Linear Algebra and Its Applications (5th Edition)
Linear Algebra and Its Applications (5th Edition)
5th Edition
ISBN: 9780321982384
Author: David C. Lay, Steven R. Lay, Judi J. McDonald
Publisher: PEARSON
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Chapter 5, Problem 1SE

Mark each statement as True or False. Justify each answer.

  1. a. If A is invertible and 1 is an eigenvalue for A, then 1 is also an eigenvalue of A−1.
  2. b. If A is row equivalent to the identity matrix I, then A is diagonalizable.
  3. c. If A contains a row or column of zeros, then 0 is an eigenvalue of A.
  4. d. Each eigenvalue of A is also an eigenvalue of A2.
  5. e. Each eigenvector of A is also an eigenvector of A2.
  6. f. Each eigenvector of an invertible matrix A is also an eigenvector of A−1.
  7. g. Eigenvalues must be nonzero scalars.
  8. h. Eigenvectors must be nonzero vectors.
  9. i. Two eigenvectors corresponding to the same eigenvalue are always linearly dependent.
  10. j. Similar matrices always have exactly the same eigenvalues.
  11. k. Similar matrices always have exactly the same eigen vectors.
  12. l. The sum of two eigenvectors of a matrix A is also an eigenvector of A.
  13. m. The eigenvalues of an upper triangular matrix A are exactly the nonzero entries on the diagonal of A.
  14. n. The matrices A and AT have the same eigenvalues, counting multiplicities.
  15. ○.      If a 5 × 5 matrix A has fewer than 5 distinct eigenvalues, then A is not diagonalizable.
  16. p. There exists a 2 × 2 matrix that has no eigenvectors in ℝ2.
  17. q. If A is diagonalizable, then the columns of A are linearly independent.
  18. r. A nonzero vector cannot correspond to two different eigenvalues of A.
  19. s. A (square) matrix A is invertible if and only if there is a coordinate system in which the transformation x i ↦ Ax is represented by a diagonal matrix.
  20. t. If each vector ej in the standard basis for ℝn is an eigenvector of A, then A is a diagonal matrix.
  21. u. If A is similar to a diagonalizable matrix B, then A is also diagonalizable.
  22. v. If A and B are invertible n × n matrices, then AB is similar to BA.
  23. w. An n × n matrix with n linearly independent eigenvectors is invertible.
  24. x. If A is an n × n diagonalizable matrix, then each vector in ℝn can be written as a linear combination of eigenvectors of A.

(a)

Expert Solution
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To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A is invertible and 1 is an eigenvalue for A, then 1 is also an eigenvalue of A1 .

Explanation:

Consider the equation as follows:

Ax=1x (1)

Left-multiply with A1 on both sides in Equation (1).

A1Ax=A11xx=A1x

Rewrite the equation as follows:

A1x=1x

In this equation, x cannot be equal to zero. That is, x0 .

Therefore, 1 is an eigenvalue of A1 .

The given statement is true.

(b)

Expert Solution
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To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

If A is a row equivalent to the identity matrix I, then A is diagonalizable.

Explanation:

If A is a row that is equivalent to the identity matrix, then A is invertible.

Refer to Example 4 in section 5.3:

The given matrix is invertible. However, matrix A is not diagonalizable.

The given statement is false.

(c)

Expert Solution
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To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A contains a row or column of zeros, then 0 is an eigenvalue of A.

Explanation:

Theorem: The Invertible Matrix Theorem

Let Abe an n×n matrix. Then,A is invertible if and only if:

s. The number 0 is not an eigenvalue of A.

t. The determinant of A is not zero.

As given in the statement, if A contains a row or column of zeros, then A is not a row equivalent to the identity matrix.

Therefore, matrix A is not invertible.

Refer to the invertible matrix theorem, o is an eigenvalue of matrix A.

The given statement is true.

(d)

Expert Solution
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To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Each eigenvalue of A is also an eigenvalue of A2 .

Explanation:

Consider a diagonal matrix whose eigenvalues are 1 and 2.

D=[1002]

The value of D2 is as follows:

D2=[1004]

Then, the diagonal entries in the matrix are squaresof the eigenvalues of matrix A.

Therefore, the eigenvalues of A2 are the squares of the eigenvalues of A.

The given statement is false.

(e)

Expert Solution
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To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Each eigenvector of A is also an eigenvector of A2 .

Explanation:

Consider a nonzero vector x.

The condition to be satisfied is Ax=λx

Left-multiply with A on both sides.

