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Linear Algebra and its Applications
Found in: Page 395
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

Question: Mark Each statement True or False. Justify each answer. In each part, A represents an \(n \times n\) matrix.

  1. If A is orthogonally diagonizable, then A is symmetric.
  2. If A is an orthogonal matrix, then A is symmetric.
  3. If A is an orthogonal matrix, then \(\left\| {A{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\) for all x in \({\mathbb{R}^n}\).
  4. The principal axes of a quadratic from \({{\bf{x}}^T}A{\bf{x}}\) can be the columns of any matrix P that diagonalizes A.
  5. If P is an \(n \times n\) matrix with orthogonal columns, then \({P^T} = {P^{ - {\bf{1}}}}\).
  6. If every coefficient in a quadratic form is positive, then the quadratic form is positive definite.
  7. If \({{\bf{x}}^T}A{\bf{x}} > {\bf{0}}\) for some x, then the quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is positive definite.
  8. By a suitable change of variable, any quadratic form can be changed into one with no cross-product term.
  9. The largest value of a quadratic form \({{\bf{x}}^T}A{\bf{x}}\), for \(\left\| {\bf{x}} \right\| = {\bf{1}}\) is the largest entery on the diagonal A.
  10. The maximum value of a positive definite quadratic form \({{\bf{x}}^T}A{\bf{x}}\) is the greatest eigenvalue of A.
  11. A positive definite quadratic form can be changed into a negative definite form by a suitable change of variable \({\bf{x}} = P{\bf{u}}\), for some orthogonal matrix P.
  12. An indefinite quadratic form is one whose eigenvalues are not definite.
  13. If P is an \(n \times n\) orthogonal matrix, then the change of variable \({\bf{x}} = P{\bf{u}}\) transforms \({{\bf{x}}^T}A{\bf{x}}\) into a quadratic form whose matrix is \({P^{ - {\bf{1}}}}AP\).
  14. If U is \(m \times n\) with orthogonal columns, then \(U{U^T}{\bf{x}}\) is the orthogonal projection of x onto ColU.
  15. If B is \(m \times n\) and x is a unit vector in \({\mathbb{R}^n}\), then \(\left\| {B{\bf{x}}} \right\| \le {\sigma _{\bf{1}}}\), where \({\sigma _{\bf{1}}}\) is the first singular value of B.
  16. A singular value decomposition of an \(m \times n\) matrix B can be written as \(B = P\Sigma Q\), where P is an \(m \times n\) orthogonal matrix and \(\Sigma \) is an \(m \times n\) diagonal matrix.
  17. If A is \(n \times n\), then A and \({A^T}A\) have the same singular values.

a. The statement is True.

b. The statement is False.

c. The statement is True.

d. The statement is False.

e. The statement is False.

f. The statement is False.

g. The statement is False.

h. The statement is True.

i. The statement is False.

j. The statement is False.

k. The statement is False

l. The statement is False.

m. The statement is True.

n. The statement is False.

o. The statement is True.

p. The statement is True.

q. The statement is False

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find an answer for part (a)

According to theorem 2, an orthogonally diagonalizable matrix then A must be a symmetric matrix.

Thus, the statement is True.

Step 2: Find an answer for part (b)

Consider the matrix A, \(A = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]\).

Matrix A not symmetric, but it is orthogonal.

Thus, the given statement is False.

Step 3: Find an answer for part (c)

According to proof of Theorem 6, if A is diagonalized, then;

\[\left\| {A{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\]

Thus, the statement is True.

Step 4: Find an answer for part (d)

The principal axes of \({{\bf{x}}^T}A{\bf{x}}\) are represented by the orthogonal matrix P. The matrix P diagonalizes A.

Thus, the statement is True.

Step 5: Find the answer for part (e)

Let a matrix P be defined as:

\(P = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\1&1\end{array}} \right]\)

The columns of P are orthogonal but not orthonormal. Therefore, \({P^T} = {P^{ - 1}}\).

Thus, the given statement is False.

Step 6: Find the answer for part (f)

According to theorem 5, if all the eigenvalues of the coefficient matrix of a quadratic equation are positive, then the quadratic equation is a positive definite.

Therefore, the statement is false.

Step 7: Find the answer for part (g)

Consider the following matrix and vector:

\(A = \left[ {\begin{array}{*{20}{c}}2&0\\0&{ - 3}\end{array}} \right]\) and \(x = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\)

Find the product \({x^T}Ax\).

\(\begin{array}{c}{x^T}Ax = \left[ {\begin{array}{*{20}{c}}1&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&0\\0&{ - 3}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2\\0\end{array}} \right]\\ = 2\\ > 0\end{array}\)

But according to theorem 6, \({x^T}Ax\) is an indefinite quadratic form.

Thus, the statement is False.

