Q8.1-9E For each of the matrices聽A聽in ... [FREE SOLUTION]

Chapter 3: Q8.1-9E (page 110)

For each of the matricesAin Exercises 7 through 11, find an orthogonal matrix S and a diagonal matrix D such that $S^{- 1} A S = D$ . Do not use technology.
9. $A = [\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}]$

Short Answer

Expert verified

The diagonal matrix is $D = [\begin{matrix} 3 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & 2 \end{matrix}]$ and the orthogonal matrix is $S = [\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}]$ .

Step by step solution

Define symmetric matrix

In linear algebra, a symmetric matrix is a square matrix that does not change when its transpose is calculated.
A symmetric matrix is defined as one whose transpose is identical to the matrix itself.
A square matrix of size n x nis symmetric if B^T= B.

Find the eigenvalues of the given matrix

Given,

$A = [\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}]$

According to theorem 7.2.1,

$\det (A - 位 I_{n}) = 0$

The formula for finding eigenspace is

$E_{位} = \ker (A - 位 I_{n}) = \{\overset{鈫�}{v} in {鈩�}^{n} : A \overset{炉}{v} = 位 \overset{炉}{v}\}$

Now we need to find an orthonormal eigenbasis for the given matrix A.

$\det (A - 位 I_{3}) = 0 鈬� d e t ([\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}] - [\begin{matrix} 位 & 0 & 0 \\ 0 & 位 & 0 \\ 0 & 0 & 位 \end{matrix}]) = 0 鈬� d e t ([\begin{matrix} 0 - 位 & 0 - 0 & 3 - 0 \\ 0 - 0 & 2 - 位 & 0 - 0 \\ 3 - 0 & 0 - 0 & 0 - 位 \end{matrix}]) = 0 鈬� d e t ([\begin{matrix} - 位 & 0 & 3 \\ 0 & 2 - 位 & 0 \\ 3 & 0 & - 位 \end{matrix}]) = 0 鈬� - 位 ((2 - 位) (- 位) - 0) - 0 + 3 (0 - (2 - 位) (3)) = 0 鈬� - 位^{3} + 2 位^{2} + 9 位 - 18 = 0 鈬� - (位 - 3) (位 + 3) (位 - 2) = 0 鈬� 位 = 2, - 3, 3$

The eigenvalues of the given matrix A are 2,-3,3.

Find the eigenspaces for λ=2

Eigenspace for $位 = 2$

$E_{2} = \ker (A - 2 I_{3}) 鈬� k e r ([\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}] - [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}]) = k e r [\begin{matrix} 0 - 2 & 0 - 0 & 3 - 0 \\ 0 - 0 & 2 - 2 & 0 - 0 \\ 3 - 0 & 0 - 0 & 0 - 2 \end{matrix}] = k e r [\begin{matrix} - 2 & 0 & 3 \\ 0 & 0 & 0 \\ 3 & 0 & - 2 \end{matrix}]$

Now find the kernel of the matrix.

$鈬� [\begin{matrix} - 2 & 0 & 3 \\ 0 & 0 & 0 \\ 3 & 0 & - 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new first row.

$(- \frac{1}{2} 脳 R_{1} 鈫� R_{1}) 鈬� [\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 0 & 0 \\ 3 & 0 & - 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new third row:

New second row:

$(\frac{2}{5} 脳 R_{2} 鈫� R_{2}) 鈬� [\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

New third row:

$(R_{3} - 3 R_{1} 鈫� R_{3}) 鈬� [\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 0 & 0 \\ 0 & 0 & \frac{5}{2} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] (R_{2} 鈫� R_{3}) 鈬� [\begin{matrix} 1 & 0 & - \frac{3}{2} \\ 0 & 0 & \frac{5}{2} \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

New first row:

$(R_{1} + \frac{3}{2} 脳 R_{2} 鈫� R_{1}) 鈬� [\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

From the above equation we get the solution as

$x_{1} + 0 + 0 = 0 鈬� x_{1} = 0 0 + 0 + x_{3} = 0 鈬� x_{3} = 0$

Therefore, the eigenvector of the given matrix can be represented as:

$鈬� [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ x_{2} \\ 0 \end{matrix}] = x_{2} [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}], x_{2} 鈭� R$

Hence the eigenspace for $位 = 2$ is $E_{2} = s p a n [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}]$ .

Find the eigenspaces for λ=-3

Eigenspace for $位 = - 3$

$E_{- 3} = \ker (A + 3 I_{3}) = k e r ([\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}] + [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]) = k e r [\begin{matrix} 0 + 3 & 0 + 0 & 3 + 0 \\ 0 + 0 & 2 + 3 & 0 + 0 \\ 3 + 0 & 0 + 0 & 0 + 3 \end{matrix}] = k e r [\begin{matrix} 3 & 0 & 3 \\ 0 & 5 & 0 \\ 3 & 0 & 3 \end{matrix}]$

Find the kernel of the matrix.

