From A First Course in Linear Algebra
Version 0.94
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
A vector space is defined as a set with two operations, meeting ten properties
(Definition VS). Just as the definition of span of a set of vectors only required
knowing how to add vectors and how to multiply vectors by scalars, so it is with
linear independence. A definition of a linear independent set of vectors in an
arbitrary vector space only requires knowing how to form linear combinations
and equating these with the zero vector. Since every vector space must
have a zero vector (Property Z), we always have a zero vector at our
disposal.
In this section we will also put a twist on the notion of the span of a set of vectors. Rather than beginning with a set of vectors and creating a subspace that is the span, we will instead begin with a subspace and look for a set of vectors whose span equals the subspace.
The combination of linear independence and spanning will be very important going forward.
Our previous definition of linear independence (Definition LI) employed a relation of linear dependence that was a linear combination on one side of an equality and a zero vector on the other side. As a linear combination in a vector space (Definition LC) depends only on vector addition and scalar multiplication, and every vector space must have a zero vector (Property Z), we can extend our definition of linear independence from the setting of to the setting of a general vector space with almost no changes. Compare these next two definitions with Definition RLDCV and Definition LICV.
Definition RLD
Relation of Linear Dependence
Suppose that is a vector
space. Given a set of vectors ,
an equation of the form
is a relation of linear dependence on . If this equation is formed in a trivial fashion, i.e. , , then we say it is a trivial relation of linear dependence on .
Definition LI
Linear Independence
Suppose that is a vector
space. The set of vectors
from
is linearly dependent if there is a relation of linear dependence on
that
is not trivial. In the case where the only relation of linear dependence on
is the trivial one, then
is a linearly independent
set of vectors.
Notice the emphasis on the word “only.” This might remind you of the definition of a nonsingular matrix, where if the matrix is employed as the coefficient matrix of a homogeneous system then the only solution is the trivial one.
Example LIP4
Linear independence in
In the vector space of polynomials with degree 4 or less,
(Example VSP) consider the set
Is this set of vectors linearly independent or dependent? Consider that
This is a nontrivial relation of linear dependence (Definition RLD) on the set and so convinces us that is linearly dependent (Definition LI).
Now, I hear you say, “Where did those scalars come from?” Do not worry about that right now, just be sure you understand why the above explanation is sufficient to prove that is linearly dependent. The remainder of the example will demonstrate how we might find these scalars if they had not been provided so readily. Let’s look at another set of vectors (polynomials) from . Let
Suppose we have a relation of linear dependence on this set,
Using our definitions of vector addition and scalar multiplication in (Example VSP), we arrive at,
Equating coefficients, we arrive at the homogeneous system of equations,
We form the coefficient matrix of this homogeneous system of equations and row-reduce to find
We expected the system to be consistent (Theorem HSC) and so can compute and Theorem CSRN tells us that the solution is unique. Since this is a homogeneous system, this unique solution is the trivial solution (Definition TSHSE), , , , . So by Definition LI the set is linearly independent.
A few observations. If we had discovered infinitely many solutions, then we could have used one of the non-trivial ones to provide a linear combination in the manner we used to show that was linearly dependent. It is important to realize that it is not interesting that we can create a relation of linear dependence with zero scalars — we can always do that — but that for , this is the only way to create a relation of linear dependence. It was no accident that we arrived at a homogeneous system of equations in this example, it is related to our use of the zero vector in defining a relation of linear dependence. It is easy to present a convincing statement that a set is linearly dependent (just exhibit a nontrivial relation of linear dependence) but a convincing statement of linear independence requires demonstrating that there is no relation of linear dependence other than the trivial one. Notice how we relied on theorems from Chapter SLE to provide this demonstration. Whew! There’s a lot going on in this example. Spend some time with it, we’ll be waiting patiently right here when you get back.
Example LIM32
Linear Independence in
Consider the two sets of vectors
and from the
vector space of all
matrices,
(Example VSM)
One set is linearly independent, the other is not. Which is which? Let’s examine first. Build a generic relation of linear dependence (Definition RLD),
Massaging the left-hand side with our definitions of vector addition and scalar multiplication in (Example VSM) we obtain,
Using our definition of matrix equality (Definition ME) and equating corresponding entries we get the homogeneous system of six equations in four variables,
Form the coefficient matrix of this homogeneous system and row-reduce to obtain
Analyzing this matrix we are led to conclude that , , , . This means there is only a trivial relation of linear dependence on the vectors of and so we call a linearly independent set (Definition LI).
