What Happens When We Select Some Vectors?
Linear independence is a condition on an entire list: the only linear combination that gives the zero vector must use zero for every coefficient. This makes it natural to ask what happens when we keep only some entries. If the original list has no nontrivial dependence relation, can a shorter list acquire one? It cannot. Conversely, if some selected entries already have a dependence relation, adding more vectors cannot remove that relation.
These statements are useful when a long list is difficult to analyze all at once. A dependent selection is enough to show that the full list is dependent. And when a full list is known to be independent, we may select any of its entries and retain independence. The key is to track which coefficients belong to selected entries and to assign zero to every omitted position.
Lists and Sublists
Throughout this tutorial the vectors and coefficients are in \(\mathbb{R}^n\) and \(\mathbb{R}\), respectively. A list records both its entries and their positions. A sublist is obtained by selecting some positions while keeping their original order. For example, from \((\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3,\mathbf{v}_4)\), the entries in positions 1 and 3 form the sublist \((\mathbf{v}_1,\mathbf{v}_3)\). Positions matter even if two entries happen to be equal.
The distinction between “all coefficients are zero” and “not all coefficients are zero” is essential. Every list has a linear combination equal to \(\mathbf{0}\): choose all coefficients to be zero. That choice is the trivial combination, and it does not make a list dependent. The empty sublist is independent as well: it has no coefficients, so there is no nontrivial choice of coefficients that could produce a dependence relation.
Every Sublist of an Independent List Is Independent
Proof. Choose any sublist, with selected positions \(i_1<\cdots<i_q\). Suppose scalars \(d_1,\dots,d_q\) satisfy \(d_1\mathbf{v}_{i_1}+\cdots+d_q\mathbf{v}_{i_q}=\mathbf{0}\). Form coefficients \(c_1,\dots,c_p\) for the full list by setting \(c_{i_k}=d_k\) for each selected position \(i_k\), and setting \(c_j=0\) at every omitted position \(j\). Then \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\), because the omitted terms contribute zero and the selected terms are exactly the assumed relation. Independence of the full list implies \(c_1=\cdots=c_p=0\). In particular, \(d_k=c_{i_k}=0\) for each \(k\). Thus the chosen sublist is independent. Since the sublist was arbitrary, every sublist is independent. \(\square\)
The proof uses only the definition of independence. A relation among selected entries becomes a relation on the full list by filling every omitted position with a zero coefficient. Independence of the full list then forces the coefficients on the selected entries to be zero too. This reasoning works whether the sublist is consecutive, whether it contains just one entry, or whether it is empty.
Worked Example: Selecting Entries from an Independent List
Consider the list in \(\mathbb{R}^3\) \(\mathbf{v}_1=(1,0,1)\), \(\mathbf{v}_2=(0,1,1)\), and \(\mathbf{v}_3=(1,1,0)\). First verify that the full list is independent. Then examine the sublist in positions 1 and 3.
Suppose \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\). Comparing coordinates gives
The first two equations give \(c_1=-c_3\) and \(c_2=-c_3\). Substituting these into the third gives \(-2c_3=0\), so \(c_3=0\), and hence \(c_1=c_2=0\). The full list is independent. The theorem therefore guarantees that \((\mathbf{v}_1,\mathbf{v}_3)\) is independent.
We can check that conclusion directly. If \(a\mathbf{v}_1+b\mathbf{v}_3=\mathbf{0}\), then the coordinates give \(a+b=0\), \(b=0\), and \(a=0\). Thus only \(a=b=0\) works.
Check: the full-list equations force all three coefficients to zero, and the sublist equation forces both of its coefficients to zero. Both lists satisfy the definition of independence.
A Dependent Sublist Makes the Full List Dependent
The theorem has an immediate contrapositive. The original implication is: if the full list is independent, then every sublist is independent. Its contrapositive is: if some sublist is dependent, then the full list is dependent. A contrapositive is logically equivalent to the original implication, so this conclusion requires no additional assumption.
Proof. Let the dependent sublist occupy positions \(i_1<\cdots<i_q\) in the full list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\). Because the sublist is dependent, there are coefficients \(d_1,\dots,d_q\), not all zero, such that \(d_1\mathbf{v}_{i_1}+\cdots+d_q\mathbf{v}_{i_q}=\mathbf{0}\). Set \(c_{i_k}=d_k\) at each selected position and \(c_j=0\) at each omitted position. These coefficients are not all zero, because the \(d_k\) are not all zero. Also, \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\), since this sum is the given sublist relation with zero terms added. Therefore the full list has a nontrivial dependence relation and is dependent. \(\square\)
This proof gives a practical way to use the corollary: keep the coefficients of the dependence relation at their original positions, and put zero everywhere else. The other entries in the full list do not affect the relation. In particular, the result does not require the added entries to be zero vectors or to have any special form.
