When Does a List Contain a Redundant Entry?
A list of vectors may describe the same span even after one of its entries is removed. The key question is whether the vector at that position can already be built from the other entries. If it can, including it adds no new linear combinations; if it cannot, deleting it makes the span smaller. This criterion works in any vector space, not only in \(\mathbb{R}^n\).
Let \(V\) be a vector space over a field \(\mathbb{F}\), where \(\mathbb{F}\) may be \(\mathbb{R}\) or \(\mathbb{C}\). A linear combination of a finite list \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) is a vector \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\), with each \(c_i\in\mathbb{F}\). Its span, written \(\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}\), is the set of all such linear combinations. The span of the empty list is defined to be \(\{\mathbf{0}\}\). A list spans a subspace \(W\) when its span equals \(W\).
We treat the entries as positions in a list. Thus, two identical vectors appearing at different positions are two entries, and removing one position leaves the other in place. This distinction matters: a set records whether a vector is present, but does not record how many times it occurs.
Proof. First suppose \(\mathbf{v}_k\in\operatorname{span}(L_k)\). Then \(\mathbf{v}_k\) is a linear combination of the entries at all the other positions. Any linear combination of the full list can therefore have its \(c_k\mathbf{v}_k\) term replaced by the corresponding combination of those other entries. It follows that every vector in the span of the full list belongs to \(\operatorname{span}(L_k)\). Conversely, every entry of \(L_k\) is also in the full list, so every linear combination of \(L_k\) belongs to the span of the full list. The spans are equal. Now suppose the two spans are equal. Since \(\mathbf{v}_k\) belongs to the span of the full list, it belongs to the equal span of \(L_k\). Thus \(\mathbf{v}_k\in\operatorname{span}(L_k)\), proving the reverse implication as well. \(\square
This result applies whether or not the list spans the entire ambient space \(V\). In particular, if the list spans a specified subspace \(W\), deleting an entry preserves the spanning property precisely when that entry is in the span of the other entries. For a one-entry list, the test also covers the edge case: deleting its entry preserves the span exactly when that vector is \(\mathbf{0}\), since the empty list spans \(\{\mathbf{0}\}\).
Testing and Removing an Entry
To use the criterion, focus on one position at a time. Try to solve an equation expressing its vector as a linear combination of the remaining entries. If the equation has a solution, the entry is removable without changing the span. If no such coefficients exist, the criterion says that deleting the entry changes the span. When the list spans a particular \(W\), a successful removal still leaves a list spanning \(W\).
Worked Example: Removing a Vector from a Spanning List in \(\mathbb{R}^2\)
Consider \(\mathbf{v}_1=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}0\\1\end{bmatrix}\), and \(\mathbf{v}_3=\begin{bmatrix}1\\1\end{bmatrix}\). Decide whether \(\mathbf{v}_3\) can be deleted without changing the span of the list.
The third vector is the sum of the first two:
So \(\mathbf{v}_3\) is in the span of the other two, and the deletion criterion guarantees that removing it preserves the span. To check the claim directly, an arbitrary vector \(\begin{bmatrix}a\\b\end{bmatrix}\in\mathbb{R}^2\) satisfies
Thus the first two vectors still span \(\mathbb{R}^2\). Check: the relation used for deletion is verified coordinate by coordinate: \(1+0=1\) in the first coordinate and \(0+1=1\) in the second.
The criterion also detects an entry that cannot be removed. For example, \(\begin{bmatrix}1\\0\end{bmatrix}\) is not a scalar multiple of \(\begin{bmatrix}0\\1\end{bmatrix}\): every scalar multiple of the latter has first coordinate \(0\). Therefore, in the two-entry list consisting of these vectors, neither entry lies in the span of the other. Deleting either one changes the span.
Worked Example: A Redundant Polynomial
In the vector space of real polynomials, let \(p_1=1+x\), \(p_2=x+x^2\), and \(p_3=1+2x+x^2\). Determine whether \(p_3\) can be deleted from the list without changing its span.
Adding the first two polynomials gives
Hence \(p_3\) is a linear combination of the other entries. To verify that the whole span is unchanged, take arbitrary scalars \(a,b,c\in\mathbb{R}\). A typical linear combination of all three polynomials is
Every combination of the three is therefore a combination of \(p_1,p_2\); combinations of \(p_1,p_2\) are already combinations of the full list. Check: the coefficient of \(1\) on the right is \(a+c\), the coefficient of \(x\) is \(a+b+2c\), and the coefficient of \(x^2\) is \(b+c\). These are exactly the coefficients obtained by expanding \(ap_1+bp_2+cp_3\).
Why Redundancy Must Be Retested
After a successful deletion, the span of the current list is unchanged. However, that does not mean every entry that was removable before the deletion will remain removable. A vector that was expressed using an entry that has since been deleted may no longer be a combination of the current remaining entries. So a repeated deletion procedure must test each candidate against the list as it stands at that stage.