So it must be that is linearly dependent. Let’s see if we can find a non-trivial relation of linear dependence on . We will begin as with , by constructing a relation of linear dependence (Definition RLD) with unknown scalars,
Massaging the left-hand side with our definitions of vector addition and scalar multiplication in (Example VSM) we obtain,
Using our definition of matrix equality (Definition ME) and equating corresponding entries we get the homogeneous system of six equations in four variables,
Form the coefficient matrix of this homogeneous system and row-reduce to obtain
Analyzing this we see that the system is consistent (we expected this since the system is homogeneous, Theorem HSC) and has free variables, namely and . This means there are infinitely many solutions, and in particular, we can find a non-trivial solution, so long as we do not pick all of our free variables to be zero. The mere presence of a nontrivial solution for these scalars is enough to conclude that is a linearly dependent set (Definition LI). But let’s go ahead and explicitly construct a non-trivial relation of linear dependence.
Choose and . There is nothing special about this choice, there are infinitely many possibilities, some “easier” than this one, just avoid picking both variables to be zero. Then we find the corresponding dependent variables to be and . So the relation of linear dependence,
is an iron-clad demonstration that is linearly dependent. Can you construct another such demonstration?
Example LIC
Linearly independent set in the crazy vector space
Is the set
linearly independent in the crazy vector space
(Example CVS)? We begin with a relation of linear independence and
massage it to a point where we can apply the definition of equality in
.
Recall the definitions of vector addition and scalar multiplication in
.
Equality in then yields the two equations,
which becomes the homogeneous system
Since the coefficient matrix of this system is nonsingular (check this!) the system has only the trivial solution . By Definition LI the set is linearly independent. Notice that even though the zero vector of is not what we might first suspected, a question about linear independence still concludes with a question about a homogeneous system of equations.
In a vector space , suppose we are given a set of vectors . Then we can immediately construct a subspace, , using Definition SS and then be assured by Theorem SSS that the construction does provide a subspace. We now turn the situation upside-down. Suppose we are first given a subspace . Can we find a set so that ? Typically is infinite and we are searching for a finite set of vectors that we can combine in linear combinations and “build” all of .
I like to think of as the raw materials that are sufficient for the construction of . If you have nails, lumber, wire, copper pipe, drywall, plywood, carpet, shingles, paint (and a few other things), then you can combine them in many different ways to create a house (or infinitely many different houses for that matter). A fast-food restaurant may have beef, chicken, beans, cheese, tortillas, taco shells and hot sauce and from this small list of ingredients build a wide variety of items for sale. Or maybe a better analogy comes from Ben Cordes — the additive primary colors (red, green and blue) can be combined to create many different colors by varying the intensity of each. The intensity is like a scalar multiple, and the combination of the three intensities is like vector addition. The three individual colors, red, yellow and blue, are the elements of the spanning set.
Because we will use terms like “spanned by” and “spanning set,” there is the potential for confusion with “the span.” Come back and reread the first paragraph of this subsection whenever you are uncertain about the difference. Here’s the working definition.
Definition TSVS
To Span a Vector Space
Suppose is a vector
space. A subset
of is a spanning
set for if
. In this case,
we also say
spans .
The definition of a spanning set requires that two sets (subspaces actually) be equal. If is a subset of , then , always. Thus it is usually only necessary to prove that . Now would be a good time to review Definition SE.
Example SSP4
Spanning set in
In Example SP4 we showed that
is a subspace of , the vector space of polynomials with degree at most (Example VSP). In this example, we will show that the set
is a spanning set for . To do this, we require that . This is an equality of sets. We can check that every polynomial in has as a root and therefore . Since is closed under addition and scalar multiplication, also.
So it remains to show that (Definition SE). To do this, begin by choosing an arbitrary polynomial in , say . This polynomial is not as arbitrary as it would appear, since we also know it must have as a root. This translates to
as a condition on .
We wish to show that is a polynomial in , that is, we want to show that can be written as a linear combination of the vectors (polynomials) in . So let’s try.
Equating coefficients (vector equality in ) gives the system of five equations in four variables,
Any solution to this system of equations will provide the linear combination we need to determine if , but we need to be convinced there is a solution for any values of that qualify to be a member of . So the question is: is this system of equations consistent? We will form the augmented matrix, and row-reduce. (We probably need to do this by hand, since the matrix is symbolic — reversing the order of the first four rows is the best way to start). We obtain a matrix in reduced row-echelon form
For your results to match our first matrix, you may find it necessary to mutiply the final row of your row-reduced matrix by the appropriate scalar, and/or add multiples of this row to some of the other rows. To obtain the second version of the matrix, the last entry of the last column has been simplified to zero according to the one condition we were able to impose on an arbitrary polynomial from . So with no leading 1’s in the last column, Theorem RCLS tells us this system is consistent. Therefore, any polynomial from can be written as a linear combination of the polynomials in , so . Therefore, and is a spanning set for by Definition TSVS.