Worked Example: Extending a Relation to the Full List
Consider the list in \(\mathbb{R}^2\) \(\mathbf{a}_1=(1,0)\), \(\mathbf{a}_2=(0,1)\), \(\mathbf{a}_3=(2,0)\), and \(\mathbf{a}_4=(1,1)\). The sublist \((\mathbf{a}_1,\mathbf{a}_3)\) occupies positions 1 and 3. Find a relation on that sublist and use it to prove that the full list is dependent.
Since \(\mathbf{a}_3=2\mathbf{a}_1\), the sublist has the relation \(2\mathbf{a}_1-\mathbf{a}_3=\mathbf{0}\). Its coefficients are not both zero. Extend them to the full list by putting zero in positions 2 and 4. The resulting coefficient list is \((2,0,-1,0)\), so the full-list relation is \(2\mathbf{a}_1+0\mathbf{a}_2-\mathbf{a}_3+0\mathbf{a}_4=\mathbf{0}\).
Check: \(2(1,0)+0(0,1)-(2,0)+0(1,1)=(2,0)-(2,0)=(0,0)\). The coefficients are not all zero, so this verifies that the full list is dependent.
What the Implications Do—and Do Not—Say
Both results concern a specific direction of implication. Independence of a full list guarantees independence of every selected sublist. Dependence of a sublist guarantees dependence of the full list. Neither statement says that every sublist of a dependent list is dependent. A particular selection may omit the entries that create the dependence relation.
Worked Example: A Dependent List Can Have an Independent Sublist
Let \(\mathbf{u}_1=(1,0)\), \(\mathbf{u}_2=(0,1)\), and \(\mathbf{u}_3=(1,1)\) in \(\mathbb{R}^2\). The full list is dependent because \(\mathbf{u}_1+\mathbf{u}_2-\mathbf{u}_3=(0,0)\), and the coefficients \((1,1,-1)\) are not all zero.
However, the sublist \((\mathbf{u}_1,\mathbf{u}_2)\) is independent. If \(a\mathbf{u}_1+b\mathbf{u}_2=\mathbf{0}\), then \(a(1,0)+b(0,1)=(a,b)=(0,0)\), which gives \(a=0\) and \(b=0\).
Check: the displayed relation verifies dependence of all three entries, while the coordinate equation verifies independence of the first two. Thus dependence of the full list does not imply dependence of each sublist.
Common Pitfall: Reversing the Implication
A common error is to reason, “The full list is dependent, so any sublist I choose must also be dependent.” The preceding example is a counterexample to that claim: the full list has the nontrivial relation \(\mathbf{u}_1+\mathbf{u}_2-\mathbf{u}_3=\mathbf{0}\), but its first two entries satisfy \(a\mathbf{u}_1+b\mathbf{u}_2=(a,b)=\mathbf{0}\) only when \(a=b=0\). The correction is to check which entries actually appear in a dependence relation. A relation on the full list may use a nonzero coefficient on an entry that is omitted from a proposed sublist.
The valid contrapositive runs the other way: if the chosen sublist is dependent, its nontrivial relation remains valid in the full list when omitted entries receive coefficient zero. It is not enough that a sublist looks repetitive or has some zero coordinates; dependence requires an actual nontrivial coefficient relation.
Using the Results Efficiently
When a list is known to be independent, there is no need to test each shorter selection from scratch: the theorem proves all of them independent. When a list is being tested for dependence, finding even one dependent sublist settles the question for the whole list. These are one-way shortcuts. An independent sublist alone does not establish that the full list is independent, and a dependent full list does not identify which smaller selections are dependent.
For a concrete check, write down the chosen positions before transferring coefficients. In the dependent-sublist argument, selected coefficients keep their values and omitted coefficients become zero. In the independent-sublist argument, begin with a putative relation on the sublist and extend it in exactly the same way; independence of the full list then forces every selected coefficient to vanish. This shared coefficient bookkeeping is the whole mechanism behind both conclusions.
Check Your Understanding
Use the definitions and coefficient-extension arguments to justify each answer.
- If \((\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3,\mathbf{v}_4)\) is independent, what can be concluded about the sublist \((\mathbf{v}_2,\mathbf{v}_4)\)?
- A sublist in positions 1 and 3 has relation \(5\mathbf{v}_1-2\mathbf{v}_3=\mathbf{0}\). Write the corresponding coefficient list for a full list of five vectors.
- Why does a dependent sublist imply that the full list is dependent, even when the omitted vectors are nonzero?
- Give an example of a dependent list with an independent proper sublist, and verify both claims from the definitions.
- Explain why the empty sublist is independent under the definition of linear independence.
