Linear Algebra/Definition and Examples of Linear Independence/Solutions

SolutionsEdit

This exercise is recommended for all readers.
Problem 1

Decide whether each subset of   is linearly dependent or linearly independent.

  1.  
  2.  
  3.  
  4.  
Answer

For each of these, when the subset is independent it must be proved, and when the subset is dependent an example of a dependence must be given.

  1. It is dependent. Considering
     
    gives rise to this linear system.
     
    Gauss' method
     
    yields a free variable, so there are infinitely many solutions. For an example of a particular dependence we can set   to be, say,  . Then we get   and  .
  2. It is dependent. The linear system that arises here
     
    has infinitely many solutions. We can get a particular solution by taking   to be, say,  , and back-substituting to get the resulting   and  .
  3. It is linearly independent. The system
     
    has only the solution   and  . (We could also have gotten the answer by inspection— the second vector is obviously not a multiple of the first, and vice versa.)
  4. It is linearly dependent. The linear system
     
    has more unknowns than equations, and so Gauss' method must end with at least one variable free (there can't be a contradictory equation because the system is homogeneous, and so has at least the solution of all zeroes). To exhibit a combination, we can do the reduction
     
    and take, say,  . Then we have that  ,  , and  .
This exercise is recommended for all readers.
Problem 2

Which of these subsets of   are linearly dependent and which are independent?

  1.  
  2.  
  3.  
  4.  
Answer

In the cases of independence, that must be proved. Otherwise, a specific dependence must be produced. (Of course, dependences other than the ones exhibited here are possible.)

  1. This set is independent. Setting up the relation   gives a linear system
     
    with only one solution:  ,  , and  .
  2. This set is independent. We can see this by inspection, straight from the definition of linear independence. Obviously neither is a multiple of the other.
  3. This set is linearly independent. The linear system reduces in this way
     
    to show that there is only the solution  ,  , and  .
  4. This set is linearly dependent. The linear system
     
    must, after reduction, end with at least one variable free (there are more variables than equations, and there is no possibility of a contradictory equation because the system is homogeneous). We can take the free variables as parameters to describe the solution set. We can then set the parameter to a nonzero value to get a nontrivial linear relation.
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  .

  1.   and  
  2.   and  
  3.   and  
Answer

Let   be the zero function  , which is the additive identity in the vector space under discussion.

  1. This set is linearly independent. Consider  . Plugging in   and   gives a linear system
     
    with the unique solution  ,  .
  2. This set is linearly independent. Consider   and plug in   and   to get
     
    which obviously gives that  ,  .
  3. This set is also linearly independent. Considering   and plugging in   and  
     
    gives that   and  .
This exercise is recommended for all readers.
Problem 4

Which of these subsets of the space of real-valued 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  .)

  1.  
  2.  
  3.  
  4.  
  5.  
  6.  
Answer

In each case, that the set is independent must be proved, and that it is dependent must be shown by exhibiting a specific dependence.

  1. This set is dependent. The familiar relation   shows that   is satisfied by   and  .
  2. This set is independent. Consider the relationship   (that " " is the zero function). Taking  ,   and   gives this system.
     
    whose only solution is  ,  , and  .
  3. By inspection, this set is independent. Any dependence   is not possible since the cosine function is not a multiple of the identity function (we are applying Corollary 1.17).
  4. By inspection, we spot that there is a dependence. Because  , we get that   is satisfied by   and  .
  5. This set is dependent. The easiest way to see that is to recall the trigonometric relationship  . (Remark. A person who doesn't recall this, and tries some  's, simply never gets a system leading to a unique solution, and never gets to conclude that the set is independent. Of course, this person might wonder if they simply never tried the right set of  's, but a few tries will lead most people to look instead for a dependence.)
  6. This set is dependent, because it contains the zero object in the vector space, the zero polynomial.
Problem 5

Does the equation   show that this set of functions   is a linearly dependent subset of the set of all real-valued functions with domain the interval   of real numbers between   and  ?

Answer

No, that equation is not a linear relationship. In fact this set is independent, as the system arising from taking   to be  ,   and   shows.

Problem 6

Why does Lemma 1.4 say "distinct"?

