Section S  Subspaces

From A First Course in Linear Algebra
Version 2.20
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections. Here’s the definition.

Definition S
Subspace
Suppose that V and W are two vector spaces that have identical definitions of vector addition and scalar multiplication, and that W is a subset of V , W V . Then W is a subspace of V .

Lets look at an example of a vector space inside another vector space.

Example SC3
A subspace of 3
We know that 3 is a vector space (Example VSCV). Consider the subset,

W = x1 x2 x3 2x1 5x2 + 7x3 = 0

It is clear that W 3, since the objects in W are column vectors of size 3. But is W a vector space? Does it satisfy the ten properties of Definition VS when we use the same operations? That is the main question. Suppose x = x1 x2 x3 and y = y1 y2 y3 are vectors from W. Then we know that these vectors cannot be totally arbitrary, they must have gained membership in W by virtue of meeting the membership test. For example, we know that x must satisfy 2x1 5x2 + 7x3 = 0 while y must satisfy 2y1 5y2 + 7y3 = 0. Our first property (Property AC) asks the question, is x + y W? When our set of vectors was 3, this was an easy question to answer. Now it is not so obvious. Notice first that

x+y = x1 x2 x3 + y1 y2 y3 = x1 + y1 x2 + y2 x3 + y3

and we can test this vector for membership in W as follows,

2(x1 + y1) 5(x2 + y2) + 7(x3 + y3) = 2x1 + 2y1 5x2 5y2 + 7x3 + 7y3 = (2x1 5x2 + 7x3) + (2y1 5y2 + 7y3) = 0 + 0 x W,y W = 0

and by this computation we see that x + y W. One property down, nine to go.

If α is a scalar and x W, is it always true that αx W? This is what we need to establish Property SC. Again, the answer is not as obvious as it was when our set of vectors was all of 3. Let’s see.

αx = α x1 x2 x3 = αx1 αx2 αx3

and we can test this vector for membership in W with

2(αx1) 5(αx2) + 7(αx3) = α(2x1 5x2 + 7x3) = α0 x W = 0

and we see that indeed αx W. Always.

If W has a zero vector, it will be unique (Theorem ZVU). The zero vector for 3 should also perform the required duties when added to elements of W. So the likely candidate for a zero vector in W is the same zero vector that we know 3 has. You can check that 0 = 0 0 0 is a zero vector in W too (Property Z).

With a zero vector, we can now ask about additive inverses (Property AI). As you might suspect, the natural candidate for an additive inverse in W is the same as the additive inverse from 3. However, we must insure that these additive inverses actually are elements of W. Given x W, is x W?

x = x1 x2 x3

and we can test this vector for membership in W with

2(x1) 5(x2) + 7(x3) = (2x1 5x2 + 7x3) = 0 x W = 0

and we now believe that x W.

Is the vector addition in W commutative (Property C)? Is x + y = y + x? Of course! Nothing about restricting the scope of our set of vectors will prevent the operation from still being commutative. Indeed, the remaining five properties are unaffected by the transition to a smaller set of vectors, and so remain true. That was convenient.

So W satisfies all ten properties, is therefore a vector space, and thus earns the title of being a subspace of 3.

Subsection TS: Testing Subspaces

In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.

Theorem TSS
Testing Subsets for Subspaces
Suppose that V is a vector space and W is a subset of V , W V . Endow W with the same operations as V . Then W is a subspace if and only if three conditions are met

  1. W is non-empty, W.
  2. If x W and y W, then x + y W.
  3. If α and x W, then αx W.

Proof   ( ) We have the hypothesis that W is a subspace, so by Definition VS we know that W contains a zero vector. This is enough to show that W. Also, since W is a vector space it satisfies the additive and scalar multiplication closure properties, and so exactly meets the second and third conditions. If that was easy, the the other direction might require a bit more work.

( ) We have three properties for our hypothesis, and from this we should conclude that W has the ten defining properties of a vector space. The second and third conditions of our hypothesis are exactly Property AC and Property SC. Our hypothesis that V is a vector space implies that Property C, Property AA, Property SMA, Property DVA, Property DSA and Property O all hold. They continue to be true for vectors from W since passing to a subset, and keeping the operation the same, leaves their statements unchanged. Eight down, two to go.

