior.

Proof. Let \(S\) be the span of a list and suppose \(\mathbf{w}\in S\). Every combination of the original list is still a combination of the enlarged list, by assigning coefficient \(0\) to \(\mathbf{w}\). Hence \(S\) is contained in the enlarged span. Conversely, any combination of the enlarged list has the form \(\mathbf{u}+c\mathbf{w}\), where \(\mathbf{u}\in S\) and \(c\in\mathbb{R}\). Since \(\mathbf{w}\in S\), we have \(c\mathbf{w}\in S\), and since \(S\) is a span, it is closed under addition. Thus \(\mathbf{u}+c\mathbf{w}\in S\). The enlarged span is contained in \(S\), so the two spans are equal. \(\square\)

A repeated entry is already in the span of the list, and the zero vector is also in every span. The proposition explains why appending either kind of entry cannot enlarge the span. It does not say the original list spans the whole ambient space; it only says the span remains unchanged.

How to Test a Claim Before Generalizing

When a statement uses words such as “every” or “always,” first identify exactly what it predicts. A claim that independence guarantees spanning is refuted by one independent list and one missing vector. A claim that spanning guarantees independence is refuted by one spanning list with a nontrivial dependence relation. For each proposed example, verify both properties directly rather than inferring one from the other.

Coordinate vectors are especially useful because their entries make both checks visible. For instance, in \(\mathbb{R}^2\), \(\mathbf{e}_1,\mathbf{e}_2\) span because \((a,b)=a\mathbf{e}_1+b\mathbf{e}_2\), and they are independent because \(c_1\mathbf{e}_1+c_2\mathbf{e}_2=(0,0)\) forces \(c_1=c_2=0\). Adding a repeated or zero entry makes the list dependent while leaving its span unchanged. Omitting a coordinate direction, as with \(\mathbf{e}_1,\mathbf{e}_2\) in \(\mathbb{R}^3\), leaves an explicit vector outside the span.

Common Pitfall: Treating Dependence as Failure to Span

A mistaken argument might say: “The list \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_1\) is dependent, so it cannot span \(\mathbb{R}^2\).” The dependence calculation is correct, but the conclusion does not follow. Directly, for every \((a,b)\in\mathbb{R}^2\),

So the list does span. The correction is to test spanning by representing an arbitrary vector, or by finding a specific vector that cannot be represented. A dependence relation answers a different question: whether some nonzero coefficients produce zero. It cannot, by itself, decide whether the list reaches every vector.

The reverse shortcut is also invalid: an independent list need not span. The list \(\mathbf{e}_1,\mathbf{e}_2\) in \(\mathbb{R}^3\) is independent, but every combination has third coordinate zero, so it misses \(\mathbf{e}_3\). For a reliable conclusion, verify the property named in the claim rather than substituting a different property.