Two Different Questions
A list of vectors can have no redundancy and still fail to produce every vector in its space. Another list can produce every vector but include unnecessary vectors. These are different failures: the first concerns spanning, and the second concerns independence. Since a basis must satisfy both requirements, checking only one cannot establish that a list is a basis.
Throughout, the scalar field is \(\mathbb{R}\). 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 spanning for a vector space \(V\) if every \(\mathbf{v}\in V\) can be written as \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\) for some real coefficients. The list is a basis for \(V\) if it is both linearly independent and spanning. Independence asks whether a combination can equal zero in a nontrivial way; spanning asks whether every vector can be reached by a combination.
The difference is easiest to see by testing each condition directly. For independence, start with an arbitrary linear combination equal to \(\mathbf{0}\) and determine whether all its coefficients must vanish. For spanning, take an arbitrary vector in the space and determine whether it can be expressed as a combination. A failed spanning test requires a vector that cannot be reached; a failed independence test requires a nontrivial dependence relation.
Independent but Not Spanning
Worked Example: Two Independent Vectors in \(\mathbb{R}^3\)
Consider the list \(\mathbf{e}_1=(1,0,0)\), \(\mathbf{e}_2=(0,1,0)\) in \(\mathbb{R}^3\). First test independence. Suppose \(c_1\mathbf{e}_1+c_2\mathbf{e}_2=\mathbf{0}\). Comparing coordinates gives \((c_1,c_2,0)=(0,0,0)\), so \(c_1=0\) and \(c_2=0\). The list is independent.
Now test spanning. Every combination of these vectors has the form \(c_1\mathbf{e}_1+c_2\mathbf{e}_2=(c_1,c_2,0)\). The vector \((0,0,1)\in\mathbb{R}^3\) has third coordinate \(1\), so it is not of this form. The list does not span \(\mathbb{R}^3\), and therefore is not a basis.
Check: the list has no nontrivial dependence relation, but the third standard basis vector cannot be made from it. This verifies both the independence claim and the failure to span.
This example illustrates why “the vectors are independent” does not imply “the vectors are a basis.” Independence rules out redundancy within the list; it does not guarantee that the list contains enough directions to reach the whole space. The missing vector in this example is explicit, so the failure to span does not depend on a visual judgment about the vectors.
Spanning but Dependent
Worked Example: A Redundant Spanning List in \(\mathbb{R}^2\)
Consider \(\mathbf{v}_1=(1,0)\), \(\mathbf{v}_2=(0,1)\), and \(\mathbf{v}_3=(1,1)\) in \(\mathbb{R}^2\). To check spanning, take an arbitrary vector \((a,b)\in\mathbb{R}^2\). It has the representation \((a,b)=a\mathbf{v}_1+b\mathbf{v}_2+0\mathbf{v}_3\). Since this works for every \(a,b\in\mathbb{R}\), the list spans \(\mathbb{R}^2\).
To check independence, test the coefficients \(1,1,-1\):
These coefficients are not all zero, so the list is dependent. It spans, but it is not a basis.
Check: the arbitrary-vector representation verifies spanning, while the displayed nontrivial relation verifies dependence. In fact, \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\), so the third vector adds no new combinations.
This is the opposite failure from the first example: the list reaches every vector, but one of its members is redundant. “The vectors span” does not imply “the vectors are a basis.” A dependence relation is enough to disprove independence; it is not necessary to solve for all possible relations.
The Two Requirements in Another Space
Worked Example: Polynomials of Degree at Most Two
Let \(P_2\) be the real vector space of polynomials of degree at most two, including the zero polynomial. Each member has the form \(a+bx+cx^2\) for real \(a,b,c\). The list \(1,x,x^2,1+x\) spans \(P_2\): for any \(a+bx+cx^2\in P_2\),
However, the list is dependent, because \(1+x\) is a combination of its first two members. Explicitly, \(1\cdot 1+1\cdot x-1\cdot(1+x)=0\), with coefficients \(1,1,-1\), not all zero. Thus this list spans \(P_2\) but is not a basis.
For contrast, \(1,x\) is independent: if \(c_1\cdot1+c_2x\) is the zero polynomial, its constant coefficient \(c_1\) and its coefficient of \(x\), \(c_2\), must both be zero. Yet it does not span \(P_2\), because no combination \(c_1\cdot1+c_2x\) equals \(x^2\). The coefficient of \(x^2\) in every such combination is zero, whereas it is one in \(x^2\).
