Equivalent Tests for a Basis

From a Basis to a Unique Representation

A basis does two jobs at once: it provides enough vectors to build every vector in a space, and it avoids redundancy among those vectors. These two requirements together have a useful consequence. Every vector in the space can be written as a linear combination of the basis vectors, and there is only one choice of coefficients that works. Conversely, if every vector has exactly one such representation, the list must both span the space and be linearly independent.

Let \(V\) be a vector space over \(\mathbb{R}\), and consider a finite ordered list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) in \(V\). A representation of \(\mathbf{x}\in V\) by this list is an equation \(\mathbf{x}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where \(c_1,\dots,c_p\in\mathbb{R}\). The coefficients are part of the representation: the question of uniqueness is whether exactly one coefficient list works, not whether the vector can be written in different-looking ways after algebraic rearrangement.

These definitions concern an ordered list, so its positions matter. For example, a repeated vector occupies two positions and can create a dependence relation. We allow \(p=0\) as well: the empty list has only the empty coefficient list, and its empty linear combination is \(\mathbf{0}\). Its span is \(\{\mathbf{0}\}\), and it is linearly independent because there are no coefficients that could violate the independence condition. This convention includes the zero vector space, which has the empty list as a basis.

Uniqueness and Independence

Before considering an entire space, focus on the vectors the list can produce. By definition, these vectors form \(\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\). The key fact is that a list is independent exactly when each vector in its span has a unique representation. Spanning the whole space is a separate requirement: it ensures that the representation property applies to every vector of \(V\).

Proof. Suppose first that the list is linearly independent. Let \(\mathbf{x}\) be in its span. By the definition of span, at least one representation exists; write two possible representations as \(\mathbf{x}=\sum_{i=1}^{p}a_i\mathbf{v}_i\) and \(\mathbf{x}=\sum_{i=1}^{p}b_i\mathbf{v}_i\). Subtracting the second equation from the first gives \(\sum_{i=1}^{p}(a_i-b_i)\mathbf{v}_i=\mathbf{0}\). Independence forces \(a_i-b_i=0\) for every \(i\), so \(a_i=b_i\) for every \(i\). Thus the representation exists and is unique. Conversely, suppose every vector in the span has exactly one representation. If \(\sum_{i=1}^{p}c_i\mathbf{v}_i=\mathbf{0}\), then the zero vector has both this representation and the representation with every coefficient zero. Uniqueness says \(c_i=0\) for every \(i\), which is independence. For the empty list, both statements hold under the conventions above: its span consists only of zero, with its one empty representation. \(\square\)

The proof shows why both parts of “exactly one” matter. Spanning supplies existence of a representation; independence supplies uniqueness. If a list is independent but does not span \(V\), uniqueness may hold for every vector it can produce while some vectors in \(V\) have no representation at all.

The Equivalent Test for a Basis

Proof. Suppose the list is a basis. Since it spans \(V\), every \(\mathbf{x}\in V\) has at least one representation. Since it is independent, the preceding theorem shows that this representation is unique. Conversely, suppose every \(\mathbf{x}\in V\) has exactly one representation. The existence part says every vector in \(V\) is in the span, so the list spans \(V\). To prove independence, suppose \(\sum_{i=1}^{p}c_i\mathbf{v}_i=\mathbf{0}\). The zero vector also has the representation \(\sum_{i=1}^{p}0\mathbf{v}_i\). By uniqueness, \(c_i=0\) for every \(i\). Hence the list is independent as well as spanning, and therefore is a basis. If \(p=0\), the condition that every vector has a representation holds precisely when \(V=\{\mathbf{0}\}\); in that case the empty list spans and is independent, so it is a basis. \(\square\)

In practice, this test can be split into two questions: can every vector be represented, and can two different coefficient lists represent the same vector? For the second question, subtracting two representations turns the issue into a homogeneous linear combination equal to zero. This is why independence is the exact condition that rules out competing coefficient lists.

Worked Applications

Common Mistake: Checking Only Uniqueness

A common error is to conclude that an independent list is a basis just because its representations are unique. For instance, the single vector \(\mathbf{u}=\begin{bmatrix}1\\0\end{bmatrix}\) is independent: if \(c\mathbf{u}=\mathbf{0}\), then \(\begin{bmatrix}c\\0\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\), so \(c=0\). Every vector in its span has a unique representation \(c\mathbf{u}\). A wrong conclusion would be: “Therefore \(\mathbf{u}\) is a basis of \(\mathbb{R}^2\).” The correction is to test spanning too: \(\begin{bmatrix}0\\1\end{bmatrix}\) cannot equal \(c\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}c\\0\end{bmatrix}\) for any \(c\). Thus the list is not a basis of \(\mathbb{R}^2\), even though representations within its span are unique.

When applying the criterion, keep the two obligations visible:

• Existence: every vector in the space under consideration can be represented. This is the spanning requirement.
• Uniqueness: if two coefficient lists represent the same vector, their coefficients agree in every position. This is the independence requirement.

The same list can be a basis of one space and fail to be a basis of a larger space. The first worked example uses its vectors as a basis of a particular subspace of \(\mathbb{R}^3\); the third example shows how an independent list can fall short when the space under consideration is larger. Always identify the space before deciding whether the list spans it.