Spanning Sets and Linear IndependenceEdit
We first characterize when a vector can be removed from a set without changing the span of that set.
 Lemma 1.1
Where is a subset of a vector space ,
for any .
 Proof
The left to right implication is easy. If then, since , the equality of the two sets gives that .
For the right to left implication assume that to show that by mutual inclusion. The inclusion is obvious. For the other inclusion , write an element of as and substitute 's expansion as a linear combination of members of the same set . This is a linear combination of linear combinations and so distributing results in a linear combination of vectors from . Hence each member of is also a member of .
 Example 1.2
In , where
the spans and are equal since is in the span .
The lemma says that if we have a spanning set then we can remove a to get a new set with the same span if and only if is a linear combination of vectors from . Thus, under the second sense described above, a spanning set is minimal if and only if it contains no vectors that are linear combinations of the others in that set. We have a term for this important property.
 Definition 1.3
A subset of a vector space is linearly independent if none of its elements is a linear combination of the others. Otherwise it is linearly dependent.
Here is an important observation:
although this way of writing one vector as a combination of the others visually sets off from the other vectors, algebraically there is nothing special in that equation about . For any with a coefficient that is nonzero, we can rewrite the relationship to set off .
When we don't want to single out any vector by writing it alone on one side of the equation we will instead say that are in a linear relationship and write the relationship with all of the vectors on the same side. The next result rephrases the linear independence definition in this style. It gives what is usually the easiest way to compute whether a finite set is dependent or independent.
 Lemma 1.4
A subset of a vector space is linearly independent if and only if for any distinct the only linear relationship among those vectors
is the trivial one: .
 Proof
This is a direct consequence of the observation above.
If the set is linearly independent then no vector can be written as a linear combination of the other vectors from so there is no linear relationship where some of the 's have nonzero coefficients. If is not linearly independent then some is a linear combination of other vectors from , and subtracting from both sides of that equation gives a linear relationship involving a nonzero coefficient, namely the in front of .
 Example 1.5
In the vector space of twowide row vectors, the twoelement set is linearly independent. To check this, set
and solving the resulting system
shows that both and are zero. So the only linear relationship between the two given row vectors is the trivial relationship.
In the same vector space, is linearly dependent since we can satisfy
with and .
 Remark 1.6
Recall the Statics example that began this book. We first set the unknownmass objects at cm and cm and got a balance, and then we set the objects at cm and cm and got a balance. With those two pieces of information we could compute values of the unknown masses. Had we instead first set the unknownmass objects at cm and cm, and then at cm and cm, we would not have been able to compute the values of the unknown masses (try it). Intuitively, the problem is that the information is a "repeat" of the information— that is, is in the span of the set — and so we would be trying to solve a twounknowns problem with what is essentially one piece of information.
 Example 1.7
The set is linearly independent in , the space of quadratic polynomials with real coefficients, because
gives
since polynomials are equal only if their coefficients are equal. Thus, the only linear relationship between these two members of is the trivial one.
 Example 1.8
In , where
the set is linearly dependent because this is a relationship
where not all of the scalars are zero (the fact that some of the scalars are zero doesn't matter).
 Remark 1.9
That example illustrates why, although Definition 1.3 is a clearer statement of what independence is, Lemma 1.4 is more useful for computations. Working straight from the definition, someone trying to compute whether is linearly independent would start by setting and concluding that there are no such and . But knowing that the first vector is not dependent on the other two is not enough. This person would have to go on to try to find the dependence , . Lemma 1.4 gets the same conclusion with only one computation.
 Example 1.10
The empty subset of a vector space is linearly independent. There is no nontrivial linear relationship among its members as it has no members.
 Example 1.11
In any vector space, any subset containing the zero vector is linearly dependent. For example, in the space of quadratic polynomials, consider the subset .
One way to see that this subset is linearly dependent is to use Lemma 1.4: we have , and this is a nontrivial relationship as not all of the coefficients are zero. Another way to see that this subset is linearly dependent is to go straight to Definition 1.3: we can express the third member of the subset as a linear combination of the first two, namely, is satisfied by taking and (in contrast to the lemma, the definition allows all of the coefficients to be zero).
(There is still another way to see that this subset is dependent that is subtler. The zero vector is equal to the trivial sum, that is, it is the sum of no vectors. So in a set containing the zero vector, there is an element that can be written as a combination of a collection of other vectors from the set, specifically, the zero vector can be written as a combination of the empty collection.)
The above examples, especially Example 1.5, underline the discussion that begins this section. The next result says that given a finite set, we can produce a linearly independent subset by discarding what Remark 1.6 calls "repeats".
