Dimension Counts the Vectors in a Basis
A basis gives a finite list of vectors that is both independent and sufficient to represent every vector in a space. Its length is therefore a natural count of the independent directions available in that space. To turn this count into a definition, however, we need to know that choosing a different basis cannot give a different length. The Invariance of Dimension Theorem provides exactly this guarantee.
Throughout, the scalar field \(\mathbb{F}\) may be \(\mathbb{R}\) or \(\mathbb{C}\). A list \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) in a vector space \(V\) is a basis if it is linearly independent and spans \(V\). The list has length \(n\). The space \(V\) is finite-dimensional if it has a basis with finite length.
Dimension is a property of the space, not a property of one particular list or one particular way of writing vectors. For example, a space might have many different bases, but the Invariance of Dimension Theorem says their lengths agree. This also means that once a basis of length \(n\) has been found, the dimension is determined: there is no need to compare it with every other possible basis.
The Empty List and the Zero Space
The zero vector space is the vector space whose only element is the zero vector, denoted \(\{\mathbf{0}\}\). It has one vector as an element. A list of basis vectors is a different kind of object: a list of length zero, called the empty list, has no entries. This distinction matters. The zero vector space is not an empty set; it contains \(\mathbf{0}\).
To use the empty list, we adopt the standard convention that the linear combination of no vectors is \(\mathbf{0}\). Consequently, the span of the empty list is \(\{\mathbf{0}\}\). Linear independence of a list means that the only linear combination equal to \(\mathbf{0}\) has all its coefficients equal to zero. For the empty list, there are no coefficients to choose, so there is no nonzero-coefficient relation. In this precise sense, the empty list is linearly independent.
Proof. Suppose first that the empty list is a basis of \(V\). A basis spans \(V\), and the span of the empty list is \(\{\mathbf{0}\}\) by the convention above. Therefore \(V=\{\mathbf{0}\}\). Conversely, suppose \(V=\{\mathbf{0}\}\). The span of the empty list is \(\{\mathbf{0}\}\), so it spans \(V\). The empty list is linearly independent because there is no choice of coefficients, not all zero, that could form a dependence relation. It is therefore a basis of \(V\). \(\square\)
This theorem makes the dimension-zero case precise: a basis of length zero must be the empty list, and that list spans only the zero vector space. We can state the resulting characterization for all finite-dimensional spaces.
Proof. Suppose \(\dim V=0\). By the definition of dimension, a basis of \(V\) has length zero, so it is the empty list. The theorem just proved says that a space with the empty list as a basis is \(\{\mathbf{0}\}\). Hence \(V=\{\mathbf{0}\}\). Conversely, suppose \(V=\{\mathbf{0}\}\). The empty list is a basis by the preceding theorem. It has length zero, so the definition of dimension gives \(\dim V=0\). Both implications hold, as required. \(\square\)
Worked Examples: Applying the Characterization
Worked Example: The Zero Vector Space Inside \(\mathbb{R}^3\)
Consider \(V=\{(0,0,0)^{\top}\}\), viewed as a vector space over \(\mathbb{R}\). Find a basis and its dimension.
The only vector in \(V\) is \(\mathbf{0}\), so \(V=\{\mathbf{0}\}\). The empty list spans \(\{\mathbf{0}\}\), since its only linear combination is the empty linear combination \(\mathbf{0}\). It is independent because it has no coefficients that could form a nontrivial relation. Thus the empty list is a basis of \(V\), and its length is zero.
Check: The basis has zero entries, and its span is exactly \(\{(0,0,0)^{\top}\}\), the given space. The dimension-zero characterization also applies because \(V=\{\mathbf{0}\}\).
Worked Example: A Nonzero Space with Two Basis Vectors
Consider \(V=\mathbb{R}^2\) over \(\mathbb{R}\), with the list \((\mathbf{e}_1,\mathbf{e}_2)\), where \(\mathbf{e}_1=(1,0)^{\top}\) and \(\mathbf{e}_2=(0,1)^{\top}\). Verify that this is a basis and determine the dimension.