Check: the first list has an explicit representation for every polynomial in \(P_2\) and an explicit nontrivial dependence relation. The shorter list has only the trivial zero combination but cannot represent \(x^2\).
The polynomial example emphasizes that the distinction is about the list and its ambient space, not about a particular way of drawing vectors in the plane or in three-dimensional space. The same pair of tests applies in any vector space: verify whether the only zero combination is the trivial one, and verify whether arbitrary vectors in the space have representations.
What Dimension and List Size Can Tell You
Suppose \(V\) is a finite-dimensional real vector space with \(\dim V=n\). The dimension theorem established in the earlier material on basis and dimension says that every basis of \(V\) has exactly \(n\) members. The Basis Extension Theorem says that any linearly independent list in a finite-dimensional space can be extended to a basis. The Spanning Set Reduction Theorem says that a finite spanning list can be reduced, by deleting redundant members, to a basis consisting of members of the original list. These results give useful size checks without replacing the two defining tests.
Proof. Let the independent list have \(k<n\) members. By the Basis Extension Theorem, it can be extended to a basis of \(V\). That basis has \(n\) members, so at least one vector was added. If the original list already spanned \(V\), every added vector would be a combination of the original list; moving that combination to one side would give a nontrivial relation among the extended list, contradicting that it is a basis. Thus the original list does not span. Now let a spanning list have \(k>n\) members. By the Spanning Set Reduction Theorem, some members can be deleted to leave a basis of \(V\), and that basis has \(n\) members. Since \(k>n\), at least one original member was deleted. A deleted member is a combination of the remaining members, which gives a nontrivial dependence relation in the original list. Therefore the original list is dependent. \(\square\)
The boundary case of exactly \(n\) members is especially useful: either one of the two properties is enough to guarantee the other. This result does not say that a list with any other size cannot be a basis by a special construction; rather, it follows from the definition of basis and the fact that a basis of \(V\) has \(n\) members.
Proof. Suppose first that the list is independent. The Basis Extension Theorem extends it to a basis. Since every basis has \(n\) members and the given list already has \(n\), the extension adds no members; hence the given list itself is a basis and spans \(V\). Suppose instead that the list spans \(V\). The Spanning Set Reduction Theorem reduces it to a basis chosen from its members. That basis has \(n\) members, equal to the number in the original list, so no member can have been deleted. The original list is therefore itself a basis, and is independent. \(\square\)
For a list of a size other than \(n\), do not conclude too much from counting alone. Fewer than \(n\) vectors cannot span if they are independent; more than \(n\) vectors cannot be independent if they span. But a short list might be dependent, and a long list might fail to span. The size result applies only together with the stated property.
Common Pitfall: Treating One Test as Both
Wrong calculation: “The vectors \((1,0,0)\) and \((0,1,0)\) are independent in \(\mathbb{R}^3\), so they form a basis of \(\mathbb{R}^3\).” The independence calculation is correct, but the conclusion does not follow. Every combination is \((c_1,c_2,0)\), so \((0,0,1)\) cannot be represented. The correction is to test spanning separately: this list is independent but not spanning, and so it is not a basis.
A related error is to say that a list must be independent because it spans. In the \(\mathbb{R}^2\) example above, every \((a,b)\) is represented by \(a\mathbf{v}_1+b\mathbf{v}_2\), but the third vector satisfies \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\). The correction is to search for a nontrivial zero combination as a separate independence test. A representation of every vector does not rule out redundancy.
When a basis claim is made, record the two questions separately:
- Independence: does \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) force every \(c_i=0\)?
- Spanning: can an arbitrary vector in the stated space be expressed as a combination of the list?
A valid basis argument must answer both questions, unless a result with appropriate hypotheses—such as the exact-size theorem above—allows one to infer the second condition from the first, or the first from the second.
Check Your Understanding
For each question, distinguish the independence test from the spanning test and state the ambient space.
- Show that \((1,0)\), \((0,1)\), and \((1,1)\) span \(\mathbb{R}^2\), then give a nontrivial dependence relation.
- In \(\mathbb{R}^3\), explain why \((1,0,0)\), \((0,1,0)\) are independent but do not span.
- In \(P_2\), is the list \(1,x,x^2\) independent, spanning, both, or neither? Verify each applicable claim.
- If \(\dim V=4\), what can you conclude about a spanning list of five vectors? State which finite-dimensional result justifies the conclusion.
- If \(\dim V=3\), explain why an independent list of three vectors must span \(V\).