 Theorem 1.12
In a vector space, any finite subset has a linearly independent subset with the same span.
 Proof
If the set is linearly independent then itself satisfies the statement, so assume that it is linearly dependent.
By the definition of dependence, there is a vector that is a linear combination of the others. Call that vector . Discard it— define the set . By Lemma 1.1, the span does not shrink .
Now, if is linearly independent then we are finished. Otherwise iterate the prior paragraph: take a vector that is a linear combination of other members of and discard it to derive such that . Repeat this until a linearly independent set appears; one must appear eventually because is finite and the empty set is linearly independent. (Formally, this argument uses induction on , the number of elements in the starting set. Problem 20 asks for the details.)
 Example 1.13
This set spans .
Looking for a linear relationship
gives a three equations/five unknowns linear system whose solution set can be parametrized in this way.
So is linearly dependent. Setting and shows that the fifth vector is a linear combination of the first two. Thus, Lemma 1.1 says that discarding the fifth vector
leaves the span unchanged . Now, the third vector of is a linear combination of the first two and we get
with the same span as , and therefore the same span as , but with one difference. The set is linearly independent (this is easily checked), and so discarding any of its elements will shrink the span.
Linear Independence and Subset RelationsEdit
Theorem 1.12 describes producing a linearly independent set by shrinking, that is, by taking subsets. We finish this subsection by considering how linear independence and dependence, which are properties of sets, interact with the subset relation between sets.
 Lemma 1.14
Any subset of a linearly independent set is also linearly independent. Any superset of a linearly dependent set is also linearly dependent.
 Proof
This is clear.
Restated, independence is preserved by subset and dependence is preserved by superset.
Those are two of the four possible cases of interaction that we can consider. The third case, whether linear dependence is preserved by the subset operation, is covered by Example 1.13, which gives a linearly dependent set with a subset that is linearly dependent and another subset that is linearly independent.
That leaves one case, whether linear independence is preserved by superset. The next example shows what can happen.
 Example 1.15
In each of these three paragraphs the subset is linearly independent.
For the set
the span is the axis. Here are two supersets of , one linearly dependent and the other linearly independent.
dependent: independent:
Checking the dependence or independence of these sets is easy.
For
the span is the plane. These are two supersets.
dependent: independent:
If
then . A linearly dependent superset is
dependent:
but there are no linearly independent supersets of . The reason is that for any vector that we would add to make a superset, the linear dependence equation
has a solution , , and .
So, in general, a linearly independent set may have a superset that is dependent. And, in general, a linearly independent set may have a superset that is independent. We can characterize when the superset is one and when it is the other.
 Lemma 1.16
Where is a linearly independent subset of a vector space ,
for any with .
 Proof
One implication is clear: if then where each and , and so is a nontrivial linear relationship among elements of .
The other implication requires the assumption that is linearly independent. With linearly dependent, there is a nontrivial linear relationship and independence of then implies that , or else that would be a nontrivial relationship among members of . Now rewriting this equation as shows that .
(Compare this result with Lemma 1.1. Both say, roughly, that is a "repeat" if it is in the span of . However, note the additional hypothesis here of linear independence.)
 Corollary 1.17
A subset of a vector space is linearly dependent if and only if some is a linear combination of the vectors , ..., listed before it.
 Proof
Consider , , , etc. Some index is the first one with linearly dependent, and there .
Lemma 1.16 can be restated in terms of independence instead of dependence: if is linearly independent and then the set is also linearly independent if and only if Applying Lemma 1.1, we conclude that if is linearly independent and then is also linearly independent if and only if . Briefly, when passing from to a superset , to preserve linear independence we must expand the span .
Example 1.15 shows that some linearly independent sets are maximal— have as many elements as possible— in that they have no supersets that are linearly independent. By the prior paragraph, a linearly independent sets is maximal if and only if it spans the entire space, because then no vector exists that is not already in the span.
This table summarizes the interaction between the properties of independence and dependence and the relations of subset and superset.
independent 


dependent 
In developing this table we've uncovered an intimate relationship between linear independence and span. Complementing the fact that a spanning set is minimal if and only if it is linearly independent, a linearly independent set is maximal if and only if it spans the space.
In summary, we have introduced the definition of linear independence to formalize the idea of the minimality of a spanning set. We have developed some properties of this idea. The most important is Lemma 1.16, which tells us that a linearly independent set is maximal when it spans the space.
ExercisesEdit
 This exercise is recommended for all readers.
 Problem 1
Decide whether each subset of is linearly dependent or linearly independent.
 This exercise is recommended for all readers.
