Linear Algebra/Basis/Solutions

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This exercise is recommended for all readers.
Problem 1

Decide if each is a basis for .


By Theorem 1.12, each is a basis if and only if each vector in the space can be given in a unique way as a linear combination of the given vectors.

  1. Yes this is a basis. The relation
    which has the unique solution , , and .
  2. This is not a basis. Setting it up as in the prior item
    gives a linear system whose solution
    is possible if and only if the three-tall vector's components , , and satisfy . For instance, we can find the coefficients and that work when , , and . However, there are no 's that work for , , and . Thus this is not a basis; it does not span the space.
  3. Yes, this is a basis. Setting up the relationship leads to this reduction
    which has a unique solution for each triple of components , , and .
  4. No, this is not a basis. The reduction
    which does not have a solution for each triple , , and . Instead, the span of the given set includes only those three-tall vectors where .
This exercise is recommended for all readers.
Problem 2

Represent the vector with respect to the basis.

  1. ,
  2. ,
  3. ,
  1. We solve
    and conclude that and so . Thus, the representation is this.
  2. The relationship is easily solved by eye to give that , , , and .
Problem 3

Find a basis for , the space of all quadratic polynomials. Must any such basis contain a polynomial of each degree:~degree zero, degree one, and degree two?


One basis is . There are bases for that do not contain any polynomials of degree one or degree zero. One is . (Every basis has at least one polynomial of degree two, though.)

Problem 4

Find a basis for the solution set of this system.


The reduction

gives that the only condition is that . The solution set is

and so the obvious candidate for the basis is this.

We've shown that this spans the space, and showing it is also linearly independent is routine.

This exercise is recommended for all readers.
Problem 5

Find a basis for , the space of matrices.


There are many bases. This is an easy one.

This exercise is recommended for all readers.
Problem 6

Find a basis for each.

  1. The subspace of
  2. The space of three-wide row vectors whose first and second components add to zero
  3. This subspace of the matrices

For each item, many answers are possible.

  1. One way to proceed is to parametrize by expressing the as a combination of the other two . Then is and
    suggests . This only shows that it spans, but checking that it is linearly independent is routine.
  2. Parametrize to get , which suggests using the sequence . We've shown that it spans, and checking that it is linearly independent is easy.
  3. Rewriting
    suggests this for the basis.
Problem 7

Check Example 1.6.


We will show that the second is a basis; the first is similar. We will show this straight from the definition of a basis, because this example appears before Theorem 1.12.

To see that it is linearly independent, we set up . Taking and gives this system

which shows that and .

The calculation for span is also easy; for any , we have that gives that and that , and so the span is the entire space.

This exercise is recommended for all readers.
Problem 8

Find the span of each set and then find a basis for that span.

  1. in
  2. in
  1. Asking which can be expressed as gives rise to three linear equations, describing the coefficients of , , and the constants.
    Gauss' method with back-substitution shows, provided that , that and . Thus, with , we can compute appropriate and for any and . So the span is the entire set of linear polynomials . Parametrizing that set suggests a basis (we've shown that it spans; checking linear independence is easy).
  2. With
    we get this system.
    Thus, the only quadratic polynomials with associated 's are the ones such that . Hence the span is . Parametrizing gives , which suggests (checking that it is linearly independent is routine).
This exercise is recommended for all readers.
Problem 9

Find a basis for each of these subspaces of the space of cubic polynomials.

  1. The subspace of cubic polynomials such that
  2. The subspace of polynomials such that and
  3. The subspace of polynomials such that , , and~
  4. The space of polynomials such that , , , and~
  1. The subspace is . Rewriting gives , which, on breaking out the parameters, suggests for the basis (it is easily verified).
  2. The given subspace is the collection of cubics such that and . Gauss' method
    gives that and that . Rewriting as suggests this for a basis . The above shows that it spans the space. Checking it is linearly independent is routine. (Comment. A worthwhile check is to verify that both polynomials in the basis have both seven and five as roots.)
  3. Here there are three conditions on the cubics, that , that ,and that . Gauss' method
    yields the single free variable , with , , and . The parametrization is this.
    Therefore, a good candidate for the basis is . It spans the space by the work above. It is clearly linearly independent because it is a one-element set (with that single element not the zero object of the space). Thus, any cubic through the three points , , and is a multiple of this one. (Comment. As in the prior question, a worthwhile check is to verify that plugging seven, five, and three into this polynomial yields zero each time.)
  4. This is the trivial subspace of . Thus, the basis is empty .

Remark. The polynomial in the third item could alternatively have been derived by multiplying out .