Answer

To emphasize that the equation   does not make the set dependent.

This exercise is recommended for all readers.
Problem 7

Show that the nonzero rows of an echelon form matrix form a linearly independent set.

Answer

We have already showed this: the Linear Combination Lemma and its corollary state that in an echelon form matrix, no nonzero row is a linear combination of the others.

This exercise is recommended for all readers.
Problem 8
  1. Show that if the set   is linearly independent set then so is the set  .
  2. What is the relationship between the linear independence or dependence of the set   and the independence or dependence of  ?
Answer
  1. Assume that the set   is linearly independent, so that any relationship   leads to the conclusion that  ,  , and  . Consider the relationship  . Rewrite it to get  . Taking   to be  , taking   to be  , and taking   to be   we have this system.
     
    Conclusion: the  's are all zero, and so the set is linearly independent.
  2. The second set is dependent
     
    whether or not the first set is independent.
Problem 9

Example 1.10 shows that the empty set is linearly independent.

  1. When is a one-element set linearly independent?
  2. How about a set with two elements?
Answer
  1. A singleton set   is linearly independent if and only if  . For the "if" direction, with  , we can apply Lemma 1.4 by considering the relationship   and noting that the only solution is the trivial one:  . For the "only if" direction, just recall that Example 1.11 shows that   is linearly dependent, and so if the set   is linearly independent then  . (Remark. Another answer is to say that this is the special case of Lemma 1.16 where  .)
  2. A set with two elements is linearly independent if and only if neither member is a multiple of the other (note that if one is the zero vector then it is a multiple of the other, so this case is covered). This is an equivalent statement: a set is linearly dependent if and only if one element is a multiple of the other. The proof is easy. A set   is linearly dependent if and only if there is a relationship   with either   or   (or both). That holds if and only if   or   (or both).
Problem 10

In any vector space  , the empty set is linearly independent. What about all of  ?

Answer

This set is linearly dependent set because it contains the zero vector.

Problem 11

Show that if   is linearly independent then so are all of its proper subsets:  ,  ,  ,  , ,  , and  . Is that "only if" also?

Answer

The "if" half is given by Lemma 1.14. The converse (the "only if" statement) does not hold. An example is to consider the vector space   and these vectors.

 
Problem 12
  1. Show that this
     
    is a linearly independent subset of  .
  2. Show that
     
    is in the span of   by finding   and   giving a linear relationship.
     
    Show that the pair   is unique.
  3. 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.
  4. Prove that it can happen when   is not linearly independent that distinct linear combinations sum to the same vector.
Answer
  1. The linear system arising from
     
    has the unique solution   and  .
  2. The linear system arising from
     
    has the unique solution   and  .
  3. Suppose that   is linearly independent. Suppose that we have both   and   (where the vectors are members of  ). Now,
     
    can be rewritten in this way.
     
    Possibly some of the  's equal some of the  's; we can combine the associated coefficients (i.e., if   then   can be rewritten as  ). That equation is a linear relationship among distinct (after the combining is done) members of the set  . We've assumed that   is linearly independent, so all of the coefficients are zero. If   is such that   does not equal any   then   is zero. If   is such that   does not equal any   then   is zero. In the final case, we have that   and so  . Therefore, the original two sums are the same, except perhaps for some   or   terms that we can neglect.
  4. This set is not linearly independent:
     
    and these two linear combinations give the same result
     
    Thus, a linearly dependent set might have indistinct sums. In fact, this stronger statement holds: if a set is linearly dependent then it must have the property that there are two distinct linear combinations that sum to the same vector. Briefly, where   then multiplying both sides of the relationship by two gives another relationship. If the first relationship is nontrivial then the second is also.
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:  .)

Answer

In this "if and only if" statement, the "if" half is clear— if the polynomial is the zero polynomial then the function that arises from the action of the polynomial must be the zero function  . For "only if" we write  . Plugging in zero   gives that  . Taking the derivative and plugging in zero   gives that  . Similarly we get that each   is zero, and   is the zero polynomial.

Problem 14

Return to Section 1.2 and redefine point, line, plane, and other linear surfaces to avoid degenerate cases.

