Lists, the Zero Vector, and the Empty Case
Some definitions that work smoothly for lists of ordinary positive length also have a meaningful case when the list has no entries. That case matters for the zero vector space: the space \(\{\mathbf{0}\}\) has a basis even though there is no nonzero vector available to put in it. At the other extreme, placing the zero vector into a list always creates a dependence relation. These two facts follow directly from the definitions once we are careful about what an empty list and an empty sum mean.
Throughout, let \(V\) be a vector space over a field \(\mathbb{F}\), with \(\mathbb{F}=\mathbb{R}\) unless stated otherwise. A list records entries in a particular order and can contain repeated entries. The empty list, written \(()\), has length zero. It is not the zero vector: \(\mathbf{0}\) is an element of \(V\), while \(()\) has no entries.
For a list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\), a linear combination is an expression \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where each \(c_i\in\mathbb{F}\). When \(p=0\), there are no terms to add. By convention, the empty sum is \(\mathbf{0}\), the additive identity of \(V\). Consequently, the only linear combination of the empty list is \(\mathbf{0}\).
For the empty list, its span is therefore the singleton set \(\{\mathbf{0}\}\), not the empty set. Its independence condition is satisfied: there is one coefficient tuple of length zero, the empty tuple, and it has no nonzero entries. There is no nontrivial coefficient choice because there are no coefficients at all. The spanning and independence parts of the definition thus behave differently: the empty list always satisfies the independence condition, but it spans only the zero vector.
The Empty List as a Basis
Proof. Suppose first that the empty list is a basis of \(V\). A basis must span \(V\). The only linear combination of the empty list is the empty sum \(\mathbf{0}\), so its span is \(\{\mathbf{0}\}\). It follows that \(V=\{\mathbf{0}\}\). Conversely, suppose \(V=\{\mathbf{0}\}\). The span of the empty list is \(\{\mathbf{0}\}=V\), so the list spans \(V\). It is also linearly independent, since its unique coefficient tuple has no nonzero entries. Thus the empty list is a basis of \(V\). \(\square\)
This argument identifies the precise role of each convention. The empty-sum convention makes the empty list span the zero vector, and the fact that there are no coefficients makes it independent. If \(V\) contains any vector besides \(\mathbf{0}\), spanning fails; if \(V\) contains only \(\mathbf{0}\), both requirements hold. In particular, the zero vector space is not the empty set: it has one vector, namely \(\mathbf{0}\), and the empty list spans that vector through the empty sum.
Worked Example: The Empty Basis of the Zero Space
Let \(V=\{\mathbf{0}\}\). Verify directly that \(()\) is a basis of \(V\).
The list has length zero, so it has no entries and no coefficients. Its only linear combination is the empty sum, which equals \(\mathbf{0}\). Therefore its span is \(\{\mathbf{0}\}=V\). To check independence, there is only one coefficient tuple of length zero: \(()\). It contains no nonzero coefficient, so the condition that only the all-zero coefficient choice produces \(\mathbf{0}\) is satisfied. The list is both spanning and independent, hence it is a basis.
Check: the set of combinations is exactly \(\{\mathbf{0}\}\), and that is exactly \(V\). There is no coefficient in the list that could form a nontrivial relation.
Worked Example: An Empty List in a Nonzero Space
Consider the empty list in \(\mathbb{R}^2\). Does it form a basis of \(\mathbb{R}^2\)?
Its only linear combination is \(\mathbf{0}=(0,0)\), so its span is \(\{(0,0)\}\). But \((1,0)\in\mathbb{R}^2\), and \((1,0)\ne(0,0)\). Thus \((1,0)\) is not in the span of the empty list. The list does not span \(\mathbb{R}^2\), so it is not a basis.
Check: no coefficients are available to produce \((1,0)\); the only permitted combination is the empty sum \((0,0)\). The failure is exactly the spanning requirement.
Why a Zero Entry Forces Dependence
Proof. Write the list as \((\mathbf{v}_1,\dots,\mathbf{v}_p)\), and suppose \(\mathbf{v}_k=\mathbf{0}\) for some index \(k\). Choose \(c_k=1\) and choose \(c_i=0\) for every \(i\ne k\). Then at least one coefficient is nonzero, because \(1\ne0\) in a field. The corresponding linear combination is \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=1\cdot\mathbf{v}_k=\mathbf{0}\). This is a nontrivial relation, so the list is linearly dependent. \(\square\)
The index \(k\) can be anywhere in the list, and other entries may be zero too. The proof uses just one zero entry: set its coefficient to \(1\) and every other coefficient to \(0\). This is why the statement is about lists, not just sets. If a list contains the zero vector twice, either position supplies a nontrivial relation. The entries’ order and any repetitions do not affect this argument.