Suppose x W. Then by the third part of our hypothesis (scalar closure), we know that (1)x W. By Theorem AISM (1)x = x, so together these statements show us that x W. x is the additive inverse of x in V , but will continue in this role when viewed as element of the subset W. So every element of W has an additive inverse that is an element of W and Property AI is completed. Just one property left.

While we have implicitly discussed the zero vector in the previous paragraph, we need to be certain that the zero vector (of V ) really lives in W. Since W is non-empty, we can choose some vector z W. Then by the argument in the previous paragraph, we know z W. Now by Property AI for V and then by the second part of our hypothesis (additive closure) we see that

0 = z + (z) W

So W contain the zero vector from V . Since this vector performs the required duties of a zero vector in V , it will continue in that role as an element of W. This gives us, Property Z, the final property of the ten required. (Sarah Fellez contributed to this proof.)

So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is “a non-empty subset (of a vector space) that is closed under vector addition and scalar multiplication.”

You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.

Example SP4
A subspace of P4
P4 is the vector space of polynomials with degree at most 4 (Example VSP). Define a subset W as

W = p(x)p P4,p(2) = 0

so W is the collection of those polynomials (with degree 4 or less) whose graphs cross the x-axis at x = 2. Whenever we encounter a new set it is a good idea to gain a better understanding of the set by finding a few elements in the set, and a few outside it. For example x2 x 2 W, while x4 + x3 7W.

Is W nonempty? Yes, x 2 W.

Additive closure? Suppose p W and q W. Is p + q W? p and q are not totally arbitrary, we know that p(2) = 0 and q(2) = 0. Then we can check p + q for membership in W,

(p + q)(2) = p(2) + q(2)  Addition in P4 = 0 + 0 p W,q W = 0

so we see that p + q qualifies for membership in W.

Scalar multiplication closure? Suppose that α and p W. Then we know that p(2) = 0. Testing αp for membership,

(αp)(2) = αp(2)  Scalar multiplication in P4 = α0 p W = 0

so αp W.

We have shown that W meets the three conditions of Theorem TSS and so qualifies as a subspace of P4. Notice that by Definition S we now know that W is also a vector space. So all the properties of a vector space (Definition VS) and the theorems of Section VS apply in full.

Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the ones presented in Subsection VS.EVS.

It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see Technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the “non-empty” condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP. Notice also that a violation need only be for a specific vector or pair of vectors.

Example NSC2Z
A non-subspace in 2, zero vector
Consider the subset W below as a candidate for being a subspace of 2

W = x1 x2 3x1 5x2 = 12

The zero vector of 2, 0 = 0 0 will need to be the zero vector in W also. However, 0W since 3(0) 5(0) = 012. So W has no zero vector and fails Property Z of Definition VS. This subspace also fails to be closed under addition and scalar multiplication. Can you find examples of this?

Example NSC2A
A non-subspace in 2, additive closure
Consider the subset X below as a candidate for being a subspace of 2

X = x1 x2 x1x2 = 0

You can check that 0 X, so the approach of the last example will not get us anywhere. However, notice that x = 1 0 X and y = 0 1 X. Yet

x+y = 1 0 + 0 1 = 1 1 X

So X fails the additive closure requirement of either Property AC or Theorem TSS, and is therefore not a subspace.

Example NSC2S
A non-subspace in 2, scalar multiplication closure
Consider the subset Y below as a candidate for being a subspace of 2

Y = x1 x2 x1 ,x2

is the set of integers, so we are only allowing “whole numbers” as the constituents of our vectors. Now, 0 Y , and additive closure also holds (can you prove these claims?). So we will have to try something different. Note that α = 1 2 and 2 3 Y , but

αx = 1 2 2 3 = 1 3 2 Y

So Y fails the scalar multiplication closure requirement of either Property SC or Theorem TSS, and is therefore not a subspace.

