Choose the Method That Matches the Information
Finding a basis is often less work when the calculation begins with the right question. If a spanning list has an obvious redundant vector, a direct dependence relation may be enough. If several vectors are given as coordinates and their relationships are not apparent, a coefficient matrix organizes the equations. If the list is already independent and the goal is to complete it, test candidate vectors against its span. These methods are connected, but they start from different information.
Throughout, scalars are real. The span of a list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) in \(\mathbb{R}^n\) is the set of all linear combinations \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). A list is independent if the only combination equal to \(\mathbf{0}\) has all coefficients zero. A basis of a space is an ordered list that is both independent and spanning. To extract a basis, it is not enough to find a list with the right number of vectors: both properties must be checked.
The word “shortest” concerns the route through the calculation, not a guarantee about the number of arithmetic operations. A visible relation can save a full row reduction; a coefficient matrix can make several relationship tests systematic; and an extension test avoids discarding vectors from a list that is already independent.
When Direct Dependence Is the Short Route
Suppose a list is known to span a space. If one vector is a combination of the others, deleting it does not change the span. The remaining list may still contain other redundancies, so repeat the check as needed. Once the list is independent, it is a basis for the same space.
Proof. Let \(S\) be the span of the full list and \(S'\) the span after \(\mathbf{v}_k\) is deleted. Every vector in \(S'\) is a combination of vectors from the full list, so \(S'\subseteq S\). By assumption, \(\mathbf{v}_k\) itself is a combination of the other vectors. Therefore any combination of the full list can have its term involving \(\mathbf{v}_k\) replaced by a combination of the remaining vectors. Every vector in \(S\) is consequently in \(S'\), so \(S\subseteq S'\). Thus \(S=S'\). \(\square\)
Worked Example: Remove a Visible Redundancy
Consider the spanning list in \(\mathbb{R}^3\) \(\mathbf{v}_1=(1,0,1)\), \(\mathbf{v}_2=(0,1,1)\), \(\mathbf{v}_3=(1,1,2)\), and \(\mathbf{v}_4=(1,0,0)\). Find a basis for \(\mathbb{R}^3\) from this list.
First compare the coordinates of \(\mathbf{v}_3\) with the first two vectors:
Thus \(\mathbf{v}_3\) is redundant. By the proposition, deleting it preserves the span. Check that the remaining three vectors span \(\mathbb{R}^3\): for an arbitrary \((x,y,z)\),
They are also independent. If \(a\mathbf{v}_1+b\mathbf{v}_2+c\mathbf{v}_4=\mathbf{0}\), then comparing coordinates gives \(a+c=0\), \(b=0\), and \(a=0\). Hence \(a=0\), \(b=0\), and \(c=0\). The list \((\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_4)\) is therefore a basis.
Check: the spanning formula uses coefficients \(z,y,x-z\), and its coordinates are \((z+x-z,\;y,\;z)=(x,y,z)\), as required.
When a Coefficient Matrix Organizes the Work
To test whether a vector is in the span of a given list, put the given vectors into the columns of a matrix. If \(\mathbf{A}=[\mathbf{a}_1\ \cdots\ \mathbf{a}_q]\), then the equation \(\mathbf{A}\mathbf{c}=\mathbf{b}\) says exactly that \(\mathbf{b}\) is the combination \(c_1\mathbf{a}_1+\cdots+c_q\mathbf{a}_q\). Thus the system is consistent if and only if \(\mathbf{b}\) belongs to the span of those columns. When testing several generators, place them as several right-hand-side columns in an augmented matrix and solve the corresponding systems.
Worked Example: Extract a Basis from Generators
Let \(\mathbf{g}_1=(1,1,0)\), \(\mathbf{g}_2=(2,1,1)\), \(\mathbf{g}_3=(3,2,1)\), and \(\mathbf{g}_4=(0,1,-1)\). Find a basis for the span of these four generators. Use \(\mathbf{g}_1,\mathbf{g}_2\) as candidate generators and test the other two by a coefficient matrix.
The coefficient matrix with the two candidate columns and the two targets is
Subtract row 1 from row 2, then multiply the new row 2 by \(-1\):
Subtract row 2 from row 1 and row 3:
The first target column now gives coefficients \(1,1\), so \(\mathbf{g}_3=\mathbf{g}_1+\mathbf{g}_2\). The second gives coefficients \(2,-1\), so \(\mathbf{g}_4=2\mathbf{g}_1-\mathbf{g}_2\). These relations show that the four generators have the same span as \(\mathbf{g}_1,\mathbf{g}_2\).
It remains to check independence of the proposed basis. If \(a\mathbf{g}_1+b\mathbf{g}_2=\mathbf{0}\), its coordinates give \(a+2b=0\), \(a+b=0\), and \(b=0\). Thus \(b=0\) and \(a=0\). Therefore \((\mathbf{g}_1,\mathbf{g}_2)\) is a basis for the span of all four generators.
Check: \(\mathbf{g}_1+\mathbf{g}_2=(1+2,1+1,0+1)=(3,2,1)=\mathbf{g}_3\), and \(2\mathbf{g}_1-\mathbf{g}_2=(2-2,2-1,0-1)=(0,1,-1)=\mathbf{g}_4\).