Worked Example: A Zero Vector in the First Position
Consider the list \(\bigl((0,0),(2,-1)\bigr)\) in \(\mathbb{R}^2\). Exhibit and check a dependence relation.
Choose coefficients \(c_1=1\) and \(c_2=0\). Then \(c_1(0,0)+c_2(2,-1)=1(0,0)+0(2,-1)=(0,0)+(0,0)=(0,0)\). The coefficient list \((1,0)\) is not all zero, so this is a dependence relation.
In coordinates, the two components are \(1\cdot0+0\cdot2=0\) and \(1\cdot0+0\cdot(-1)=0\). Check: both components of the combination are zero, and the coefficient on the zero vector is nonzero. The list is dependent.
Including a zero entry does not add any new vectors to the span. This is another direct consequence of the empty-sum and linear-combination definitions, and it helps explain why a zero entry can affect independence while leaving the span unchanged.
Proof. Let the list be \((\mathbf{v}_1,\dots,\mathbf{v}_p)\), with \(\mathbf{v}_k=\mathbf{0}\), and compare it with the list formed by deleting that entry. Any linear combination of the shorter list is also a linear combination of the original list: assign coefficient \(0\) to the deleted entry. Thus every vector in the shorter list’s span is in the original list’s span. In the other direction, take any linear combination of the original list. Its term involving the deleted entry is \(c_k\mathbf{v}_k=c_k\mathbf{0}=\mathbf{0}\). Removing that term leaves the same vector, expressed as a linear combination of the shorter list. The two spans therefore contain each other and are equal. \(\square\)
Worked Example: A Zero Vector in the Middle
In \(\mathbb{R}^2\), consider \(\bigl((2,-1),(0,0),(1,3)\bigr)\). Verify both that the list is dependent and that its span is unchanged when the middle entry is deleted.
For dependence, choose coefficients \(c_1=0\), \(c_2=1\), \(c_3=0\). The combination is \(0(2,-1)+1(0,0)+0(1,3)=(0,0)+(0,0)+(0,0)=(0,0)\). The coefficient \(c_2=1\) is nonzero, so the relation is nontrivial.
For the span calculation, let \(a,b,c\in\mathbb{R}\). A general combination of the three entries is \(a(2,-1)+b(0,0)+c(1,3)=(2a+c,-a+3c)\). Deleting the middle entry gives \(a(2,-1)+c(1,3)=(2a+c,-a+3c)\), the same expression for every \(a,c\). Conversely, every combination after deletion is obtained in the original list by choosing \(b=0\). Check: the zero entry contributes \((0,0)\) for every coefficient \(b\), and the explicit nontrivial relation confirms dependence.
Common Mistakes
Mistake: treating the empty sum as undefined or as an empty set. A student might claim that the empty list has empty span because it has no entries. The correction is to apply the definition: its combinations consist of the empty sum, and the empty sum equals \(\mathbf{0}\). Thus the span is the singleton set \(\{\mathbf{0}\}\). It is not \(\varnothing\).
Mistake: requiring the coefficient on a zero vector to be zero. For the list \(((0,0),(2,-1))\), a mistaken argument says the first coefficient must be zero because the first vector is zero. But choosing \(c_1=1\) and \(c_2=0\) gives \(1(0,0)+0(2,-1)=(0,0)\), with a nonzero coefficient. The definition permits this choice, so the list is dependent.
Mistake: confusing the zero vector space with the empty list. The space \(\{\mathbf{0}\}\) contains one vector; the list \(()\) contains no entries. The empty list is a basis of \(\{\mathbf{0}\}\) because its only combination is that vector. It is not a basis of a nonzero space, because it cannot produce any nonzero vector.
Check Your Understanding
For each question, use the definitions of span, independence, and basis, including the empty-sum convention.
- What is the span of the empty list in a vector space \(V\), and why?
- Prove that the empty list is not a basis of \(\mathbb{R}\).
- Give a nontrivial dependence relation for the list \(\bigl((1,0),(0,0),(0,1)\bigr)\) in \(\mathbb{R}^2\).
- Does inserting a zero vector into a finite list change its span? Justify your answer.
- Explain why a coefficient equal to \(1\) on a zero entry proves dependence, even though its term in the sum is zero.