There are two examples of subspaces that are trivial. Suppose that V is any vector space. Then V is a subset of itself and is a vector space. By Definition S, V qualifies as a subspace of itself. The set containing just the zero vector Z = 0 is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.

Definition TS
Trivial Subspaces
Given the vector space V , the subspaces V and 0 are each called a trivial subspace.

We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem.

Theorem NSMS
Null Space of a Matrix is a Subspace
Suppose that A is an m × n matrix. Then the null space of A, NA, is a subspace of n.

Proof   We will examine the three requirements of Theorem TSS. Recall that NA = x nAx = 0.

First, 0 NA, which can be inferred as a consequence of Theorem HSC. So NA.

Second, check additive closure by supposing that x NA and y NA. So we know a little something about x and y: Ax = 0 and Ay = 0, and that is all we know. Question: Is x + y NA? Let’s check.

A(x + y) = Ax + Ay  Theorem MMDAA = 0 + 0 x NA,y NA = 0  Theorem VSPCV

So, yes, x + y qualifies for membership in NA.

Third, check scalar multiplication closure by supposing that α and x NA. So we know a little something about x: Ax = 0, and that is all we know. Question: Is αx NA? Let’s check.

A(αx) = α(Ax)  Theorem MMSMM = α0 x NA = 0  Theorem ZVSM

So, yes, αx qualifies for membership in NA.

Having met the three conditions in Theorem TSS we can now say that the null space of a matrix is a subspace (and hence a vector space in its own right!).

Here is an example where we can exercise Theorem NSMS.

Example RSNS
Recasting a subspace as a null space
Consider the subset of 5 defined as

W = x1 x2 x3 x4 x5 3x1 + x2 5x3 + 7x4 + x5 = 0, 4x1 + 6x2 + 3x3 6x4 5x5 = 0, 2x1 + 4x2 + 7x4 + x5 = 0

It is possible to show that W is a subspace of 5 by checking the three conditions of Theorem TSS directly, but it will get tedious rather quickly. Instead, give W a fresh look and notice that it is a set of solutions to a homogeneous system of equations. Define the matrix

A = 3 15 7 1 4 6 3 6 5 24 0 7 1

and then recognize that W = NA. By Theorem NSMS we can immediately see that W is a subspace. Boom!

Subsection TSS: The Span of a Set

The span of a set of column vectors got a heavy workout in Chapter V and Chapter M. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you haven’t already, compare them with Definition LCCV and Definition SSCV.

Definition LC
Linear Combination
Suppose that V is a vector space. Given n vectors u1,u2,u3,,un and n scalars α1,α2,α3,,αn, their linear combination is the vector

α1u1 + α2u2 + α3u3 + + αnun.

Example LCM
A linear combination of matrices
In the vector space M23 of 2 × 3 matrices, we have the vectors

x = 132 2 0 7 y = 312 5 5 1 z = 424 1 1 1

and we can form linear combinations such as

2x + 4y + (1)z = 2 132 2 0 7 + 4 312 5 5 1 + (1) 424 1 1 1 = 264 4 0 14 + 1248 20 20 4 + 42 4 1 1 1 = 10 0 8 23 19 17  or, 4x 2y + 3z = 4 132 2 0 7 2 312 5 5 1 + 3 424 1 1 1 = 4128 8 0 28 + 6 2 410 10 2 + 12612 3 3 3 = 102024 1 7 29

When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.

Definition SS
Span of a Set
Suppose that V is a vector space. Given a set of vectors S = {u1,u2,u3,,ut}, their span, S, is the set of all possible linear combinations of u1,u2,u3,,ut. Symbolically,

S = α1u1 + α2u2 + α3u3 + + αtutαi ,1 i t = i=1tα iuiαi ,1 i t

Theorem SSS
Span of a Set is a Subspace
Suppose V is a vector space. Given a set of vectors S = {u1,u2,u3,,ut} V , their span, S, is a subspace.

Proof   We will verify the three conditions of Theorem TSS. First,

0 = 0 + 0 + 0 + + 0  Property Z for V = 0u1 + 0u2 + 0u3 + + 0ut  Theorem ZSSM

So we have written 0 as a linear combination of the vectors in S and by Definition SS, 0 S and therefore S.