The coefficient matrix is useful here because it tests multiple generator relationships in one organized calculation. Its role is to solve the stated combination equations; the basis conclusion still requires checking that the retained list is independent and that every discarded generator is in its span.
When Basis Extension Is the Short Route
If a list is already independent, deleting vectors is unnecessary and may destroy the information that must be retained. Instead, test candidate vectors one at a time. A candidate inside the current span would make the enlarged list dependent; a candidate outside the span keeps it independent.
Proof. Suppose first that \(\mathbf{w}\) belongs to the span, so \(\mathbf{w}=\sum_{i=1}^{p}a_i\mathbf{v}_i\). Then \(\sum_{i=1}^{p}a_i\mathbf{v}_i-\mathbf{w}=\mathbf{0}\) is a nontrivial relation on the enlarged list, since the coefficient of \(\mathbf{w}\) is \(-1\). Thus the enlarged list is dependent. Conversely, suppose the enlarged list is dependent. There are scalars \(c_1,\dots,c_p,d\), not all zero, such that \(\sum_{i=1}^{p}c_i\mathbf{v}_i+d\mathbf{w}=\mathbf{0}\). If \(d=0\), then \(\sum_{i=1}^{p}c_i\mathbf{v}_i=\mathbf{0}\); independence of the original list forces every \(c_i=0\), contradicting that the coefficients are not all zero. Hence \(d\ne0\), and rearranging gives \(\mathbf{w}=-\sum_{i=1}^{p}(c_i/d)\mathbf{v}_i\). Therefore \(\mathbf{w}\) is in the original span. This proves both implications. \(\square\)
Worked Example: Extend an Independent List
Extend \(\mathbf{u}_1=(1,1,0)\), \(\mathbf{u}_2=(0,1,1)\) to a basis of \(\mathbb{R}^3\), using \(\mathbf{e}_1=(1,0,0)\) as a candidate.
First, \(\mathbf{e}_1\) is not in the span of \(\mathbf{u}_1,\mathbf{u}_2\). If \(a\mathbf{u}_1+b\mathbf{u}_2=\mathbf{e}_1\), the coordinates require \(a=1\), \(a+b=0\), and \(b=0\). The first and third equations give \(a=1\) and \(b=0\), which contradicts \(a+b=0\). By the extension test, the enlarged list is independent.
It also spans \(\mathbb{R}^3\). For any \((x,y,z)\), choose coefficients \(a=y-z\), \(b=z\), and \(c=x-y+z\):
Indeed, the first coordinate is \(y-z+x-y+z=x\), the second is \(y-z+z=y\), and the third is \(z\). Thus the list is spanning as well as independent, so it is a basis.
Check: at \((x,y,z)=(0,1,0)\), the coefficients are \(1,0,-1\), and \(1\mathbf{u}_1-1\mathbf{e}_1=(1,1,0)-(1,0,0)=(0,1,0)\).
Common Mistakes and a Practical Choice
Do not confuse a relation among generators with a complete basis proof. In the second example, finding \(\mathbf{g}_3=\mathbf{g}_1+\mathbf{g}_2\) alone does not show that \(\mathbf{g}_1,\mathbf{g}_2\) are independent. That required a separate check. Likewise, testing that an appended vector is outside the span proves the enlarged list is independent, but a basis of the intended space also needs to span that space.
A common coefficient error is to match only some coordinates. For example, in the extension example, the incorrect claim \((x,y,z)=y\mathbf{u}_1+z\mathbf{u}_2+(x-y-z)\mathbf{e}_1\) gives the first coordinate \(y+x-y-z=x-z\), not \(x\). The correct coefficient of \(\mathbf{e}_1\) is \(x-y+z\), because the first two terms contribute \(y-z\) to the first coordinate. Substitution into every coordinate is a reliable check.
Use the following comparison as a starting point, not as a substitute for verifying the result:
| Information given | Efficient first move | Final checks |
|---|---|---|
| A spanning list with an apparent redundancy | Write the dependence relation and delete a redundant vector | Preserved span and independence of the retained list |
| Several generators with unclear relationships | Use their coordinate columns in a coefficient matrix | Each discarded generator is in the retained span; retained vectors are independent |
| An independent list to be completed | Test candidate vectors for membership in the current span | Independence after extension and spanning of the target space |
For small examples, begin with the cheapest test that answers the immediate question. A visible linear combination is usually quicker than row reduction. When no relation is apparent, a coefficient matrix lays out the equations without guesswork. When preserving an independent list matters, the extension test tells exactly what a candidate must satisfy. In every case, the basis definition remains the final standard: independence and spanning are both required.
Check Your Understanding
For each question, identify the method that fits the given information and verify the conclusion.
- In a spanning list, suppose \(\mathbf{v}_4=2\mathbf{v}_1-\mathbf{v}_2\). Why may \(\mathbf{v}_4\) be deleted without changing the span?
- For \(\mathbf{a}_1=(1,0)\), \(\mathbf{a}_2=(1,1)\), test whether \((3,2)\) is in their span by solving a coefficient system.
- Explain why an independent list remains independent when a vector outside its span is appended.
- Find coefficients expressing \((x,y,z)\) as a combination of \((1,1,0)\), \((0,1,1)\), and \((1,0,0)\), and verify all three coordinates.
- A calculation shows that one generator is in the span of two retained generators. What additional property must be checked before calling the retained list a basis for the original span?
