Linear Algebra and Its Applications, Exercise 2.3.23

Posted on September 27, 2011 by

Exercise 2.3.23. Let v_1 through v_9 be vectors in \mathbf{R}^7. Answer the following questions:

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

b) Do the nine vectors span \mathbf{R}^7? Not span \mathbf{R}^7? Might span \mathbf{R}^7?

c) Suppose the nine vectors are the columns of a matrix A. Does Ax = b 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 \mathbf{R}^7 that has dimension 7. So the vectors are not linearly independent.

b) The vectors might or might not span \mathbf{R}^7. For example, consider the set of vectors (1, 0, 0, 0, 0, 0, 0), (2, 0, 0, 0, 0, 0, 0), through (9, 0, 0, 0, 0, 0, 0). The nine vectors do not span \mathbf{R}^7 but rather span a subspace of dimension 1.

c) The matrix A 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 Ax = 0 is guaranteed to have a solution (and in fact would have an infinite number of them) but the system Ax = b might or might not have a solution depending on the value of b.

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.

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Linear Algebra and Its Applications, Exercise 2.3.22

Posted on September 26, 2011 by

Exercise 2.3.22. Given a vector space V of dimension 7 and a subspace W of V of dimension 4, state whether the following are true or false:

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

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

Answer: 1) True. Suppose w_1, w_2, w_3, and w_4 are a basis for W. Per 2L (page 86) any linearly independent set in V can be extended to a basis for V by adding more vectors if necessary. The four vectors w_1 through w_4 are already linearly independent (since they are a basis) and hence can be extended by adding additional vectors to form a basis for V.

More specifically, we can find three vectors v_1, v_2, and v_3 such that a) the three vectors are not in W (and hence are linearly independent of w_1 through w_4), and b) the three vectors are linearly independent of each other. The resulting seven vectors are linearly independent. Since the dimension of V is 7 these seven linearly independent vectors must be a basis for V. (See exercise 2.3.15.)

2) False. Consider the vectors v_1 = (1, 0, 0, 0, 0, 0, 0) through v_7 = (0, 0, 0, 0, 0, 0, 1) with v_i having a one in the i^{th} position and zeros elsewhere. These vectors are linearly independent and span V and hence are a basis for it.

Now suppose W is the subspace of all vectors of the form (a, a, b, b, c, c, d). The vectors v_1 through v_6 are not in the subspace W and hence cannot be part of a basis for it. Thus it is not possible to remove three vectors from the basis set v_1 through v_7 and form a basis for W.

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.

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Linear Algebra and Its Applications, Exercise 2.3.21

Posted on September 25, 2011 by

Exercise 2.3.21. Suppose A is a 64 by 17 matrix and has rank 11. How many independent vectors are solutions to the system Ax = 0? What about the system A^Ty = 0?

Answer: If the rank of A is 11 then performing elimination on A produces an matrix U with 11 pivots and thus 11 basic variables. Since U (like A) 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 Ax = 0. The nullspace of A (i.e., the set of all vectors satisfying Ax = 0) therefore has dimension 6, and any basis for the nullspace has 6 linearly independent vectors each of which satisfy Ax = 0.

Since A is 64 by 17 the matrix A^T is 17 by 64. The original matrix A had 11 pivots and 11 linearly independent rows. The rows of A become columns in A^T and thus A^T has 11 linearly independent columns and also has rank 11. Since A^T has 64 columns there are 64 – 11 or 53 free variables when considering the system A^Ty = 0. The nullspace of A^T (i.e., the set of all vectors satisfying A^Ty = 0) therefore has dimension 53, and any basis for the nullspace has 53 linearly independent vectors each of which satisfy A^Ty = 0.

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.

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Linear Algebra and Its Applications, Exercise 2.3.20

Posted on September 24, 2011 by

Exercise 2.3.20.Consider the set of all 2 by 2 matrices that have the sum of their rows equal to the sum of their columns. What is a basis for this subspace? Consider the analogous set of 3 by 3 matrices with equal row and column sums. List five linearly independent matrices from this set.

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

A = \begin{bmatrix} a&b \\ b&a \end{bmatrix}

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

A = a \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} + b \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}

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:

\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \quad \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{bmatrix} \quad \begin{bmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0 \end{bmatrix}

\begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} \quad \begin{bmatrix} 1&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix}

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.

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Linear Algebra and Its Applications, Exercise 2.3.19

Posted on September 23, 2011 by

Exercise 2.3.19. Suppose A is an m by n matrix, with n columns taken from \mathbf{R}^m. What is the rank of A if its column vectors are linearly independent? What is the rank of A if its column vectors span \mathbf{R}^m? What is the rank of A if its column vectors are a basis for \mathbf{R}^m?

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

If the n columns of A span \mathbf{R}^m then we must have n \ge m. There can be no more than m linearly independent vectors in \mathbf{R}^m so out of the n columns of A only m columns can have pivots. Therefore the rank r = m.

If the n columns of A are a basis for \mathbf{R}^m then they are linearly independent, which means the rank r = n, and they also span \mathbf{R}^m so we must also have r = m. We thus have r = m = n.

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.

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Linear Algebra and Its Applications, Exercise 2.3.18

Posted on September 22, 2011 by

Exercise 2.3.18. Indicate whether the following statements are true or false:

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

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

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

A = \begin{bmatrix} 1&0 \\ 0&1 \\ 0&1 \end{bmatrix} \quad \rm and \quad b = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

corresponding to the system

\setlength\arraycolsep{0.2em}\begin{array}{rcrcl}x_1&&&=&0 \\ &&x_2&=&0 \\ &&x_2&=&1 \end{array}

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

b) True. If A is a 5 by 7 matrix then it has seven columns, each of which is an element of \mathbf{R}^5. But it is impossible to have more than five linearly independent vectors in \mathbf{R}^5 so the columns of A 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.