Second, suppose x S and y S. Can we conclude that x + y S? What do we know about x and y by virtue of their membership in S? There must be scalars from , α1,α2,α3,,αt and β1,β2,β3,,βt so that

x = α1u1 + α2u2 + α3u3 + + αtut y = β1u1 + β2u2 + β3u3 + + βtut

Then

x + y = α1u1 + α2u2 + α3u3 + + αtut + β1u1 + β2u2 + β3u3 + + βtut = α1u1 + β1u1 + α2u2 + β2u2 + α3u3 + β3u3 + + αtut + βtut  Property AAProperty C = (α1 + β1)u1 + (α2 + β2)u2 + (α3 + β3)u3 + + (αt + βt)ut  Property DSA

Since each αi + βi is again a scalar from we have expressed the vector sum x + y as a linear combination of the vectors from S, and therefore by Definition SS we can say that x + y S.

Third, suppose α and x S. Can we conclude that αx S? What do we know about x by virtue of its membership in S? There must be scalars from , α1,α2,α3,,αt so that

x = α1u1 + α2u2 + α3u3 + + αtut

Then

αx = α α1u1 + α2u2 + α3u3 + + αtut = α(α1u1) + α(α2u2) + α(α3u3) + + α(αtut)  Property DVA = (αα1)u1 + (αα2)u2 + (αα3)u3 + + (ααt)ut  Property SMA

Since each ααi is again a scalar from we have expressed the scalar multiple αx as a linear combination of the vectors from S, and therefore by Definition SS we can say that αx S.

With the three conditions of Theorem TSS met, we can say that S is a subspace (and so is also vector space, Definition VS). (See Exercise SS.T20, Exercise SS.T21, Exercise SS.T22.)

Example SSP
Span of a set of polynomials
In Example SP4 we proved that

W = p(x)p P4,p(2) = 0

is a subspace of P4, the vector space of polynomials of degree at most 4. Since W is a vector space itself, let’s construct a span within W. First let

S = x4 4x3 + 5x2 x 2,2x4 3x3 6x2 + 6x + 4

and verify that S is a subset of W by checking that each of these two polynomials has x = 2 as a root. Now, if we define U = S, then Theorem SSS tells us that U is a subspace of W. So quite quickly we have built a chain of subspaces, U inside W, and W inside P4.

Rather than dwell on how quickly we can build subspaces, let’s try to gain a better understanding of just how the span construction creates subspaces, in the context of this example. We can quickly build representative elements of U,

3(x44x3+5x2x2)+5(2x43x36x2+6x+4) = 13x427x315x2+27x+14

and

(2)(x44x3+5x2x2)+8(2x43x36x2+6x+4) = 14x416x358x2+50x+36

and each of these polynomials must be in W since it is closed under addition and scalar multiplication. But you might check for yourself that both of these polynomials have x = 2 as a root.

I can tell you that y = 3x4 7x3 x2 + 7x 2 is not in U, but would you believe me? A first check shows that y does have x = 2 as a root, but that only shows that y W. What does y have to do to gain membership in U = S? It must be a linear combination of the vectors in S, x4 4x3 + 5x2 x 2 and 2x4 3x3 6x2 + 6x + 4. So let’s suppose that y is such a linear combination,

y = 3x4 7x3 x2 + 7x 2 = α1(x4 4x3 + 5x2 x 2) + α 2(2x4 3x3 6x2 + 6x + 4) = (α1 + 2α2)x4 + (4α 1 3α2)x3 + (5α 1 6α2)x2 + (α 1 + 6α2)x (2α1 + 4α2)

Notice that operations above are done in accordance with the definition of the vector space of polynomials (Example VSP). Now, if we equate coefficients, which is the definition of equality for polynomials, then we obtain the system of five linear equations in two variables

α1 + 2α2 = 3 4α1 3α2 = 7 5α1 6α2 = 1 α1 + 6α2 = 7 2α1 + 4α2 = 2

Build an augmented matrix from the system and row-reduce,

1 2 3 4 3 7 5 611 6 7 2 4 2  RREF 100 0 0 0 01 0 0 0 0 00

With a leading 1 in the final column of the row-reduced augmented matrix, Theorem RCLS tells us the system of equations is inconsistent. Therefore, there are no scalars, α1 and α2, to establish y as a linear combination of the elements in U. So yU.