Notice that an alternative to row-reducing the augmented matrix by hand would be to appeal to Theorem FS by expressing the column space of the coefficient matrix as a null space, and then verifying that the condition on guarantees that is in the column space, thus implying that the system is always consistent. Give it a try, we’ll wait. This has been a complicated example, but worth studying carefully.
Given a subspace and a set of vectors, as in Example SSP4 it can take some work to determine that the set actually is a spanning set. An even harder problem is to be confronted with a subspace and required to construct a spanning set with no guidance. We will now work an example of this flavor, but some of the steps will be unmotivated. Fortunately, we will have some better tools for this type of problem later on.
Example SSM22
Spanning set in
In the space of all
matrices,
consider the subspace
and find a spanning set for .
We need to construct a limited number of matrices in so that every matrix in can be expressed as a linear combination of this limited number of matrices. Suppose that is a matrix in . Then we can form a column vector with the entries of and write
Row-reducing this matrix and applying Theorem REMES we obtain the equivalent statement,
We can then express the subspace in the following equal forms,
So the set
spans by Definition TSVS.
Example SSC
Spanning set in the crazy vector space
In Example LIC we determined that the set
is linearly independent in the crazy vector space
(Example CVS).
We now show that is
a spanning set for .
Given an arbitrary vector we desire to show that it can be written as a linear combination of the elements of . In other words, are there scalars and so that
We will act as if this equation is true and try to determine just what and would be (as functions of and ).
Equality in then yields the two equations,
which becomes the linear system with a matrix representation
The coefficient matrix of this system is nonsingular, hence invertible (Theorem NI), and we can employ its inverse to find a solution (Theorem TTMI, Theorem SNCM),
We could chase through the above implications backwards and take the existence of these solutions as sufficient evidence for being a spanning set for . Instead, let us view the above as simply scratchwork and now get serious with a simple direct proof that is a spanning set. Ready? Suppose is any vector from , then compute the following linear combination using the definitions of the operations in ,
This final sequence of computations in is sufficient to demonstrate that any element of can be written (or expressed) as a linear combination of the two vectors in , so . Since the reverse inclusion is trivially true, and we say spans (Definition TSVS). Notice that this demonstration is no more or less valid if we hide from the reader our scratchwork that suggested and .
In Chapter R we will take up the matter of representations fully, where Theorem VRRB will be critical for Definition VR. We will now motivate and prove a critical theorem that tells us how to “represent” a vector. This theorem could wait, but working with it now will provide some extra insight into the nature of linearly independent spanning sets. First an example, then the theorem.
Example AVR
A vector representation
Consider the set
from the vector space . Let be the matrix whose columns are the set , and verify that is nonsingular. By Theorem NMLIC the elements of form a linearly independent set. Suppose that . Then has a (unique) solution (Theorem NMUS) and hence is consistent. By Theorem SLSLC, . Since is arbitrary, this is enough to show that , and therefore is a spanning set for (Definition TSVS). (This set comes from the columns of the coefficient matrix of Archetype B.)
Now examine the situation for a particular choice of , say . Because is a spanning set for , we know we can write as a linear combination of the vectors in ,
The nonsingularity of the matrix tells that the scalars in this linear combination are unique. More precisely, it is the linear independence of that provides the uniqueness. We will refer to the scalars , , as a “representation of relative to .” In other words, once we settle on as a linearly independent set that spans , the vector is recoverable just by knowing the scalars , , (use these scalars in a linear combination of the vectors in ). This is all an illustration of the following important theorem, which we prove in the setting of a general vector space.
Theorem VRRB
Vector Representation Relative to a Basis
Suppose that is a vector
space and is a linearly
independent set that spans .
Let be any vector
in . Then there
exist unique scalars
such that
Proof That can be written as a linear combination of the vectors in follows from the spanning property of the set (Definition TSVS). This is good, but not the meat of this theorem. We now know that for any choice of the vector there exist some scalars that will create as a linear combination of the basis vectors. The real question is: Is there more than one way to write as a linear combination of ? Are the scalars unique? (Technique U)
Assume there are two ways to express as a linear combination of . In other words there exist scalars and so that
Then notice that
But this is a relation of linear dependence on a linearly independent set of vectors (Definition RLD)! Now we are using the other assumption about , that is a linearly independent set. So by Definition LI it must happen that the scalars are all zero. That is,
And so we find that the scalars are unique.
This is a very typical use of the hypothesis that a set is linearly independent — obtain a relation of linear dependence and then conclude that the scalars must all be zero. The result of this theorem tells us that we can write any vector in a vector space as a linear combination of the vectors in a linearly independent spanning set, but only just. There is only enough raw material in the spanning set to write each vector one way as a linear combination. So in this sense, we could call a linearly independent spanning set a “minimal spanning set.” These sets are so important that we will give them a simpler name (“basis”) and explore their properties further in the next section.
is linearly independent and spans . Write the vector a linear combination of the elements of . How many ways are there to answer this question?