A2x=A(Ax)=A(λx)=λAx=λ2x

The derived equation shows that x is also an eigenvector of A2 .

The given statement is true.

(f)

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To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Each eigenvector of an invertible matrix A is also an eigenvector of A1 .

Explanation:

Consider a nonzero vector x.

The condition to be satisfied is Ax=λx

Left-multiply with A1 on both sides.

AA1x=A1(λx)=λA1x

Matrix A is invertible and the eigenvalue of the matrix is nonzero.

λ1x=A1x

The derived equation shows that x is also an eigenvector of A1 .

The given statement is true.

(g)

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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Eigenvalues must be nonzero scalars.

Explanation:

For each singular square matrix, 0 will be the eigenvalue.

The given statement is false.

(h)

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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Eigenvectors must be nonzero vectors.

Explanation:

Definition:

An eigenvector of an n×n matrix A is a nonzero vector x, that is, Ax=λx for some scalar λ . Scalar λ is called an eigenvalue of A if there is a nontrivial solution of Ax=λx ; such an x is called an eigenvector corresponding to λ1 .

Refer to the definition; an eigenvector must be nonzero.

The given statement is true.

(i)

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To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Two eigenvectors corresponding to the same eigenvalue are always linearly independent.

Explanation:

Refer to example 4 in section 5.1.

The eigenvalue for the given matrix is 2.

However, the calculated eigenvector for the eigenvalue is not linearly independent.

The given statement is false.

(j)

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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Similar matrices always have exactly the same eigenvalues.

Explanation:

Theorem 4 (Section 5.2):

If n×n matrices A and B are similar, then they have the same characteristic polynomial, and hence, the same eigenvalues (with the same multiplicities).

Refer to theorem 4; the given statement is correct.

The given statement is true.

(k)

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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Similar matrices always have exactly the same eigenvectors.

Explanation:

Refer to example 3 in section 5.3.

Consider 3×3 matrix as A=[133353331] .

The eigenvalues of the matrix are λ=1,2 .

The diagonalizable matrix is D=[100020002] .

Consider the matrix A is similar to the matrix D.

Suppose the eigenvalues of matrix D to be in column I3 of matrix A. However, in the given problem, it is not the case.

Therefore, the given statement is incorrect.

The given statement is false.

(l)

Expert Solution
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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

The sum of two eigenvectors of matrix A is also an eigenvector of A.

Explanation:

Consider a 2×2 matrix as A=[4006]

The value of e1 is given as follows:

e1=[10] .

The value of e2 is given as follows:

e2=[01] .

Here, e1 and e2 are eigenvectors of matrix A. However, the sum of e1 and e2 are not the eigenvector of matrix A.

Therefore, the given statement is incorrect.

The given statement is false.

(m)

Expert Solution
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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

The eigenvalues of upper triangular matrix A are the exact nonzero entries on the diagonal of A.

Explanation:

Theorem 1: (Section 5.1):

The eigenvalues of a triangular matrix are the entries on its main diagonal.

In the given problem, the upper triangular matrix A has exactly nonzero entries. In reality, it is not necessary for that to be the case. Zero entries can also be available.

The given statement is false.

(n)

Expert Solution
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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Matrices A and AT have the same eigenvalues, counting multiplicities.

Explanation:

Apply the concept of determinant transpose property.

Show the determinant calculation of matrix A as follows:

det(ATλI)=det(AλI)T=det(AλI)

Therefore, matrices A and AT have a polynomial with the same characteristics.

The given statement is true.

(o)

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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

If a 5×5 matrix A has fewer than 5 distinct eigenvalues, then A is not diagonalizable.

Explanation:

Refer to practice problem 3 in section 5.3.

Consider an identity matrix A that is 5×5 . Even for lesser eigenvalues, the given matrix A is diagonalizable.

The given statement is false.

(p)

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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

A 2×2 matrix that has no eigenvectors in 2 exists.

Explanation:

Consider matrix A.

Matrix A rotates the vectors through π/2 radians about the origin.

The value of Ax is not a multiple of x when x is a nonzero value.

The given statement is true.

(q)

Expert Solution
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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

If A is diagonalizable, then the columns of A are linearly independent.

Explanation:

Consider a diagonal matrix A with 0 as diagonals.

Then, the columns of the diagonal matrix A are not linearly independent.

The given statement is false.

(r)

Expert Solution
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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

A nonzero vector cannot correspond to two different eigenvalues of A.