Let’s again examine membership in a span.

Example SM32
A subspace of M32
The set of all 3 × 2 matrices forms a vector space when we use the operations of matrix addition (Definition MA) and scalar matrix multiplication (Definition MSM), as was show in Example VSM. Consider the subset

S = 3 1 4 2 5 5 , 1 1 2 1 141 , 3 1 1 2 1911 , 4 2 1 2 142 , 3 1 4 0 177

and define a new subset of vectors W in M32 using the span (Definition SS), W = S. So by Theorem SSS we know that W is a subspace of M32. While W is an infinite set, and this is a precise description, it would still be worthwhile to investigate whether or not W contains certain elements.

First, is

y = 9 3 7 3 1011

in W? To answer this, we want to determine if y can be written as a linear combination of the five matrices in S. Can we find scalars, α1,α2,α3,α4,α5 so that

9 3 7 3 1011 = α1 3 1 4 2 5 5 + α2 1 1 2 1 141 + α3 3 1 1 2 1911 + α4 4 2 1 2 142 + α5 3 1 4 0 177 = 3α1 + α2 + 3α3 + 4α4 + 3α5 α1 + α2 α3 + 2α4 + α5 4α1 + 2α2 α3 + α4 4α5 2α1 α2 + 2α3 2α4 5α1 + 14α2 19α3 + 14α4 17α55α1 α2 11α3 2α4 + 7α5

Using our definition of matrix equality (Definition ME) we can translate this statement into six equations in the five unknowns,

3α1 + α2 + 3α3 + 4α4 + 3α5 = 9 α1 + α2 α3 + 2α4 + α5 = 3 4α1 + 2α2 α3 + α4 4α5 = 7 2α1 α2 + 2α3 2α4 = 3 5α1 + 14α2 19α3 + 14α4 17α5 = 10 5α1 α2 11α3 2α4 + 7α5 = 11

This is a linear system of equations, which we can represent with an augmented matrix and row-reduce in search of solutions. The matrix that is row-equivalent to the augmented matrix is

1000 5 8 2 0 10019 4 1 0 010 7 8 0 0 001 17 8 1 0 0 0 0 0 0 0 000 0 0

So we recognize that the system is consistent since there is no leading 1 in the final column (Theorem RCLS), and compute n r = 5 4 = 1 free variables (Theorem FVCS). While there are infinitely many solutions, we are only in pursuit of a single solution, so let’s choose the free variable α5 = 0 for simplicity’s sake. Then we easily see that α1 = 2, α2 = 1, α3 = 0, α4 = 1. So the scalars α1 = 2, α2 = 1, α3 = 0, α4 = 1, α5 = 0 will provide a linear combination of the elements of S that equals y, as we can verify by checking,

9 3 7 3 1011 = 2 3 1 4 2 5 5 + (1) 1 1 2 1 141 + (1) 4 2 1 2 142

So with one particular linear combination in hand, we are convinced that y deserves to be a member of W = S. Second, is

x = 2 1 3 1 4 2

in W? To answer this, we want to determine if x can be written as a linear combination of the five matrices in S. Can we find scalars, α1,α2,α3,α4,α5 so that

2 1 3 1 4 2 = α1 3 1 4 2 5 5 + α2 1 1 2 1 141 + α3 3 1 1 2 1911 + α4 4 2 1 2 142 + α5 3 1 4 0 177 = 3α1 + α2 + 3α3 + 4α4 + 3α5 α1 + α2 α3 + 2α4 + α5 4α1 + 2α2 α3 + α4 4α5 2α1 α2 + 2α3 2α4 5α1 + 14α2 19α3 + 14α4 17α55α1 α2 11α3 2α4 + 7α5

Using our definition of matrix equality (Definition ME) we can translate this statement into six equations in the five unknowns,

