The Question Behind Appending a Vector
Suppose \(\mathbf{v}_1,\dots,\mathbf{v}_p\) are linearly independent vectors in \(\mathbb{R}^n\), and we add a vector \(\mathbf{w}\). The existing vectors have no redundancy, but the enlarged list might: the new vector could already be made from the old ones. The decisive question is therefore whether \(\mathbf{w}\) belongs to their span.
A linear combination of \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is a vector \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), where the \(c_i\) are real numbers. Their span, written \(\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\), is the set of all such combinations. A list is linearly independent when the equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\) has only the solution in which every coefficient is zero. We use the usual convention that the empty list is independent and its span is \(\{\mathbf{0}\}\).
Proof. First suppose that \(\mathbf{w}\in\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\). By the definition of span, there are scalars \(a_1,\dots,a_p\) such that \(\mathbf{w}=a_1\mathbf{v}_1+\cdots+a_p\mathbf{v}_p\). Rearranging gives \(a_1\mathbf{v}_1+\cdots+a_p\mathbf{v}_p-1\mathbf{w}=\mathbf{0}\). This is a nontrivial relation because the coefficient of \(\mathbf{w}\) is \(-1\), so the appended list is dependent.
For the other direction, suppose the appended list is dependent. Then there are scalars \(c_1,\dots,c_p,d\), not all zero, such that \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p+d\mathbf{w}=\mathbf{0}\). If \(d=0\), this becomes a relation among the original list. Its independence forces \(c_1=\cdots=c_p=0\), contradicting that the coefficients are not all zero. Thus \(d\ne0\). Dividing by \(d\) and rearranging gives \(\mathbf{w}=-\frac{c_1}{d}\mathbf{v}_1-\cdots-\frac{c_p}{d}\mathbf{v}_p\). Therefore \(\mathbf{w}\) belongs to the span of the original list. We have proved both implications. \(\square\)
The proof explains why the hypothesis on the original list matters. If a dependence is present after appending, the coefficient of the new vector must be nonzero: otherwise the dependence would already have been present before appending. That nonzero coefficient can be divided out, expressing the new vector using the old ones.
Testing the Criterion with Coordinates
Put the original vectors into the columns of \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_p\end{bmatrix}\), and let \(\mathbf{x}=(x_1,\dots,x_p)^{\top}\). By matrix multiplication, \(\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_p\mathbf{v}_p\). Consequently, \(\mathbf{w}\) is in the span of the original list exactly when the system \(\mathbf{A}\mathbf{x}=\mathbf{w}\) is consistent.
This is an application of the theorem, not a different criterion: consistency means that the new vector is a combination of the old columns, and inconsistency means it is outside their span. Row operations preserve the solutions of the system, so the reduced augmented matrix gives the same consistency answer as the original one. When the original columns are independent and a solution exists, that solution is unique: if \(\mathbf{A}\mathbf{x}=\mathbf{w}\) and \(\mathbf{A}\mathbf{y}=\mathbf{w}\), then \(\mathbf{A}(\mathbf{x}-\mathbf{y})=\mathbf{0}\); independence forces \(\mathbf{x}-\mathbf{y}=\mathbf{0}\), hence \(\mathbf{x}=\mathbf{y}\).
Worked Example: The New Vector Is Outside the Span
Let \(\mathbf{v}_1=(1,0,1)^{\top}\) and \(\mathbf{v}_2=(0,1,1)^{\top}\), and append \(\mathbf{w}=(1,1,3)^{\top}\). First, the original vectors are independent: \(c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}\) gives \((c_1,c_2,c_1+c_2)^{\top}=\mathbf{0}\), so the first two coordinates give \(c_1=c_2=0\).
Test whether \(\mathbf{w}\) is in their span by solving \(\mathbf{A}\mathbf{x}=\mathbf{w}\). The augmented matrix reduces as follows:
The last row represents \(0=1\), so the system is inconsistent. Hence \(\mathbf{w}\) is outside the span, and the theorem says the appended list is independent. The coordinate equations show the obstruction directly: to match the first two coordinates requires \(x_1=1\) and \(x_2=1\), but then the third coordinate of \(x_1\mathbf{v}_1+x_2\mathbf{v}_2\) is \(2\), not \(3\).
Check: If a combination matched \(\mathbf{w}\), its first two coordinates would force \(x_1=x_2=1\); that combination is \((1,1,2)^{\top}\ne(1,1,3)^{\top}\). There is no representation.
Worked Example: The New Vector Is in the Span
Use the same independent vectors \(\mathbf{v}_1=(1,0,1)^{\top}\) and \(\mathbf{v}_2=(0,1,1)^{\top}\), but now append \(\mathbf{w}=(2,-1,1)^{\top}\). Solve \(\mathbf{A}\mathbf{x}=\mathbf{w}\):
The system is consistent, with \(x_1=2\) and \(x_2=-1\). Thus \(\mathbf{w}=2\mathbf{v}_1-\mathbf{v}_2\), so the appended list is dependent. The same equality gives a dependence relation: \(2\mathbf{v}_1-\mathbf{v}_2-\mathbf{w}=\mathbf{0}\).