Answer

The work in this section suggests that an  -dimensional non-degenerate linear surface should be defined as the span of a linearly independent set of   vectors.

Problem 15
  1. Show that any set of four vectors in   is linearly dependent.
  2. Is this true for any set of five? Any set of three?
  3. What is the most number of elements that a linearly independent subset of   can have?
Answer
  1. For any  , ...,  ,
     
    yields a linear system
     
    that has infinitely many solutions (Gauss' method leaves at least two variables free). Hence there are nontrivial linear relationships among the given members of  .
  2. Any set five vectors is a superset of a set of four vectors, and so is linearly dependent. With three vectors from  , the argument from the prior item still applies, with the slight change that Gauss' method now only leaves at least one variable free (but that still gives infinitely many solutions).
  3. The prior item shows that no three-element subset of   is independent. We know that there are two-element subsets of   that are independent— one is
     
    and so the answer is two.
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?

Answer

Yes; here is one.

 
Problem 17

Must every linearly dependent set have a subset that is dependent and a subset that is independent?

Answer

Yes. The two improper subsets, the entire set and the empty subset, serve as examples.

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.)

Answer

In   the biggest linearly independent set has four vectors. There are many examples of such sets, this is one.

 

To see that no set with five or more vectors can be independent, set up

 

and note that the resulting linear system

 

has four equations and five unknowns, so Gauss' method must end with at least one   variable free, so there are infinitely many solutions, and so the above linear relationship among the four-tall vectors has more solutions than just the trivial solution.

The smallest linearly independent set is the empty set.

The biggest linearly dependent set is  . The smallest is  .

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.

  1. How does linear independence relate to intersection: can an intersection of linearly independent sets be independent? Must it be?
  2. How does linear independence relate to complementation?
  3. Show that the union of two linearly independent sets need not be linearly independent.
  4. Characterize when the union of two linearly independent sets is linearly independent, in terms of the intersection of the span of each.
Answer
  1. The intersection of two linearly independent sets   must be linearly independent as it is a subset of the linearly independent set   (as well as the linearly independent set   also, of course).
  2. The complement of a linearly independent set is linearly dependent as it contains the zero vector.
  3. We must produce an example. One, in  , is
     
    since the linear dependence of   is easily seen.
  4. The union of two linearly independent sets   is linearly independent if and only if their spans have a trivial intersection  . To prove that, assume that   and   are linearly independent subsets of some vector space. For the "only if" direction, assume that the intersection of the spans is trivial  . Consider the set  . Any linear relationship   gives  . The left side of that equation sums to a vector in  , and the right side is a vector in  . Therefore, since the intersection of the spans is trivial, both sides equal the zero vector. Because   is linearly independent, all of the  's are zero. Because   is linearly independent, all of the  's are zero. Thus, the original linear relationship among members of   only holds if all of the coefficients are zero. That shows that   is linearly independent. For the "if" half we can make the same argument in reverse. If the union   is linearly independent, that is, if the only solution to   is the trivial solution  , ...,  , then any vector   in the intersection of the spans   must be the zero vector because each scalar is zero.
This exercise is recommended for all readers.
Problem 20

For Theorem 1.12,

  1. fill in the induction for the proof;
  2. 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.
Answer
  1. We do induction on the number of vectors in the finite set  . The base case is that   has no elements. In this case   is linearly independent and there is nothing to check— a subset of   that has the same span as   is   itself. For the inductive step assume that the theorem is true for all sets of size  ,  , ...,   in order to prove that it holds when   has   elements. If the  -element set   is linearly independent then the theorem is trivial, so assume that it is dependent. By Corollary 1.17 there is an   that is a linear combination of other vectors in  . Define   and note that   has the same span as   by Lemma 1.1. The set   has   elements and so the inductive hypothesis applies to give that it has a linearly independent subset with the same span. That subset of   is the desired subset of  .
  2. Here is a sketch of the argument. The induction argument details have been left out. If the finite set   is empty then there is nothing to prove. If   then the empty subset will do. Otherwise, take some nonzero vector   and define  . If   then this proof is finished by noting that   is linearly independent. If not, then there is a nonzero vector   (if every   is in   then  ). Define  . If   then this proof is finished by using Theorem 1.17 to show that   is linearly independent. Repeat the last paragraph until a set with a big enough span appears. That must eventually happen because   is finite, and   will be reached at worst when every vector from   has been used.
Problem 21

With a little calculation we can get formulas to determine whether or not a set of vectors is linearly independent.