3α1 + α2 + 3α3 + 4α4 + 3α5 = 2 α1 + α2 α3 + 2α4 + α5 = 1 4α1 + 2α2 α3 + α4 4α5 = 3 2α1 α2 + 2α3 2α4 = 1 5α1 + 14α2 19α3 + 14α4 17α5 = 4 5α1 α2 11α3 2α4 + 7α5 = 2

This is a linear system of equations, which we can represent with an augmented matrix and row-reduce in search of solutions. The matrix that is row-equivalent to the augmented matrix is

1000 5 8 0 0 10038 8 0 0 010 7 8 0 0 00117 8 0 0 0 0 0 0 1 0 000 0 0

With a leading 1 in the last column Theorem RCLS tells us that the system is inconsistent. Therefore, there are no values for the scalars that will place x in W, and so we conclude that xW.

Notice how Example SSP and Example SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.

Subsection SC: Subspace Constructions

Several of the subsets of vectors spaces that we worked with in Chapter M are also subspaces — they are closed under vector addition and scalar multiplication in m.

Theorem CSMS
Column Space of a Matrix is a Subspace
Suppose that A is an m × n matrix. Then CA is a subspace of m.

Proof   Definition CSM shows us that CA is a subset of m, and that it is defined as the span of a set of vectors from m (the columns of the matrix). Since CA is a span, Theorem SSS says it is a subspace.

That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.

Theorem RSMS
Row Space of a Matrix is a Subspace
Suppose that A is an m × n matrix. Then A is a subspace of n.

Proof   Definition RSM says A = CAt, so the row space of a matrix is a column space, and every column space is a subspace by Theorem CSMS. That’s enough.

One more.

Theorem LNSMS
Left Null Space of a Matrix is a Subspace
Suppose that A is an m × n matrix. Then A is a subspace of m.

Proof   Definition LNS says A = NAt, so the left null space is a null space, and every null space is a subspace by Theorem NSMS. Done.

So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition VS and in the basic theorems presented in Section VS. We have worked with these objects as just sets in Chapter V and Chapter M, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.

Subsection READ: Reading Questions

  1. Summarize the three conditions that allow us to quickly test if a set is a subspace.
  2. Consider the set of vectors W = a b c 3a 2b + c = 5

    Is the set W a subspace of 3? Explain your answer.

  3. Name five general constructions of sets of column vectors (subsets of m) that we now know as subspaces.

Subsection EXC: Exercises

C15 Working within the vector space 3, determine if b = 4 3 1 is in the subspace W,

W = 3 2 3 , 1 0 3 , 1 1 0 , 2 1 3

 
Contributed by Chris Black Solution [930]

C16 Working within the vector space 4, determine if b = 1 1 0 1 is in the subspace W,

W = 1 2 1 1 , 1 0 3 1 , 2 1 1 2

 
Contributed by Chris Black Solution [930]

C17 Working within the vector space 4, determine if b = 2 1 2 1 is in the subspace W,

W = 1 2 0 2 , 1 0 3 1 , 0 1 0 2 , 1 1 2 0

 
Contributed by Chris Black Solution [931]

C20 Working within the vector space P3 of polynomials of degree 3 or less, determine if p(x) = x3 + 6x + 4 is in the subspace W below.

W = x3 + x2 + x,x3 + 2x 6,x2 5

 
Contributed by Robert Beezer Solution [932]

C21 Consider the subspace

W = 2 1 3 1 , 40 2 3 , 31 2 1

of the vector space of 2 × 2 matrices, M22. Is C = 3 3 6 4 an element of W?  
Contributed by Robert Beezer Solution [934]

C25 Show that the set W = x1 x2 3x1 5x2 = 12 from Example NSC2Z fails Property AC and Property SC.  
Contributed by Robert Beezer

C26 Show that the set Y = x1 x2 x1 ,x2 from Example NSC2S has Property AC.  
Contributed by Robert Beezer

M20 In 3, the vector space of column vectors of size 3, prove that the set Z is a subspace.