C20 In the vector space of matrices, , determine if the set below is linearly independent.
Contributed by Robert Beezer Solution [879]
C21 In the crazy vector space
(Example CVS), is the set
linearly independent?
Contributed by Robert Beezer Solution [880]
C22 In the vector space of polynomials , determine if the set is linearly independent or linearly dependent.
Contributed by Robert Beezer Solution [882]
C23 Determine if the set
is linearly independent in the crazy vector space
(Example CVS).
Contributed by Robert Beezer Solution [883]
C30 In Example LIM32, find another nontrivial relation of linear dependence on the linearly
dependent set of
matrices, .
Contributed by Robert Beezer
C40 Determine if the set
spans the vector space of polynomials with degree 4 or less,
.
Contributed by Robert Beezer Solution [884]
C41 The set is a subspace of , the vector space of all matrices. Prove that is a spanning set for .
Contributed by Robert Beezer Solution [884]
C42 Determine if the set
spans the crazy vector space
(Example CVS).
Contributed by Robert Beezer Solution [885]
M10 Halfway through Example SSP4, we need to show that the system of equations
is consistent for every choice of the vector of constants for which .
Express the column space of the coefficient matrix of this system as a null
space, using Theorem FS. From this use Theorem CSCS to establish that
the system is always consistent. Notice that this approach removes from
Example SSP4 the need to row-reduce a symbolic matrix.
Contributed by Robert Beezer Solution [887]
C20 Contributed by Robert Beezer Statement [875]
Begin with a relation of linear dependence on the vectors in
and
massage it according to the definitions of vector addition and scalar multiplication
in ,
By our definition of matrix equality (Definition ME) we arrive at a homogeneous system of linear equations,
The coefficient matrix of this system row-reduces to the matrix,
and from this we conclude that the only solution is . Since the relation of linear dependence (Definition RLD) is trivial, the set is linearly independent (Definition LI).
C21 Contributed by Robert Beezer Statement [875]
We begin with a relation of linear dependence using unknown scalars
and
. We
wish to know if these scalars must both be zero. Recall that the zero vector in
is
and
that the definitions of vector addition and scalar multiplication are not what we
might expect.
From this we obtain two equalities, which can be converted to a homogeneous system of equations,
This homogeneous system has a singular coefficient matrix (Theorem SMZD), and so has more than just the trivial solution (Definition NM). Any nontrivial solution will give us a nontrivial relation of linear dependence on . So is linearly dependent (Definition LI).
C22 Contributed by Robert Beezer Statement [875]
Begin with a relation of linear dependence (Definition RLD),
Massage according to the definitions of scalar multiplication and vector addition in the definition of (Example VSP) and use the zero vector dro this vector space,
The definition of the equality of polynomials allows us to deduce the following four equations,
Row-reducing the coefficient matrix of this homogeneous system leads to the unique solution . So the only relation of linear dependence on is the trivial one, and this is linear independence for (Definition LI).
C23 Contributed by Robert Beezer Statement [876]
Notice, or discover, that the following gives a nontrivial relation of linear dependence
on in
, so by Definition LI,
the set
is linearly dependent.
C40 Contributed by Robert Beezer Statement [876]
The vector space
has dimension 5 by Theorem DP. Since
contains only 3 vectors,
and , Theorem G
tells us that
does not span .
C41 Contributed by Robert Beezer Statement [876]
We want to show that
(Definition TSVS), which is an equality of sets (Definition SE).
First, show that . Begin by checking that each of the three matrices in is a member of the set . Then, since is a vector space, the closure properties (Property AC, Property SC) guarantee that every linear combination of elements of remains in .
Second, show that . We want to convince ourselves that an arbitrary element of is a linear combination of elements of . Choose
The values of are not totally arbitary, since membership in requires that . Now, rewrite as follows,
C42 Contributed by Robert Beezer Statement [877]
We will try to show that
spans . Let
be an arbitrary
element of and
search for scalars
and
such that
Equality in leads to the system
This system has a singular coefficient matrix whose column space is simply . So any choice of and that causes the column vector to lie outside the column space will lead to an inconsistent system, and hence create an element that is not in the span of . So does not span .
For example, choose and , and then we can see that and we know that cannot be written as a linear combination of the vectors in . A shorter solution might begin by asserting that is not in and then establishing this claim alone.
M10 Contributed by Robert Beezer Statement [877]
Theorem FS provides the matrix
and so if denotes the coefficient matrix of the system, then . The single homogeneous equation in is equivalent to the condition on the vector of constants (use as variables and then multiply by 16).