Explanation:

Consider the following example:

Ax=λ1x and Ax=λ2x .

Equate both the values of Ax .

λ1x=λ2x(λ1λ2)x=0

Suppose x cannot be equal to zero, that is, x0 .

λ1=λ2

Therefore, for a nonzero vector, two different eigenvalues are the same.

The given statement is true.

(s)

Expert Solution
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Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

A (square) matrix A is invertible if and only if there is a coordinate system in which the transformation xAx is represented by a diagonal matrix.

Explanation:

Theorem 8 (Section 5.4):

Diagonal Matrix Representation:

Suppose A=PDP1 , where D is a diagonal n×n matrix. If B is the basis for n (formed from the columns of P), then D is the B-matrix for the transformation xAx .

Consider a singular matrix A that is diagonalizable.

Refer to the theorem; the transformation xAx is represented by a diagonal matrix relative to a coordinate system, which is determined by the eigenvectors of matrix A.

The given statement is false.

(t)

Expert Solution
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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If each vector ej in the standard basis for n is an eigenvector of A, then A is a diagonal matrix.

Explanation:

Refer to the definition of matrix multiplication.

Show the multiplication of matrix A with I as follows:

A=AI=A[e1e2en]=[Ae1Ae2Aen]

If the values are considered as Aej=djej for j=1,,n , and so on.

Then, matrix A becomes a diagonal matrix with the diagonal entries of d1,,dn .

Therefore, the given statement is correct.

The given statement is true.

(u)

Expert Solution
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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A is similar to diagonalizable matrix B, then A is also diagonalizable.

Explanation:

Consider matrix B.

For a diagonalizable matrix:

B=PDP1

Here, D is a diagonalizable matrix.

Consider matrix A. Write the equation as follows:

A=QBQ1 (2)

Substitute PDP1 for B in Equation (2).

A=Q(PDP1)Q1=(QP)D(PQ)1

Therefore, matrix A is diagonalizable.

The given statement is true.

(v)

Expert Solution
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Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A and B are invertible n×n matrices, then AB is similar to BA.

Explanation:

If the matrix B is invertible, the value of AB is similar to B(AB)B1 .

This calculated value is equal to BA.

The given statement is true.

(w)

Expert Solution
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To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

An n×n matrix with n linearly independent eigenvectors is invertible.

Explanation:

Theorem 5 (Section 5.3):

The diagonalization theorem:

An n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

In fact, A=PDP1 , with D is a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors on P.

Refer to the theorem; the n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors, but it is not compulsory that it should be invertible. Any of the eigenvalues can be zero.

The given statement is false.

(x)

Expert Solution
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To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A is an n×n diagonalizable matrix, then each vector in n can be written as a linear combination of the eigenvectors of A.

Explanation:

Theorem 5 (Section 5.3):

The diagonalization theorem:

An n×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

In fact, A=PDP1 , with D as a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors on P.

A is diagonalizable if and only if A has n linearly independent eigenvectors v1,,vn in n .

Refer to basis theorem:

{v1,,vn} spans n .

That is, each vector in n can be written as a linear combination of v1,,vn

The given statement is true.

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Chapter 5 Solutions

Linear Algebra and Its Applications (5th Edition)