Z = x1 x2 x3 4x1 x2 + 5x3 = 0

 
Contributed by Robert Beezer Solution [935]

T20 A square matrix A of size n is upper triangular if Aij = 0 whenever i > j. Let UTn be the set of all upper triangular matrices of size n. Prove that UTn is a subspace of the vector space of all square matrices of size n, Mnn.  
Contributed by Robert Beezer Solution [938]

T30 Let P be the set of all polynomials, of any degree. The set P is a vector space. Let E be the subset of P consisting of all polynomials with only terms of even degree. Prove or disprove: the set E is a subspace of P.  
Contributed by Chris Black Solution [940]

T31 Let P be the set of all polynomials, of any degree. The set P is a vector space. Let F be the subset of P consisting of all polynomials with only terms of odd degree. Prove or disprove: the set F is a subspace of P.  
Contributed by Chris Black Solution [942]

Subsection SOL: Solutions

C15 Contributed by Chris Black Statement [925]
For b to be an element of W = S there must be linear combination of the vectors in S that equals b (Definition SSCV). The existence of such scalars is equivalent to the linear system SA,b being consistent, where A is the matrix whose columns are the vectors from S (Theorem SLSLC).

31124 2 0 1 1 3 3 3031  RREF 10 12 120 0 112120 0 0 0 0 1

So by Theorem RCLS the system is inconsistent, which indicates that b is not an element of the subspace W.

C16 Contributed by Chris Black Statement [925]
For b to be an element of W = S there must be linear combination of the vectors in S that equals b (Definition SSCV). The existence of such scalars is equivalent to the linear system SA,b being consistent, where A is the matrix whose columns are the vectors from S (Theorem SLSLC).

1 121 2 0 1 1 1310 1 1 2 1  RREF 10013 0 10 0 0 0113 0 00 0

So by Theorem RCLS the system is consistent, which indicates that b is in the subspace W.

C17 Contributed by Chris Black Statement [926]
For b to be an element of W = S there must be linear combination of the vectors in S that equals b (Definition SSCV). The existence of such scalars is equivalent to the linear system SA,b being consistent, where A is the matrix whose columns are the vectors from S (Theorem SLSLC).

1101 2 0 1 1 0 302 2 1 2 0  RREF 1000 32 0 100 1 0 01032 0 00112

So by Theorem RCLS the system is consistent, which indicates that b is in the subspace W.

C20 Contributed by Robert Beezer Statement [926]
The question is if p can be written as a linear combination of the vectors in W. To check this, we set p equal to a linear combination and massage with the definitions of vector addition and scalar multiplication that we get with P3 (Example VSP)

p(x) = a1(x3 + x2 + x) + a 2(x3 + 2x 6) + a 3(x2 5) x3 + 6x + 4 = (a 1 + a2)x3 + (a 1 + a3)x2 + (a 1 + 2a2)x + (6a2 5a3)

Equating coefficients of equal powers of x, we get the system of equations,

a1 + a2 = 1 a1 + a3 = 0 a1 + 2a2 = 6 6a2 5a3 = 4

The augmented matrix of this system of equations row-reduces to

1000 0 0 0 0 010 0 0 0 1

There is a leading 1 in the last column, so Theorem RCLS implies that the system is inconsistent. So there is no way for p to gain membership in W, so pW.

C21 Contributed by Robert Beezer Statement [927]
In order to belong to W, we must be able to express C as a linear combination of the elements in the spanning set of W. So we begin with such an expression, using the unknowns a,b,c for the scalars in the linear combination.

C = 3 3 6 4 = a 2 1 3 1 +b 40 2 3 +c 31 2 1

Massaging the right-hand side, according to the definition of the vector space operations in M22 (Example VSM), we find the matrix equality,

3 3 6 4 = 2a + 4b 3c a + c 3a + 2b + 2c a + 3b + c

Matrix equality allows us to form a system of four equations in three variables, whose augmented matrix row-reduces as follows,

2 433 1 0 1 3 3 2 2 6 1 3 1 4  RREF 100 2 0 0 1 0 01 1 0 0 0 0

Since this system of equations is consistent (Theorem RCLS), a solution will provide values for a,b and c that allow us to recognize C as an element of W.