Step 8: Find the answer for part (h)

According to the principal axes theorem, a quadratic form can be represented by the expression \({x^T}Ax\), where A is a symmetric matrix.

Therefore, the statement is True.

Step 9: Find an answer for part (i)

From example 3 of section 7.3, it can be observed that if \(\left\| x \right\| = 1\), the largest eigenvalue of the quadratic equation \({{\bf{x}}^T}A{\bf{x}}\) is the largest eigenvalue of A.

Thus, the statement is False.

Step 10: Find an answer for part (j)

The maximum value of the quadratic expression \({{\bf{x}}^T}A{\bf{x}}\) can be made as large as possible, and it depends upon the set of unit vectors.

Thus, the statement is False.

Step 11: Find the answer for part (k)

If there is the orthogonal change for \({\bf{x}} = P{\bf{y}}\), then the positive definite quadratic changes into another positive quadratic.

Thus, the given statement is False.

Step 12: Find the answer for part (l)

As every square matrix has a characteristic equation, therefore its root, i.e., eigenvalues, always exist.

So, for matrix A the eigenvalues exist.

Thus, the statement is False.

Step 13: Find the answer for part (m)

If \({\bf{x}} = P{\bf{u}}\), then

\(\begin{array}{c}{{\bf{x}}^T}A{\bf{x}} = {\left( {P{\bf{y}}} \right)^T}A\left( {P{\bf{y}}} \right)\\ = {{\bf{y}}^T}{P^T}AP{\bf{y}}\\ = {{\bf{y}}^T}{P^{ - 1}}AP{\bf{y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{P^T} = {P^{ - 1}}} \right)\end{array}\)

Thus, the given statement is True.

Step 14: Find the answer for part (n)

Let the matrix \(U = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\1&{ - 1}\end{array}} \right]\).

\(U{U^T}{\bf{x}}\) represents the orthogonal projection of x onto ColU, if an only if U is orthonormal.

Thus, the given statement is False.

Step 15: Find the answer for part (o)

As \(\left\| {B{\bf{x}}} \right\|\) shows the length of vectors and \({\sigma _1}\) is the greatest singular value, then the inequality \[\left\| {B{\bf{x}}} \right\| \le {\sigma _1}\] is true.

Therefore, the statement is True.

Step 16: Find the answer for part (p)

According to theorem 10, a matrix \(m \times n\) matrix A can be expressed as\(U\Sigma {V^T}\), where U and V are orthogonal.

Hence, \({V^T}\) is also orthogonal.

Thus, the statement is True.

Step 17: Find an answer for part (q)

Consider the matrix:

\(A = \left[ {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right]\)

The singular values for A are 2 and 1, and the singular values of product \({A^T}A\) are 4 and 1.

Thus, the given statement is True.

Most popular questions for Math Textbooks

Question 11: Prove that any \(n \times n\) matrix A admits a polar decomposition of the form \(A = PQ\), where P is a \(n \times n\) positive semidefinite matrix with the same rank as A and where Q is an \(n \times n\) orthogonal matrix. (Hint: Use a singular value decomposition, \(A = U\sum {V^T}\), and observe that \(A = \left( {U\sum {U^T}} \right)\left( {U{V^T}} \right)\).) This decomposition is used, for instance, in mechanical engineering to model the deformation of a material. The matrix P describe the stretching or compression of the material (in the directions of the eigenvectors of P), and Q describes the rotation of the material in space.

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

24. Using the notation of Exercise 23, show that \({A^T}{u_j} = {\sigma _j}{v_j}\) for \({\bf{1}} \le {\bf{j}} \le {\bf{r}} = {\bf{rank}}\;{\bf{A}}\)

Question: 12. Exercises 12–14 concern an \(m \times n\) matrix \(A\) with a reduced singular value decomposition, \(A = {U_r}D{V_r}^T\), and the pseudoinverse \({A^ + } = {U_r}{D^{ - 1}}{V_r}^T\).

Verify the properties of \({A^ + }\):

a. For each \({\rm{y}}\) in \({\mathbb{R}^m}\), \(A{A^ + }{\rm{y}}\) is the orthogonal projection of \({\rm{y}}\) onto \({\rm{Col}}\,A\).

b. For each \({\rm{x}}\) in \({\mathbb{R}^n}\), \({A^ + }A{\rm{x}}\) is the orthogonal projection of \({\rm{x}}\) onto \({\rm{Row}}\,A\).

c. \(A{A^ + }A = A\)and \({A^ + }A{A^ + } = {A^ + }\).

Question: Let \({x_1}\,,{x_2}\) denote the variables for the two-dimensional data in Exercise 1. Find a new variable \({y_1}\) of the form \({y_1} = {c_1}{x_1} + {c_2}{x_2}\), with\(c_1^2 + c_2^2 = 1\), such that \({y_1}\) has maximum possible variance over the given data. How much of the variance in the data is explained by \({y_1}\)?

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).
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