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The determinant of A is the product of the...Ch. 5.2 - a. If A is 3 3, with columns a1, a2, and a3, then...Ch. 5.2 - A widely used method for estimating eigenvalues of...Ch. 5.2 - Show that if A and B are similar, then det A = det...Ch. 5.2 - Let A = [.6.3.4.7], v1 = [3/74/7], x0 = [.5.5]....Ch. 5.2 - Let A = [abcd]. Use formula (1) for a determinant...Ch. 5.2 - Let A = [.5.2.3.3.8.3.20.4], v1 = [.3.6.1], v2 =...Ch. 5.3 - Compute A8, where A = [4321].Ch. 5.3 - Let A = [31227], v1 = [31], and v2 = [21]. 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The text has...Ch. 5.4 - Let B = b1,b2,b3 and D = d1,d2 be bases for vector...Ch. 5.4 - Let D = {d1, d2} and B = {b1, b2} be bases for...Ch. 5.4 - Let = e1,e2,e3 be the standard basis for 3, B =...Ch. 5.4 - Let B = b1,b2,b3 be a basis for a vector space V...Ch. 5.4 - Let T : 2 3 be the transformation that maps a...Ch. 5.4 - Let T : 2 4 be the transformation that maps a...Ch. 5.4 - Assume the mapping T : 2 2 defined by T(a0 + a1t...Ch. 5.4 - Let B = {b1, b2, b3} be a basis for a vector space...Ch. 5.4 - Define T :2 3 = by T (p) = [p(-1)p(0)p(1)]. a....Ch. 5.4 - Define T : 3 4 by T(p) = [p(-3)p(-1)p(1)p(3)]. a....Ch. 5.4 - In Exercises 11 and 12, find the B-matrix for the...Ch. 5.4 - In Exercises 11 and 12, find the B-matrix for the...Ch. 5.4 - In Exercises 1316, define T : 2 2 by T(x) = Ax....Ch. 5.4 - In Exercises 1316, define T : 2 2 by T(x) = Ax....Ch. 5.4 - In Exercises 1316, define T : 2 2 by T(x) = Ax....Ch. 5.4 - In Exercises 1316, define T : 2 2 by T(x) = Ax....Ch. 5.4 - Let A = [1113] and B = {b1, b2}, for b1 = [11], b2...Ch. 5.4 - Define T : 3 3 by T (x) = Ax, where A is a 3 3...Ch. 5.4 - Verify the statements in Exercises 1924. The...Ch. 5.4 - Verify the statements in Exercises 1924. The...Ch. 5.4 - Verify the statements in Exercises 1924. The...Ch. 5.4 - Verify the statements in Exercises 1924. The...Ch. 5.4 - Verify the statements in Exercises 1924. The...Ch. 5.4 - Verify the statements in Exercises 1924. The...Ch. 5.4 - The trace of a square matrix A is the sum of the...Ch. 5.4 - It can be shown that the trace of a matrix A...Ch. 5.4 - Let V be n with a basis B = {b1 ,, bn}; let W be n...Ch. 5.4 - Let V be a vector space with a basis B = {b1, ,...Ch. 5.4 - Let V be a vector space with a basis B = {b1, ...Ch. 5.4 - [M] In Exercises 30 and 31, find the B-matrix for...Ch. 5.4 - [M] In Exercises 30 and 31, find the B-matrix for...Ch. 5.5 - Show that if a and b are real, then the...Ch. 5.5 - Let each matrix in Exercises 16 act on 2. Find the...Ch. 5.5 - Let each matrix in Exercises 16 act on 2. Find the...Ch. 5.5 - Let each matrix in Exercises 16 act on 2. Find the...Ch. 5.5 - Let each matrix in Exercises 16 act on 2. Find the...Ch. 5.5 - Let each matrix in Exercises 16 act on 2. Find the...Ch. 5.5 - Let each matrix in Exercises 16 act on 2. Find the...Ch. 5.5 - In Exercises 712, use Example 6 to list the...Ch. 5.5 - In Exercises 712, use Example 6 to list the...Ch. 5.5 - In Exercises 712, use Example 6 to list the...Ch. 5.5 - In Exercises 712, use Example 6 to list the...Ch. 5.5 - In Exercises 712, use Example 6 to list the...Ch. 5.5 - In Exercises 712, use Example 6 to list the...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Exercises 1320, find an invertible matrix P and...Ch. 5.5 - In Example 2, solve the first equation in (2) for...Ch. 5.5 - Let A be a complex (or real) n n matrix, and let...Ch. 5.5 - Chapter 7 will focus on matrices A with the...Ch. 5.5 - Let A be an n n real matrix with the property...Ch. 5.5 - Let A be a real n n matrix, and let x be a vector...Ch. 5.5 - Let A be a real 2 2 matrix with a complex...Ch. 5.6 - The matrix A below has eigenvalues 1, 23, and 13,...Ch. 5.6 - What happens to the sequence {xk } in Practice...Ch. 5.6 - Let A be a 2 2 matrix with eigenvalues 3 and 1/3...Ch. 5.6 - Suppose the eigenvalues of a 3 3 matrix A are 3,...Ch. 5.6 - In Exercises 36, assume that any initial vector x0...Ch. 5.6 - Determine