M20 Contributed by Robert Beezer Statement [928]
The membership criteria for Z is a single linear equation, which comprises a homogeneous system of equations. As such, we can recognize Z as the solutions to this system, and therefore Z is a null space. Specifically, Z = N415 . Every null space is a subspace by Theorem NSMS.

A less direct solution appeals to Theorem TSS.

First, we want to be certain Z is non-empty. The zero vector of 3, 0 = 0 0 0 , is a good candidate, since if it fails to be in Z, we will know that Z is not a vector space. Check that

4(0) (0) + 5(0) = 0

so that 0 Z.

Suppose x = x1 x2 x3 and y = y1 y2 y3 are vectors from Z. Then we know that these vectors cannot be totally arbitrary, they must have gained membership in Z by virtue of meeting the membership test. For example, we know that x must satisfy 4x1 x2 + 5x3 = 0 while y must satisfy 4y1 y2 + 5y3 = 0. Our second criteria asks the question, is x + y Z? Notice first that

x+y = x1 x2 x3 + y1 y2 y3 = x1 + y1 x2 + y2 x3 + y3

and we can test this vector for membership in Z as follows,

4(x1 + y1) 1(x2 + y2) + 5(x3 + y3) = 4x1 + 4y1 x2 y2 + 5x3 + 5y3 = (4x1 x2 + 5x3) + (4y1 y2 + 5y3) = 0 + 0 x Z,y Z = 0

and by this computation we see that x + y Z.

If α is a scalar and x Z, is it always true that αx Z? To check our third criteria, we examine

αx = α x1 x2 x3 = αx1 αx2 αx3

and we can test this vector for membership in Z with

4(αx1) (αx2) + 5(αx3) = α(4x1 x2 + 5x3) = α0 x Z = 0

and we see that indeed αx Z. With the three conditions of Theorem TSS fulfilled, we can conclude that Z is a subspace of 3.

T20 Contributed by Robert Beezer Statement [928]
Apply Theorem TSS.

First, the zero vector of Mnn is the zero matrix, O, whose entries are all zero (Definition ZM). This matrix then meets the condition that Oij = 0 for i > j and so is an element of UTn.

Suppose A,B UTn. Is A + B UTn? We examine the entries of A + B “below” the diagonal. That is, in the following, assume that i > j.

A + Bij = Aij + Bij  Definition MA = 0 + 0  A,B UTn = 0

which qualifies A + B for membership in UTn.

Suppose α and A UTn. Is αA UTn? We examine the entries of αA “below” the diagonal. That is, in the following, assume that i > j.

αAij = α Aij  Definition MSM = α0  A UTn = 0

which qualifies αA for membership in UTn.

Having fulfilled the three conditions of Theorem TSS we see that UTn is a subspace of Mnn.

T30 Contributed by Chris Black Statement [928]
Proof: Let E be the subset of P comprised of all polynomials with all terms of even degree. Clearly the set E is non-empty, as z(x) = 0 is a polynomial of even degree. Let p(x) and q(x) be arbitrary elements of E. Then there exist nonnegative integers m and n so that

p(x) = a0 + a2x2 + a 4x4 + + a 2nx2n q(x) = b0 + b2x2 + b 4x4 + + b 2mx2m

for some constants a0,a2,,a2n and b0,b2,,b2m. Without loss of generality, we can assume that m n. Thus, we have

p(x) + q(x) = (a0 + b0) + (a2 + b2)x2 + + (a 2m + b2m)x2m + a 2m+2x2m+2 + + a 2nx2n

so p(x) + q(x) has all even terms, and thus p(x) + q(x) E. Similarly, let α be a scalar. Then

αp(x) = α(a0 + a2x2 + a 4x4 + + a 2nx2n) = αa0 + (αa2)x2 + (αa 4)x4 + + (αa 2n)x2n

so that αp(x) also has only terms of even degree, and αp(x) E. Thus, E is a subspace of P.

T31 Contributed by Chris Black Statement [929]
This conjecture is false. We know that the zero vector in P is the polynomial z(x) = 0, which does not have odd degree. Thus, the set F does not contain the zero vector, and cannot be a vector space.