Why One Deletion May Not Be Enough
A dependent list has at least one nontrivial dependence relation, but that relation may not be the list’s only redundancy. Removing a vector can eliminate one relation while leaving another among the vectors that remain. To decide whether a particular deletion works, we must ask not just whether the original list is dependent, but whether the remaining vectors admit any nontrivial relation.
Throughout, the vectors are columns in \(\mathbb{R}^n\), and a list keeps track of positions as well as vector values. Thus two equal vectors in different positions are still two entries of the list. For vectors \(\mathbf{v}_1,\dots,\mathbf{v}_p\), a relation is an equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\), where each \(c_i\in\mathbb{R}\). The relation is nontrivial if at least one coefficient is nonzero. The list is dependent if it has a nontrivial relation, and independent if its only relation has every coefficient zero.
For an index \(k\), deleting \(\mathbf{v}_k\) means keeping the other entries, in their original order. A relation among the remaining entries is exactly a relation for the full list in which the coefficient \(c_k\) is set to zero. This observation gives a direct test for the chosen deletion.
Proof. Suppose first that the list after deleting \(\mathbf{v}_k\) is independent. Any relation for the full list with \(c_k=0\) is a relation among precisely the entries that remain. Independence of those entries forces every coefficient on them to be zero. The deleted coefficient is already zero, so all \(c_i=0\). Conversely, suppose every full-list relation with \(c_k=0\) has all coefficients zero. Any relation among the entries after deletion can be made into a relation for the full list by assigning coefficient \(0\) to \(\mathbf{v}_k\). By the supposition, all its coefficients are zero. Therefore the remaining list is independent. \(\square\)
When the original list is dependent, the criterion has a useful equivalent form: deleting \(\mathbf{v}_k\) gives an independent list exactly when every nontrivial relation for the original list has \(c_k\ne0\). Indeed, if a nontrivial relation has \(c_k=0\), it remains a nontrivial relation after deletion. If no such relation exists, the deletion criterion applies. The condition concerns every relation, not just one relation we happen to find.
Examples: What the Coefficients Tell Us
Worked Example: Deleting the Wrong Vector Leaves a Duplicate
Consider the list \(\mathbf{v}_1=(1,0)\), \(\mathbf{v}_2=(1,0)\), \(\mathbf{v}_3=(0,1)\) in \(\mathbb{R}^2\). The first two entries are equal, so the full list has the relation \(\mathbf{v}_1-\mathbf{v}_2=\mathbf{0}\), with coefficient vector \((1,-1,0)\). Compare deleting \(\mathbf{v}_3\) with deleting \(\mathbf{v}_2\).
If we delete \(\mathbf{v}_3\), the remaining list is \(\mathbf{v}_1,\mathbf{v}_2\). For coefficients \(a,b\), its relation equation is
This equation says \(a+b=0\), which has the nontrivial solution \(a=1,\ b=-1\). Thus deleting \(\mathbf{v}_3\) leaves a dependent list. Substitution checks the relation: \((1,0)-(1,0)=(0,0)\).
If instead we delete \(\mathbf{v}_2\), the remaining entries are \(\mathbf{v}_1,\mathbf{v}_3\). Their relation equation is
so \(a=0\) and \(b=0\). The remaining list is independent. The original relation has a nonzero coefficient on \(\mathbf{v}_2\), so that particular relation is eliminated by deleting \(\mathbf{v}_2\); the equation for the remaining list confirms that no other relation survives.
Worked Example: Every Single Deletion Works
Let \(\mathbf{u}_1=(1,0)\), \(\mathbf{u}_2=(0,1)\), and \(\mathbf{u}_3=(1,1)\). The list is dependent because
To see what happens for all possible deletions, write a general relation:
Thus \(c_1=-c_3\) and \(c_2=-c_3\). Every relation therefore has the form \((-t,-t,t)\), where \(t\in\mathbb{R}\). If the relation is nontrivial, then \(t\ne0\), so all three coefficients are nonzero. In particular, for each chosen index \(k\), there is no nontrivial relation with \(c_k=0\). The deletion criterion proves that deleting any one of the three vectors leaves an independent list.
For a direct check, deleting \(\mathbf{u}_1\) leaves \(\mathbf{u}_2,\mathbf{u}_3\), whose relation equation is \(a(0,1)+b(1,1)=(b,a+b)=(0,0)\). It gives \(b=0\) and then \(a=0\). Deleting \(\mathbf{u}_2\) gives \(a(1,0)+b(1,1)=(a+b,b)=(0,0)\), so \(b=a=0\). Deleting \(\mathbf{u}_3\) gives \(a(1,0)+b(0,1)=(a,b)=(0,0)\), again only the trivial solution. The original relation also checks by addition: \((-1,0)+(0,-1)+(1,1)=(0,0)\).