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Linear Algebra and Its Applications, Exercise 2.3.17

Posted on September 21, 2011 by

Exercise 2.3.17. Suppose that V and W are subspaces of \mathbf{R}^5, each with dimension 3. Show that V and W must have at least one vector in common other than the zero vector.

Answer: Since V and W each have dimension 3 their respective bases each contain three vectors. Let v_1, v_2, and v_3 be a basis for V and w_1, w_2, and w_3 be a basis for W.

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 w_3 is dependent on the other five vectors, so that we have

w_3 = c_1v_1 + c_2v_2 + c_3v_3 + c_4w_1 + c_5w_2

for some set of weights c_1 through c_5. We can rearrange the above equation as follows:

c_1v_1 + c_2v_2 + c_3v_3 = -c_4w_1 - c_5w_2 + w_3

Now consider the vector u = c_1v_1 + c_2v_2 + c_3v_3. Since u is a linear combination of the basis vectors v_1, v_2, and v_3 it is in the subspace V. But from the above equation we also have u = -c_4w_1 - c_5w_2 + w_3 so that u is a linear combination of the basis vectors w_1, w_2, and w_3 and thus is also in the subspace W.

So u is a member of both V and W. Now suppose u = 0. We then have -c_4w_1 - c_5w_2 + w_3 = 0 or w_3 = c_4w_1 + c_5w_2 so that w_3 is a linear combination of w_1 and w_2 and the set of vectors w_1, w_2, and w_3 is linearly dependent. But this contradicts the assumption that w_1, w_2, and w_3 form a basis for W and are thus linearly independent. Since the assumption u = 0 leads to a contradiction we conclude that u \ne 0.

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

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.

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Linear Algebra and Its Applications, Exercise 2.3.16

Posted on September 20, 2011 by

Exercise 2.3.16. What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it?

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 a_{12} = a_{21}, a_{13} = a_{31}, and a_{23} = a_{32}. Any symmetric matrix

A = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{12}&a_{22}&a_{13} \\ a_{13}&a_{23}&a_{33} \end{bmatrix}

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

A = a_{11} \begin{bmatrix} 1&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} + a_{22} \begin{bmatrix} 0&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix} + a_{33} \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&1 \end{bmatrix}

+ a_{12} \begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&0 \end{bmatrix} + a_{13} \begin{bmatrix} 0&0&1 \\ 0&0&0 \\ 1&0&0 \end{bmatrix} + a_{23} \begin{bmatrix} 0&0&0 \\ 0&0&1 \\ 0&1&0 \end{bmatrix}

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.

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Linear Algebra and Its Applications, Exercise 2.3.15

Posted on September 19, 2011 by

Exercise 2.3.15. f the vector space V has dimension k show that

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

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

Answer: a) Assume that we have a set of linearly independent vectors v_1, \dotsc, v_k in V. Suppose that this set does not span V. By theorem 2L (page 86) we can extend this set to form a basis by adding additional vectors v_{k+1}, \dotsc, v_{k+p}. But if this expanded set of vectors is a basis for V then the dimension of V is (by definition) k+p > k which contradicts the assumption that the dimension of V is k. Therefore the linearly independent set v_1, \dotsc, v_k must span V and be a basis for it.

b) Assume that we have a set of vectors w_1, \dotsc, w_k in V that span V. 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 l vectors where l < k. But since the dimension of V is k then there must exist some set of k vectors that is a basis for V. But if the reduced set of l vectors is a basis for V and the other set of k vectors is also a basis then by theorem 2K we must have l = k not l < k. We therefore conclude that the spanning set w_1, \dotsc, w_k is in fact linearly independent and is thus a basis for V.

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.

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Linear Algebra and Its Applications, Exercise 2.3.14

Posted on September 18, 2011 by

Exercise 2.3.14. Suppose we have the following matrix

A = \begin{bmatrix} 1&2&1 \\ 0&0&4 \end{bmatrix}

How can you extend the rows of A to create a basis for \mathbf{R}^3? How can you reduce the columns of A to create a basis for \mathbf{R}^2?

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

A' = \begin{bmatrix} 1&2&1 \\ 0&1&0 \\ 0&0&4 \end{bmatrix}

Since A' is in echelon form and has pivots in every column, the columns are linearly independent. Since A' also has pivots in every row the rows are also linearly independent. (See exercise 2.3.5.) The three rows also span \mathbf{R}^3 with any vector v = (v_1, v_2, v_3) in \mathbf{R}^3 expressible as

v = v_1 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} + (v_2 - 2v_1) \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \frac{1}{4} (v_3 - v_1) \begin{bmatrix} 0 \\ 0 \\ 4 \end{bmatrix}

Since the three rows of A' are linearly independent and span \mathbf{R}^3 they form a basis for \mathbf{R}^3.

Turning to the columns of A, since A 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:

A'' = \begin{bmatrix} 1&1 \\ 0&4 \end{bmatrix}

Since A'' is in echelon form and has pivots in all columns the columns are linearly independent. They also span \mathbf{R}^2 with any vector w = (w_1, w_2) expressible as

w = (w_1 - \frac{1}{4}w_2) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \frac{1}{4}w_2 \begin{bmatrix} 1 \\ 4 \end{bmatrix}

Since the two columns of A'' are linearly independent and span \mathbf{R}^2 they form a basis for \mathbf{R}^2.

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.

Posted in linear algebra | Leave a comment