  1. Show that this subset of  
     
    is linearly independent if and only if  .
  2. Show that this subset of  
     
    is linearly independent iff  .
  3. When is this subset of  
     
    linearly independent?
  4. 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.)
Answer
  1. Assuming first that  ,
     
    gives
     
    which has a solution if and only if   (we've assumed in this case that  , and so back substitution yields a unique solution). The   case is also not hard— break it into the   and   subcases and note that in these cases  . Comment. An earlier exercise showed that a two-vector set is linearly dependent if and only if either vector is a scalar multiple of the other. That can also be used to make the calculation.
  2. The equation
     
    gives rise to a homogeneous linear system. We proceed by writing it in matrix form and applying Gauss' method. We first reduce the matrix to upper-triangular. Assume that  .
     
    (where we've assumed for the moment that   in order to do the row reduction step). Then, under the assumptions, we get this.
     
    shows that the original system is nonsingular if and only if the   entry is nonzero. This fraction is defined because of the   assumption, and it will equal zero if and only if its numerator equals zero. We next worry about the assumptions. First, if   but   then we swap
     
    and conclude that the system is nonsingular if and only if either   or  . That's the same as asking that their product be zero:
     
    (in going from the first line to the second we've applied the case assumption that   by substituting   for  ). Since we are assuming that  , we have that  . With   we can rewrite this to fit the form we need: in this   and   case, the given system is nonsingular when  , as required. The remaining cases have the same character. Do the   but   case and the   and   but   case by first swapping rows and then going on as above. The  ,  , and   case is easy— a set with a zero vector is linearly dependent, and the formula comes out to equal zero.
  3. It is linearly dependent if and only if either vector is a multiple of the other. That is, it is not independent iff
     
    (or both) for some scalars   and  . Eliminating   and   in order to restate this condition only in terms of the given letters  ,  ,  ,  ,  ,  , we have that it is not independent— it is dependent— iff  .
  4. Dependence or independence is a function of the indices, so there is indeed a formula (although at first glance a person might think the formula involves cases: "if the first component of the first vector is zero then ...", this guess turns out not to be correct).
This exercise is recommended for all readers.
Problem 22
  1. Prove that a set of two perpendicular nonzero vectors from   is linearly independent when  .
  2. What if  ?  ?
  3. Generalize to more than two vectors.
Answer

Recall that two vectors from   are perpendicular if and only if their dot product is zero.

  1. Assume that   and   are perpendicular nonzero vectors in  , with  . With the linear relationship  , apply   to both sides to conclude that  . Because   we have that  . A similar application of   shows that  .
  2. Two vectors in   are perpendicular if and only if at least one of them is zero. We define   to be a trivial space, and so both   and   are the zero vector.
  3. The right generalization is to look at a set   of vectors that are mutually orthogonal (also called pairwise perpendicular): if   then   is perpendicular to  . Mimicking the proof of the first item above shows that such a set of nonzero vectors is linearly independent.
Problem 23

Consider the set of functions from the open interval   to  .

  1. Show that this set is a vector space under the usual operations.
  2. 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.)
  3. 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.

Answer
  1. This check is routine.
  2. The summation is infinite (has infinitely many summands). The definition of linear combination involves only finite sums.
  3. No nontrivial finite sum of members of   adds to the zero object: assume that
     
    (any finite sum uses a highest power, here  ). Multiply both sides by   to conclude that each coefficient is zero, because a polynomial describes the zero function only when it is the zero polynomial.
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"?

Answer

It is both "if" and "only if".

Let   be a subset of the subspace   of the vector space  . The assertion that any linear relationship   among members of   must be the trivial relationship  , ...,   is a statement that holds in   if and only if it holds in  , because the subspace   inherits its addition and scalar multiplication operations from  .