Problem 10

We've seen that it is possible for a basis to remain a basis when it is reordered. Must it always remain a basis?


Yes. Linear independence and span are unchanged by reordering.

Problem 11

Can a basis contain a zero vector?


No linearly independent set contains a zero vector.

This exercise is recommended for all readers.
Problem 12

Let be a basis for a vector space.

  1. Show that is a basis when . What happens when at least one is ?
  2. Prove that is a basis where .
  1. To show that it is linearly independent, note that gives that , which in turn implies that each is zero. But with that means that each is zero. Showing that it spans the space is much the same; because is a basis, and so spans the space, we can for any write , and then . If any of the scalars are zero then the result is not a basis, because it is not linearly independent.
  2. Showing that is linearly independent is easy. To show that it spans the space, assume that . Then, we can represent the same with respect to in this way .
Problem 13

Find one vector that will make each into a basis for the space.

  1. in
  2. in
  3. in

Each forms a linearly independent set if is omitted. To preserve linear independence, we must expand the span of each. That is, we must determine the span of each (leaving out), and then pick a lying outside of that span. Then to finish, we must check that the result spans the entire given space. Those checks are routine.

  1. Any vector that is not a multiple of the given one, that is, any vector that is not on the line will do here. One is .
  2. By inspection, we notice that the vector is not in the span of the set of the two given vectors. The check that the resulting set is a basis for is routine.
  3. For any member of the span , the coefficient of equals the constant term. So we expand the span if we add a quadratic without this property, say, . The check that the result is a basis for is easy.
This exercise is recommended for all readers.
Problem 14

Where is a basis, show that in this equation

each of the 's is zero. Generalize.


To show that each scalar is zero, simply subtract . The obvious generalization is that in any equation involving only the 's, and in which each appears only once, each scalar is zero. For instance, an equation with a combination of the even-indexed basis vectors (i.e., , , etc.) on the right and the odd-indexed basis vectors on the left also gives the conclusion that all of the coefficients are zero.

Problem 15

A basis contains some of the vectors from a vector space; can it contain them all?


No; no linearly independent set contains the zero vector.

Problem 16

Theorem 1.12 shows that, with respect to a basis, every linear combination is unique. If a subset is not a basis, can linear combinations be not unique? If so, must they be?


Here is a subset of that is not a basis, and two different linear combinations of its elements that sum to the same vector.

Thus, when a subset is not a basis, it can be the case that its linear combinations are not unique.

But just because a subset is not a basis does not imply that its combinations must be not unique. For instance, this set

does have the property that

implies that . The idea here is that this subset fails to be a basis because it fails to span the space; the proof of the theorem establishes that linear combinations are unique if and only if the subset is linearly independent.

This exercise is recommended for all readers.
Problem 17

A square matrix is symmetric if for all indices and , entry equals entry .

  1. Find a basis for the vector space of symmetric matrices.
  2. Find a basis for the space of symmetric matrices.
  3. Find a basis for the space of symmetric matrices.
  1. Describing the vector space as
    suggests this for a basis.
    Verification is easy.
  2. This is one possible basis.
  3. As in the prior two questions, we can form a basis from two kinds of matrices. First are the matrices with a single one on the diagonal and all other entries zero (there are of those matrices). Second are the matrices with two opposed off-diagonal entries are ones and all other entries are zeros. (That is, all entries in are zero except that and are one.)
This exercise is recommended for all readers.
Problem 18

We can show that every basis for contains the same number of vectors.

  1. Show that no linearly independent subset of contains more than three vectors.
  2. Show that no spanning subset of contains fewer than three vectors. (Hint. Recall how to calculate the span of a set and show that this method, when applied to two vectors, cannot yield all of .)
  1. Any four vectors from are linearly related because the vector equation
    gives rise to a linear system
    that is homogeneous (and so has a solution) and has four unknowns but only three equations, and therefore has nontrivial solutions. (Of course, this argument applies to any subset of with four or more vectors.)
  2. Given , ..., ,
    to decide which vectors
    are in the span of , set up
    and row reduce the resulting system.
    There are two variables and but three equations, so when Gauss' method finishes, on the bottomrow there will be some relationship of the form . Hence, vectors in the span of the two-element set must satisfy some restriction. Hence the span is not all of .
Problem 19

One of the exercises in the Subspaces subsection shows that the set

is a vector space under these operations.

Find a basis.


We have (using these peculiar operations with care)

and so a good candidate for a basis is this.

To check linear independence we set up

(the vector on the right is the zero object in this space). That yields the linear system

with only the solution and . Checking the span is similar.