a) Are the nine vectors linearly independent? Not linearly independent? Might be linearly independent?

b) Do the nine vectors span ? Not span ? Might span ?

c) Suppose the nine vectors are the columns of a matrix . Does have a solution? Not have a solution? Might have a solution?

Answer: a) We cannot have a set of 9 linearly independent vectors in a space like that has dimension 7. So the vectors are not linearly independent.

b) The vectors might or might not span . For example, consider the set of vectors , , through . The nine vectors do not span but rather span a subspace of dimension 1.

c) The matrix would have nine columns but only seven rows, and would correspond to a system of seven linear equations with nine unknowns. This system could not have more than seven basic variables and thus would have at least two free variables. Since the free variables can take on any value the system is guaranteed to have a solution (and in fact would have an infinite number of them) but the system might or might not have a solution depending on the value of .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

1) You can create a basis for by adding three vectors to any set of vectors that is a basis for .

2) You can create a basis for by removing three vectors from any set of vectors that is a basis for .

Answer: 1) True. Suppose , , , and are a basis for . Per 2L (page 86) any linearly independent set in can be extended to a basis for by adding more vectors if necessary. The four vectors through are already linearly independent (since they are a basis) and hence can be extended by adding additional vectors to form a basis for .

More specifically, we can find three vectors , , and such that a) the three vectors are not in (and hence are linearly independent of through ), and b) the three vectors are linearly independent of each other. The resulting seven vectors are linearly independent. Since the dimension of is 7 these seven linearly independent vectors must be a basis for . (See exercise 2.3.15.)

2) False. Consider the vectors through with having a one in the position and zeros elsewhere. These vectors are linearly independent and span and hence are a basis for it.

Now suppose is the subspace of all vectors of the form . The vectors through are not in the subspace and hence cannot be part of a basis for it. Thus it is not possible to remove three vectors from the basis set through and form a basis for .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Answer: If the rank of is 11 then performing elimination on produces an matrix with 11 pivots and thus 11 basic variables. Since (like ) has 17 columns this means that there are 17 – 11 or 6 free variables that can be set to arbitrary values in solving the system . The nullspace of (i.e., the set of all vectors satisfying ) therefore has dimension 6, and any basis for the nullspace has 6 linearly independent vectors each of which satisfy .

Since is 64 by 17 the matrix is 17 by 64. The original matrix had 11 pivots and 11 linearly independent rows. The rows of become columns in and thus has 11 linearly independent columns and also has rank 11. Since has 64 columns there are 64 – 11 or 53 free variables when considering the system . The nullspace of (i.e., the set of all vectors satisfying ) therefore has dimension 53, and any basis for the nullspace has 53 linearly independent vectors each of which satisfy .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Answer: Any 2 by 2 matrix in the set will have the form

with the sum of every row and every column being . Any such matrix can be represented as a linear combination of two matrices as follows:

Since the two matrices are linearly independent and span the subspace they are a basis for the subspace.

The following five matrices are linearly independent members of the analogous set for 3 by 3 matrices:

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Answer: If the column vectors of are linearly independent then there must be a pivot in every one of the columns, so that the rank .

If the columns of span then we must have . There can be no more than linearly independent vectors in so out of the columns of only columns can have pivots. Therefore the rank .

If the columns of are a basis for then they are linearly independent, which means the rank , and they also span so we must also have . We thus have .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

a) given a matrix whose columns are linearly independent, the system has one and only solution for any right-hand side

b) if is a 5 by 7 matrix then the columns of cannot be linearly independent

Answer: a) False. If has fewer columns than rows then the system has more equations than unknowns and may not have a solution. For example, if

corresponding to the system

then the columns of are linearly independent but the system has no solution since the second and the third equations result in a contradiction.

b) True. If is a 5 by 7 matrix then it has seven columns, each of which is an element of . But it is impossible to have more than five linearly independent vectors in so the columns of must be linearly dependent.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Answer: Since and each have dimension 3 their respective bases each contain three vectors. Let , , and be a basis for and , , and be a basis for .

Now consider the combined set of six vectors. Since we have six vectors in a vector space of dimension 5 the combined set of vectors is linearly dependent, with at least one vector expressible as a linear combination of the other five vectors. Without loss of generality assume that is dependent on the other five vectors, so that we have

for some set of weights through . We can rearrange the above equation as follows:

Now consider the vector . Since is a linear combination of the basis vectors , , and it is in the subspace . But from the above equation we also have so that is a linear combination of the basis vectors , , and and thus is also in the subspace .

So is a member of both and . Now suppose . We then have or so that is a linear combination of and and the set of vectors , , and is linearly dependent. But this contradicts the assumption that , , and form a basis for and are thus linearly independent. Since the assumption leads to a contradiction we conclude that .

We have thus shown that there must exist a nonzero vector that is a member of both and .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

Answer: There are nine possible entries that can be set in a 3 b 3 matrix, but if the matrix is symmetric then only six of them can be set independently, since we must have , , and . Any symmetric matrix

can be represented as a linear combination of six linearly independent matrices as follows:

Since the above set of six linearly independent matrices spans the space of 3 by 3 symmetric matrices it is a basis for the space, and the dimension of the space is therefore six.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

a) if a set of vectors in is linearly independent then that set forms a basis

b) if a set of vectors in spans then that set forms a basis

Answer: a) Assume that we have a set of linearly independent vectors in . Suppose that this set does not span . By theorem 2L (page 86) we can extend this set to form a basis by adding additional vectors . But if this expanded set of vectors is a basis for then the dimension of is (by definition) which contradicts the assumption that the dimension of is . Therefore the linearly independent set must span and be a basis for it.

b) Assume that we have a set of vectors in that span . Suppose that this set is not linearly independent and thus not a basis. By theorem 2L (page 86) we can reduce this set to form a basis by removing one or more vectors so that the new set has vectors where . But since the dimension of is then there must exist some set of vectors that is a basis for . But if the reduced set of vectors is a basis for and the other set of vectors is also a basis then by theorem 2K we must have not . We therefore conclude that the spanning set is in fact linearly independent and is thus a basis for .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

How can you extend the rows of to create a basis for ? How can you reduce the columns of to create a basis for ?

Answer: As defined is in echelon form but has only two pivots, in the first and third columns. We can add another row to to provide a pivot for the second column:

Since is in echelon form and has pivots in every column, the columns are linearly independent. Since also has pivots in every row the rows are also linearly independent. (See exercise 2.3.5.) The three rows also span with any vector in expressible as

Since the three rows of are linearly independent and span they form a basis for .

Turning to the columns of , since has three columns but only two pivots (in the first and third columns) the three columns must be linearly dependent, and in fact the second column is twice the first. We can therefore remove the second column to form the following 2 by 2 matrix:

Since is in echelon form and has pivots in all columns the columns are linearly independent. They also span with any vector expressible as

Since the two columns of are linearly independent and span they form a basis for .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.