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.

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Linear Algebra and Its Applications, Exercise 2.3.13

Posted on September 17, 2011 by

Exercise 2.3.13.What are the dimensions of the following spaces?

a) vectors in \mathbf{R}^4 with components that sum to zero

b) the nullspace associated with the 4 by 4 identity matrix

c) the space of all 4 by 4 matrices

Answer: a) If the components of a vector v sum to zero then we have

v_1 + v_2 + v_3 + v_4 = 0 \rightarrow v_4 = -v_1 - v_2 - v_3

Such a vector v can be expressed as the linear combination of three vectors as follows:

v = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ -v_1 - v_2 - v_3 \end{bmatrix} = v_1 \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix} + v_2 \begin{bmatrix} 0 \\ 1 \\ 0 \\ -1 \end{bmatrix} + v_3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ -1 \end{bmatrix}

The three vectors are linearly independent (as can be easily seen by doing elimination on a matrix whose columns are the vectors) and since they span the space they are a basis for it. The dimension of the space is therefore 3.

b) The nullspace of the 4 by 4 identity matrix I contains solutions to Ix = 0 or

\begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}

This is equivalent to

x_1 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}

But since these vectors are linearly independent the only solution is x_1 = x_2 = x_3 = x_4 = 0 or x = (0, 0, 0, 0). The nullspace consists only of the zero vector and (by convention) its dimension is zero.

c) The space of all 4 by 4 matrices has as a basis the set of matrices in which one element is one and the rest are zero; for example

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

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

There are 16 such matrixes in this basis, so the dimension of the space is 16.

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 2.3.12

Posted on September 16, 2011 by

Exercise 2.3.12.Suppose that the set of vectors v_1, v_2, v_3,  and v_4,  is a basis for \mathbf{R}^4 and that W is a subspace of \mathbf{R}^4. Provide a counterexample to the conjecture that some subset of v_1, v_2, v_3,  and v_4 is necessarily a basis for W.

Answer: Suppose that v_1, v_2, v_3,  and v_4 are equal to the vectors (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1) and suppose that W is the subspace consisting of all vectors whose first two elements are equal to each other and whose last two elements are equal to each other; i.e., vectors in W are of the form (a, a, c, c). (It is fairly simple to verify that W is in fact a subspace, so I omit that here.) In this case none of the vectors v_1, v_2, v_3,  and v_4 are in the subspace W and thus cannot be part of a basis for W.

Instead we could use, for example, the vectors (1, 1, 0, 0) and (0, 0, 1, 1) as a basis for the subspace W, with any vector w = (a, a, c, c) in the subspace expressible as

w = a \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + c \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \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 2.3.11

Posted on September 15, 2011 by

Exercise 2.3.11. Consider the subspace of \mathbf{R}^3 consisting of all vectors whose first two components are equal. Find two different bases for this subspace.

Answer: All vectors in the subspace are of the form (a, a, c). One basis for the subspace consists of the vectors (1, 1, 0) and (0, 0, 1) with any vector v = (a, a, c) in the subspace expressible as

v = a \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + c \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

Another basis for the subspace consists of the vectors (1, 1, 1) and (0, 0, -1) with any vector v = (a, a, c) in the subspace expressible as

v = a \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + (a - c) \begin{bmatrix} 0 \\ 0 \\ -1 \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 2.3.10

Posted on September 14, 2011 by

Exercise 2.3.10. The set of all 2 by 2 matrices forms a vector space under the standard rules for multiplying two matrices and multiplying a matrix by a scalar. Find a basis for the space and describe the subspace spanned by the set of all matrices U in echelon form.

Answer: Any 2 by 2 matrix

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

can be represented as a linear combination of four matrices as follows:

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

These four matrices are linearly independent and form a basis for the space of all 2 by 2 matrices.

The set of 2 by 2 echelon matrices consists of all matrices U of the form

\begin{bmatrix} a&b \\ 0&d \end{bmatrix}

where any of a, b, or d may be zero. The subspace spanned is the set of all upper triangular matrices, and the three matrices

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

serve as a basis for the subspace.

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