the evolution of the dynamical system in...Ch. 5.6 - In old-growth forests of Douglas fir, the spotted...Ch. 5.6 - Show that if the predation parameter p in Exercise...Ch. 5.6 - Let A have the properties described in Exercise 1....Ch. 5.6 - Prob. 8ECh. 5.6 - In Exercises 914, classify the origin as an...Ch. 5.6 - In Exercises 914, classify the origin as an...Ch. 5.6 - In Exercises 914, classify the origin as an...Ch. 5.6 - In Exercises 914, classify the origin as an...Ch. 5.6 - In Exercises 914, classify the origin as an...Ch. 5.6 - In Exercises 914, classify the origin as an...Ch. 5.6 - Let A = [.40.2.3.8.3.3.2.5]. The vector v1 = [163]...Ch. 5.7 - A real 3 3 matrix A has eigenvalues .5, .2 + .3i,...Ch. 5.7 - A real 3 3 matrix A has eigenvalues .5, .2 + .3i....Ch. 5.7 - A real 3 3 matrix A has eigenvalues 5, .2 + .3i,...Ch. 5.7 - A panicle moving in a planar force field has a...Ch. 5.7 - Let A be a 2 2 matrix with eigenvalues 3 and 1...Ch. 5.7 - In Exercises 36, solve the initial value problem...Ch. 5.7 - In Exercises 36, solve the initial value problem...Ch. 5.7 - In Exercises 36, solve the initial value problem...Ch. 5.7 - In Exercises 36, solve the initial value problem...Ch. 5.7 - In Exercises 7 and 8, make a change of variable...Ch. 5.7 - In Exercises 7 and 8, make a change of variable...Ch. 5.7 - In Exercises 918, construct the general solution...Ch. 5.7 - In Exercises 918, construct the general solution...Ch. 5.7 - In Exercises 918, construct the general solution...Ch. 5.7 - In Exercises 918, construct the general solution...Ch. 5.7 - In Exercises 918, construct the general solution...Ch. 5.7 - In Exercises 918, construct the general solution...Ch. 5.7 - [M] Find formulas for the voltages v1 and v2 (as...Ch. 5.7 - [M] Find formulas for the voltages v1 and v2 for...Ch. 5.7 - [M] Find formulas for the current it and the...Ch. 5.7 - [M] The circuit in the figure is described by the...Ch. 5.8 - How can you tell if a given vector x is a good...Ch. 5.8 - In Exercises 14, the matrix A is followed by a...Ch. 5.8 - In Exercises 14, the matrix A is followed by a...Ch. 5.8 - In Exercises 14, the matrix A is followed by a...Ch. 5.8 - In Exercises 14, the matrix A is followed by a...Ch. 5.8 - Let A = [15162021]. The vectors x, , A5x are...Ch. 5.8 - Let A = [2367]. Repeat Exercise 5, using the...Ch. 5.8 - Exercises 13 and 14 apply to a 3 3 matrix A whose...Ch. 5.8 - Exercises 13 and 14 apply to a 3 3 matrix A whose...Ch. 5.8 - Suppose Ax = x with x 0. Let or be a scalar...Ch. 5.8 - Suppose n is an eigenvalue of the B in Exercise...Ch. 5.8 - A common misconception is that if A has a strictly...Ch. 5 - Mark each statement as True or False. Justify each...Ch. 5 - Show that if x is an eigenvector of the matrix...Ch. 5 - Suppose x is an eigenvector of A corresponding to...Ch. 5 - Use mathematical induction to show that if is an...Ch. 5 - If p(t) = c0 + c1t + c2t2 + + cntn, define p(A)...Ch. 5 - Suppose A = PDP1, where P is 2 2 and D = [2007]....Ch. 5 - Suppose A is diagonalizable and p(t) is the...Ch. 5 - a. Let A be a diagonalizable n n matrix. Show...Ch. 5 - Show that I A is invertible when all the...Ch. 5 - Show that if A is diagonalizable, with all...Ch. 5 - Let u be an eigenvector of A corresponding to an...Ch. 5 - Let G = [AX0B] Use formula (1) for the determinant...Ch. 5 - Use Exercise 12 to find the eigenvalues of the...Ch. 5 - Use Exercise 12 to find the eigenvalues of the...Ch. 5 - Let J be the n n matrix of all 1s, and consider A...Ch. 5 - Apply the result of Exercise 15 to find the...Ch. 5 - Let A = [a11a12a21a22]. Recall from Exercise 25 in...Ch. 5 - Let A = [.4.3.41.2]. Explain why Ak approaches...Ch. 5 - Exercises 1923 concern the polynomial p(t) = a0 +...Ch. 5 - Exercises 1923 concern the polynomial p(t) = a0 +...Ch. 5 - Use mathematical induction to prove that for n 2,...Ch. 5 - Exercises 1923 concern the polynomial p(t) = a0 +...Ch. 5 - Exercises 1923 concern the polynomial p(t) = a0 +...
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