Check: \(2(1,0,1)^{\top}-(0,1,1)^{\top}=(2,-1,1)^{\top}=\mathbf{w}\). Substituting into the relation gives \((2,0,2)^{\top}-(0,1,1)^{\top}-(2,-1,1)^{\top}=(0,0,0)^{\top}\).
A Parameter and the Boundary Between the Cases
Worked Example: Finding the Parameter Values
Let \(\mathbf{v}_1=(1,0,1)^{\top}\) and \(\mathbf{v}_2=(0,1,1)^{\top}\). For which real values of \(t\) does appending \(\mathbf{w}=(t,1,2)^{\top}\) preserve independence? The original pair is independent by the coordinate argument in the previous examples. Reduce the augmented matrix:
The system is inconsistent exactly when \(1-t\ne0\). Therefore, the appended list is independent for \(t\ne1\), and dependent for \(t=1\). At \(t=1\), the new vector is \((1,1,2)^{\top}=\mathbf{v}_1+\mathbf{v}_2\), which supplies the dependence relation \(\mathbf{v}_1+\mathbf{v}_2-\mathbf{w}=\mathbf{0}\).
Check: At \(t=1\), the displayed combination has coordinates \((1+0,0+1,1+1)^{\top}=(1,1,2)^{\top}\). If \(t\ne1\), the last reduced row says \(0=1-t\), which cannot hold; so there is no combination representing \(\mathbf{w}\).
Why the Original Independence Hypothesis Cannot Be Dropped
The span test by itself does not decide whether a list is independent unless the original list is already known to be independent. Consider \(\mathbf{u}_1=(1,0)^{\top}\) and \(\mathbf{u}_2=(2,0)^{\top}\) in \(\mathbb{R}^2\). The original list is dependent because \(\mathbf{u}_2=2\mathbf{u}_1\). Append \(\mathbf{w}=(0,1)^{\top}\). This new vector is not in the span of \(\mathbf{u}_1,\mathbf{u}_2\), since every combination of those vectors has second coordinate \(0\). But the enlarged list is still dependent: \(2\mathbf{u}_1-\mathbf{u}_2+0\mathbf{w}=\mathbf{0}\).
The tempting but incorrect argument is: “The new vector is outside the old span, so the enlarged list is independent.” The correction is to check that the old list is independent first. In the theorem’s proof, the coefficient of the new vector must be nonzero only because a relation involving just the old vectors is ruled out by that hypothesis. Without it, appending a vector cannot repair an existing dependence.
A Practical Decision Procedure
When applying the criterion, keep the role of each calculation clear. The original list is checked for independence; the new vector is then tested for membership in its span. If the original list is independent, this procedure gives an exact yes-or-no answer about the enlarged list.
Confirm that its columns are independent, for example by solving the homogeneous system \(\mathbf{A}\mathbf{x}=\mathbf{0}\).
Form \(\mathbf{A}\mathbf{x}=\mathbf{w}\), or equivalently the augmented matrix \([\mathbf{A}\mid\mathbf{w}]\).
A contradictory row means \(\mathbf{w}\) lies outside the span, so appending preserves independence. A consistent system means \(\mathbf{w}\) lies in the span, so the appended list is dependent.
If the system is consistent, substitute its coefficients into the linear combination to exhibit a dependence relation. If it is inconsistent, identify the incompatible coordinate equations.
This criterion is useful whenever an independent list is being enlarged. It gives a conceptual description—avoid adding a vector already produced by the list—and a coordinate method for deciding whether that has happened. The method also identifies the reason for dependence when it occurs: a solution of \(\mathbf{A}\mathbf{x}=\mathbf{w}\) explicitly writes the new vector as a combination of the old vectors.
Check Your Understanding
For each problem, distinguish the assumption that the original list is independent from the test of whether the new vector belongs to its span.
- Let \(\mathbf{v}_1=(1,1,0)^{\top}\), \(\mathbf{v}_2=(0,1,1)^{\top}\), and \(\mathbf{w}=(2,3,1)^{\top}\). Is the appended list independent? Set up and solve the membership system.
- Suppose \(\mathbf{v}_1,\mathbf{v}_2\) are independent and \(\mathbf{w}=a\mathbf{v}_1+b\mathbf{v}_2\). Write a nontrivial dependence relation for the appended list.
- For independent \(\mathbf{v}_1,\mathbf{v}_2\), explain what a row \([0\ 0\mid 5]\) in the reduced augmented matrix tells you.
- Give an example of a dependent original list and a vector outside its span, and explain why appending that vector does not make the full list independent.
- If \(\mathbf{A}\mathbf{x}=\mathbf{w}\) has a solution and the columns of \(\mathbf{A}\) are independent, prove that this solution is unique.