For any \(\mathbf{x}=(a,b)^{\top}\in\mathbb{R}^2\), its coordinates give the representation \(\mathbf{x}=a\mathbf{e}_1+b\mathbf{e}_2\). Hence the list spans \(\mathbb{R}^2\). To check independence, suppose \(c_1\mathbf{e}_1+c_2\mathbf{e}_2=\mathbf{0}\). Computing the coordinates gives \((c_1,c_2)^{\top}=(0,0)^{\top}\), so \(c_1=c_2=0\). The list is independent as well as spanning, so it is a basis.
Check: For example, \((3,-2)^{\top}=3\mathbf{e}_1-2\mathbf{e}_2=(3,-2)^{\top}\). Also \(\mathbb{R}^2\ne\{\mathbf{0}\}\), since \(\mathbf{e}_1\ne\mathbf{0}\); accordingly, the dimension-zero characterization rules out dimension zero, in agreement with the basis length \(2\).
Worked Example: The Space of Polynomials of Degree at Most One
Let \(P_1(\mathbb{R})\) be the real vector space of polynomials \(a+bx\), where \(a,b\in\mathbb{R}\). Determine its dimension using the list \((1,x)\).
Every polynomial in \(P_1(\mathbb{R})\) has the form \(a+bx=a\cdot1+b\cdot x\), so \((1,x)\) spans the space. For independence, suppose \(c_1\cdot1+c_2\cdot x\) is the zero polynomial. The zero polynomial has constant coefficient \(0\) and coefficient of \(x\) equal to \(0\); comparing coefficients gives \(c_1=0\) and \(c_2=0\). Therefore the list is independent and is a basis.
Check: For \(p(x)=4-3x\), the basis representation gives \(4\cdot1-3\cdot x=4-3x\). Also \(1\) is a nonzero polynomial, so this space is not \(\{\mathbf{0}\}\). Its dimension is \(2\), consistent with the characterization.
Do Not Confuse the Zero Vector with the Zero Space
The zero vector \(\mathbf{0}\) is an element of every vector space. The zero vector space \(\{\mathbf{0}\}\) is a particular space with no other elements. For instance, \(\mathbb{R}^2\) contains its zero vector \((0,0)^{\top}\), but also contains \((1,0)^{\top}\), so \(\mathbb{R}^2\ne\{\mathbf{0}\}\). Its dimension is \(2\), not \(0\).
A related distinction is that the empty list and a list containing the zero vector are not the same. The empty list is a basis of \(\{\mathbf{0}\}\). The singleton list \((\mathbf{0})\) is not independent, because the coefficient \(1\) gives the nontrivial relation \(1\cdot\mathbf{0}=\mathbf{0}\). So the singleton list cannot be used as a basis, even though it contains the only vector in the zero space.
Common mistake. A student might write, “The zero space has the basis \((\mathbf{0})\), which contains one vector, so its dimension is \(1\).” The proposed basis fails the independence condition: \(1\cdot\mathbf{0}=\mathbf{0}\) uses a nonzero coefficient. The correction is to use the empty list. It spans \(\{\mathbf{0}\}\) and is independent, so the dimension is \(0\).
Why the Zero-Dimension Case Is Useful
The characterization provides a quick interpretation of a dimension calculation. A finite-dimensional space has dimension zero exactly when there are no nonzero vectors in it. Conversely, finding even one nonzero vector proves that a finite-dimensional space cannot have dimension zero. This can be useful when the space is described by a rule or by a list of vectors: before searching for a basis of positive length, check whether the space contains anything beyond \(\mathbf{0}\).
Dimension zero is not a special exception to the definition of dimension. It follows from the same rule as every other finite dimension: count the vectors in a basis. What is special is that the basis has no entries. Remembering the distinction between an empty list, the zero vector, and the zero vector space prevents many errors in the smallest case.
Check Your Understanding
Use the definition of a basis and the dimension-zero characterization to answer the following.
- What is the length of the empty list, and what space does it span?
- Explain why the list \((\mathbf{0})\) is not a basis of the zero vector space.
- A finite-dimensional vector space \(V\) contains a vector \(\mathbf{v}\ne\mathbf{0}\). What can you conclude about \(\dim V\)?
- Show that the list \((1)\) is a basis of the real vector space of constant polynomials, and state its dimension.
- Can a vector space with dimension zero contain two distinct vectors? Explain using the theorem.