 Problem 2
Which of these subsets of are linearly dependent and which are independent?
 This exercise is recommended for all readers.
 Problem 3
Prove that each set is linearly independent in the vector space of all functions from to .
 and
 and
 and
 This exercise is recommended for all readers.
 Problem 4
Which of these subsets of the space of realvalued functions of one real variable is linearly dependent and which is linearly independent? (Note that we have abbreviated some constant functions; e.g., in the first item, the " " stands for the constant function .)
 Problem 5
Does the equation show that this set of functions is a linearly dependent subset of the set of all realvalued functions with domain the interval of real numbers between and ?
 Problem 6
Why does Lemma 1.4 say "distinct"?
 This exercise is recommended for all readers.
 Problem 7
Show that the nonzero rows of an echelon form matrix form a linearly independent set.
 This exercise is recommended for all readers.
 Problem 8
 Show that if the set is linearly independent set then so is the set .
 What is the relationship between the linear independence or dependence of the set and the independence or dependence of ?
 Problem 9
Example 1.10 shows that the empty set is linearly independent.
 When is a oneelement set linearly independent?
 How about a set with two elements?
 Problem 10
In any vector space , the empty set is linearly independent. What about all of ?
 Problem 11
Show that if is linearly independent then so are all of its proper subsets: , , , , , , and . Is that "only if" also?
 Problem 12
 Show that this
 Show that
 Assume that is a subset of a vector space and that is in , so that is a linear combination of vectors from . Prove that if is linearly independent then a linear combination of vectors from adding to is unique (that is, unique up to reordering and adding or taking away terms of the form ). Thus as a spanning set is minimal in this strong sense: each vector in is "hit" a minimum number of times— only once.
 Prove that it can happen when is not linearly independent that distinct linear combinations sum to the same vector.
 Problem 13
Prove that a polynomial gives rise to the zero function if and only if it is the zero polynomial. (Comment. This question is not a Linear Algebra matter, but we often use the result. A polynomial gives rise to a function in the obvious way: .)
 Problem 14
Return to Section 1.2 and redefine point, line, plane, and other linear surfaces to avoid degenerate cases.
 Problem 15
 Show that any set of four vectors in is linearly dependent.
 Is this true for any set of five? Any set of three?
 What is the most number of elements that a linearly independent subset of can have?
 This exercise is recommended for all readers.
 Problem 16
Is there a set of four vectors in , any three of which form a linearly independent set?
 Problem 17
Must every linearly dependent set have a subset that is dependent and a subset that is independent?
 Problem 18
In , what is the biggest linearly independent set you can find? The smallest? The biggest linearly dependent set? The smallest? ("Biggest" and "smallest" mean that there are no supersets or subsets with the same property.)
 This exercise is recommended for all readers.
 Problem 19
Linear independence and linear dependence are properties of sets. We can thus naturally ask how those properties act with respect to the familiar elementary set relations and operations. In this body of this subsection we have covered the subset and superset relations. We can also consider the operations of intersection, complementation, and union.
 How does linear independence relate to intersection: can an intersection of linearly independent sets be independent? Must it be?
 How does linear independence relate to complementation?
 Show that the union of two linearly independent sets need not be linearly independent.
 Characterize when the union of two linearly independent sets is linearly independent, in terms of the intersection of the span of each.
 This exercise is recommended for all readers.
 Problem 20
For Theorem 1.12,
 fill in the induction for the proof;
 give an alternate proof that starts with the empty set and builds a sequence of linearly independent subsets of the given finite set until one appears with the same span as the given set.
 Problem 21
With a little calculation we can get formulas to determine whether or not a set of vectors is linearly independent.
 Show that this subset of
 Show that this subset of
 When is this subset of
 This is an opinion question: for a set of four vectors from , must there be a formula involving the sixteen entries that determines independence of the set? (You needn't produce such a formula, just decide if one exists.)
 This exercise is recommended for all readers.
 Problem 22
 Prove that a set of two perpendicular nonzero vectors from is linearly independent when .
 What if ? ?
 Generalize to more than two vectors.
 Problem 23
Consider the set of functions from the open interval to .
 Show that this set is a vector space under the usual operations.
 Recall the formula for the sum of an infinite geometric series: for all . Why does this not express a dependence inside of the set (in the vector space that we are considering)? (Hint. Review the definition of linear combination.)
 Show that the set in the prior item is linearly independent.
This shows that some vector spaces exist with linearly independent subsets that are infinite.
 Problem 24
Show that, where is a subspace of , if a subset of is linearly independent in then is also linearly independent in . Is that "only if"?