Worked Example: Two Separate Redundancies
Consider \(\mathbf{w}_1=(1,0)\), \(\mathbf{w}_2=(1,0)\), \(\mathbf{w}_3=(0,1)\), and \(\mathbf{w}_4=(0,1)\). For coefficients \(c_1,c_2,c_3,c_4\), the relation equation is
It requires \(c_1+c_2=0\) and \(c_3+c_4=0\). For example, \((1,-1,0,0)\) gives the checked relation \((1,0)-(1,0)=(0,0)\), while \((0,0,1,-1)\) gives \((0,1)-(0,1)=(0,0)\). These are two distinct sources of dependence.
Delete \(\mathbf{w}_1\). The remaining list \(\mathbf{w}_2,\mathbf{w}_3,\mathbf{w}_4\) still has the relation \(\mathbf{w}_3-\mathbf{w}_4=\mathbf{0}\). In the original four-coefficient notation, this is \((0,0,1,-1)\), whose coefficient on the deleted vector is zero. The deletion criterion therefore predicts dependence, and substitution confirms it: \((0,1)-(0,1)=(0,0)\). In fact, no single deletion can remove both duplicate pairs: deleting one entry leaves at least one of the two displayed relations among the remaining entries.
This example shows why the existence of a relation involving the vector we delete is not sufficient. The relation \((1,-1,0,0)\) involves \(\mathbf{w}_1\), but eliminating that relation does not eliminate the separate relation \((0,0,1,-1)\).
When a Chosen Deletion Makes the List Independent
The coefficient criterion gives a precise interpretation of a successful deletion from a dependent list. If deleting \(\mathbf{v}_k\) leaves an independent list, then every nontrivial relation for the full list must use \(\mathbf{v}_k\) with a nonzero coefficient. In particular, at least one such relation exists, since the full list is assumed dependent. Solving that relation for the deleted vector expresses it as a linear combination of the entries that remain.
Proof. Dependence supplies a nontrivial relation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). If \(c_k=0\), this would be a nontrivial relation among the remaining entries, contradicting their independence. Hence \(c_k\ne0\). Rearranging and dividing by \(c_k\) gives
This proves that \(\mathbf{v}_k\) is a linear combination of the remaining vectors. To prove uniqueness, suppose also that \(\mathbf{v}_k=\sum_{i\ne k}d_i\mathbf{v}_i\). Subtracting the two expressions gives \(\sum_{i\ne k}(d_i+c_i/c_k)\mathbf{v}_i=\mathbf{0}\). This is a relation among the remaining entries. Their independence forces \(d_i+c_i/c_k=0\) for each \(i\ne k\). Thus the coefficients in the two expressions agree, proving uniqueness. \(\square\)
The reverse implication also holds when the list remaining after deletion is already known to be independent: if \(\mathbf{v}_k\) is a linear combination of those vectors, moving that combination to one side gives a nontrivial relation for the full list. The coefficient on \(\mathbf{v}_k\) is \(1\), so the full list is dependent. This explains the distinction between two questions: a representation of the deleted vector confirms dependence of the full list, but the remaining list must still be checked for independence.
Edge Case and Common Pitfall
Deleting the only vector is consistent with the same criterion. If \(p=1\), the coefficient \(c_1\) is required to be zero when testing a relation among the entries after deletion. The resulting equation has only the coefficient \(c_1=0\), so it is trivial. Therefore the empty list is independent, whether the original single vector was zero or nonzero. In particular, if the original vector is \(\mathbf{0}\), the original list is dependent because \(1\cdot\mathbf{0}=\mathbf{0}\); its nontrivial relation has coefficient \(1\), not coefficient \(0\).
Common mistake: “The deleted vector appears in a dependence relation, so the remaining list is independent.” The first part may be true, but it says nothing about other relations. In the four-vector example, \((1,-1,0,0)\) is a relation with a nonzero coefficient on \(\mathbf{w}_1\). Deleting \(\mathbf{w}_1\) removes that particular relation, but the remaining list still satisfies \((0,0,1,-1)\), or \(\mathbf{w}_3-\mathbf{w}_4=\mathbf{0}\). The correction is to test whether any nontrivial relation has coefficient zero on the chosen vector. If one does, that relation survives the deletion; if none does, the deletion criterion proves independence.
For practical work, keep the chosen index fixed while examining relations. A relation with zero coefficient at that index is direct evidence that deletion fails. To establish that deletion succeeds, show that every relation with that coefficient set to zero is trivial, either by solving the coefficient equations or by row-reducing the matrix whose columns are the remaining vectors. A single relation with a nonzero deleted coefficient is not enough to establish success.
Check Your Understanding
For each question, distinguish the relation you have found from all possible relations that may remain after deletion.
- For \((1,0),(1,0),(0,1)\), which deletion leaves a dependent list, and which deletion leaves an independent list?
- For \((1,0),(0,1),(1,1)\), write all coefficient vectors of relations in terms of one parameter. What does this show about deleting any one entry?
- For \((1,0),(1,0),(0,1),(0,1)\), give a nontrivial relation that survives deleting the first vector.
- State the deletion criterion using the coefficient \(c_k\) in a relation for the full list.
- If a list consists only of the zero vector, what is a nontrivial relation for it, and is the list after deletion independent?