Worked Example: A Removable Entry Can Cease to Be Removable
Let \(\mathbf{v}_1=\begin{bmatrix}1\\0\end{bmatrix}\), \(\mathbf{v}_2=\begin{bmatrix}0\\1\end{bmatrix}\), and \(\mathbf{v}_3=\begin{bmatrix}1\\1\end{bmatrix}\). Initially, \(\mathbf{v}_1\) is removable because
Delete \(\mathbf{v}_1\). The remaining list \((\mathbf{v}_2,\mathbf{v}_3)\) still spans the same space as the original list, by the deletion criterion. But now test whether \(\mathbf{v}_3\) can be removed. That would require a scalar \(c\) with \(\mathbf{v}_3=c\mathbf{v}_2\), or
The first coordinates would require \(1=0\), which is impossible. Thus \(\mathbf{v}_3\) is not removable from the current list. Check: although \(\mathbf{v}_3\) was originally removable using \(\mathbf{v}_1+\mathbf{v}_2\), that expression uses \(\mathbf{v}_1\), which is no longer present. The current test correctly uses only \(\mathbf{v}_2\).
There is a useful contrast. If an entry is not in the span of the other entries at some stage, deleting more of those other entries cannot make it removable. The later list has fewer vectors available for a linear combination, not more.
Proof. Every linear combination of the entries left after the additional deletions is also a linear combination of the original other entries: it is obtained by assigning coefficient \(0\) to each deleted entry. Therefore the span of the entries left is contained in the span of the original other entries. If \(\mathbf{v}_k\) belonged to the new, smaller span, it would belong to the original span as well, contrary to the assumption. Hence it remains outside the new span. \(\square
Together, these observations give the right logic for repeated removal: each deletion that passes the current test preserves the span; candidates must be retested after the list changes; and a failed test cannot turn into a successful one merely by deleting other entries.
A Finite Removal Procedure
For a finite list, the criterion gives a straightforward procedure for removing entries while preserving the span. At each stage, choose an entry and test whether it belongs to the span of the other entries currently present. If it does, delete it. If not, keep it and test another position. Continue only with the current list, not with an earlier version of it.
Record the list that exists now and identify the entry to test.
Try to express the chosen vector as a linear combination of all the other current entries.
The deletion criterion ensures that this step preserves the span.
A successful expression from an earlier stage may have used an entry that is now gone.
This process must terminate if we keep making deletions: each one reduces the finite number of entries by one, and that number cannot become negative. At termination, no entry tested for deletion has passed the current membership test. Every span-preserving deletion is justified at the stage where it is made. Thus, if the initial list spans \(W\), the final list still spans \(W\). The procedure does not promise that a particular deletion order will produce the same final list; it only guarantees that each accepted deletion preserves the span.
Worked Example: Removing a Matrix from a Spanning List
In the vector space of \(2\times2\) real matrices, let
Test whether \(\mathbf{C}\) can be deleted from the list \((\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})\). Direct addition and scalar multiplication give
Thus \(\mathbf{C}\) is in the span of the other entries and may be removed without changing the span. More explicitly, for any \(a,b,c,d\in\mathbb{R}\),
Check: both sides equal \(\begin{bmatrix}a+c&b+2c\\d&0\end{bmatrix}\). The calculation confirms that every combination using \(\mathbf{C}\) can be rewritten using the remaining entries.
Common Mistakes
Mistake: assuming an entry stays removable after one of the vectors used to express it is deleted. In the preceding example, \(\mathbf{v}_1\) can initially be written as \(\mathbf{v}_3-\mathbf{v}_2\). A wrong follow-up calculation might claim that \(\mathbf{v}_3\) is removable after deleting \(\mathbf{v}_1\), because \(\mathbf{v}_3=\mathbf{v}_1+\mathbf{v}_2\). That calculation uses \(\mathbf{v}_1\), which is no longer an available entry. The corrected current-list test asks whether \(\mathbf{v}_3=c\mathbf{v}_2\); comparing first coordinates gives \(1=0\), so no such scalar exists. Retest using only the entries that remain.
Mistake: confusing a list of entries with a set of distinct vectors. If a list is \((\mathbf{u},\mathbf{u})\), either position can be deleted and one copy remains. Treating the list as the set \(\{\mathbf{u}\}\) loses the information that there were two entries. The deletion criterion applies to a chosen position, with the other positions left available for combinations.
Mistake: deleting an entry just because it looks similar to another. A vector can be removed only if it is a linear combination of the other entries. For example, the vectors \(\begin{bmatrix}1\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) are not scalar multiples, so neither can be removed from that two-entry list without changing its span. In longer lists, test the full set of other entries rather than comparing only pairs.
Check Your Understanding
For each question, identify the current list precisely and use the deletion criterion.
- In \(\mathbb{R}^2\), can \(\begin{bmatrix}2\\3\end{bmatrix}\) be deleted from the list \(\left(\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}1\\2\end{bmatrix},\begin{bmatrix}2\\3\end{bmatrix}\right)\)? Give the combination or explain why none exists.
- For \(q_1=1\), \(q_2=x\), and \(q_3=1-x\), which entries can be removed initially without changing the span?
- If \(\mathbf{v}_k\) is not in the span of the other entries, can deleting one of those other entries make \(\mathbf{v}_k\) removable? Explain using span containment.
- For the list \((\mathbf{u},\mathbf{u})\), what is the span after deleting either one of the two positions?
- Explain why a successful deletion preserves the span of a list that spans a subspace \(W\).