$鈬� [\begin{matrix} 3 & 0 & 3 \\ 0 & 5 & 0 \\ 3 & 0 & 3 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new first row.

$(\frac{1}{3} 脳 R_{1} 鈫� R_{1}) 鈬� [\begin{matrix} 1 & 0 & 1 \\ 0 & 5 & 0 \\ 3 & 0 & 3 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new third row:

$(R_{3} - 3 R_{1} 鈫� R_{3}) 鈬� [\begin{matrix} 1 & 0 & 1 \\ 0 & 5 & 0 \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

New second row:

$(\frac{1}{5} 脳 R_{2} 鈫� R_{2}) 鈬� [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

From equation (1) we get,

$x_{1} + 0 + x_{3} = 0 鈬� x_{1} = - x_{3} 0 + x_{2} + 0 = 0 鈬� x_{2} = 0$

Therefore the eigenvector is given as

$鈬� [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} - x_{3} \\ 0 \\ x_{3} \end{matrix}] = x_{3} [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}], x_{3} 鈭� R$

Hence the eigenspace for $位 = - 3$ is $E_{- 3} = s p a n [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}]$ .

Find the eigenspaces for λ=3

Eigenspace for $位 = 3$

$E_{3} = \ker (A - 3 I_{3}) = k e r ([\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}] - [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]) = k e r [\begin{matrix} 0 - 3 & 0 - 0 & 3 - 0 \\ 0 - 0 & 2 - 3 & 0 - 0 \\ 3 - 0 & 0 - 0 & 0 - 3 \end{matrix}] = k e r [\begin{matrix} - 3 & 0 & 3 \\ 0 & - 1 & 0 \\ 3 & 0 & - 3 \end{matrix}]$

Finding the kernel:

$鈬� [\begin{matrix} - 3 & 0 & 3 \\ 0 & - 1 & 0 \\ 3 & 0 & - 3 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new first row:

$(- \frac{1}{3} 脳 R_{1} 鈫� R_{1}) 鈬� [\begin{matrix} 1 & 0 & - 1 \\ 0 & - 1 & 0 \\ 3 & 0 & - 3 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new third row:

$(R_{3} - 3 R_{1} 鈫� R_{3}) 鈬� [\begin{matrix} 1 & 0 & - 1 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

To get new second row:

$(- 1 脳 R_{2} 鈫� R_{2}) 鈬� [\begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}]$

From the above equation we get,

$x_{1} + 0 - x_{3} = 0 鈬� x_{1} = x_{3} 0 + x_{2} + 0 = 0 鈬� x_{2} = 0$

The eigenvectors are:

$鈬� [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} x_{3} \\ 0 \\ x_{3} \end{matrix}] = x_{3} [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}], x_{3} 鈭� R$

Hence the eigenspace for $位 = 3$ is $E_{3} = s p a n [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}]$ .

Find the orthogonal matrix

The eigenspaces are perpendicular to each other. The orthonormal eigenbasis can be calculated as follows:

$\overset{鈫�}{v_{1}} = \frac{1}{\sqrt{1^{2} + 0^{2} + 1^{2}}} [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] \overset{鈫�}{v_{1}} = \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] \overset{鈫�}{v_{2}} = \frac{1}{\sqrt{{(- 1)}^{2} + 0^{2} + 1^{2}}} [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] \overset{鈫�}{v_{2}} = \frac{1}{\sqrt{2}} [\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}] \overset{鈫�}{v_{3}} = \frac{1}{\sqrt{0^{2} + 1^{2} + 0^{2}}} [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] \overset{鈫�}{v_{3}} = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}]$

$S = [\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}]$

Find the inverse of orthogonal matrix:

$S^{- 1} = {[\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}]}^{- 1} = \frac{1}{- 1 (\frac{1}{\sqrt{2}} 脳 \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} 脳 (- \frac{1}{\sqrt{2}}))} a d j [\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}] = \frac{1}{- 1 (1)} [\begin{matrix} - \frac{1}{\sqrt{2}} & 0 & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & - \frac{1}{\sqrt{2}} \\ 0 & - 1 & 0 \end{matrix}] = S^{- 1} [\begin{matrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \end{matrix}]$

Hence the orthogonal matrix is $S = [\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}]$ .

Find the diagonal matrix

Now we have to find the diagonal matrix as below:

$D = S^{- 1} A S = [\begin{matrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}] = [\begin{matrix} \frac{3}{\sqrt{2}} & 0 & \frac{3}{\sqrt{2}} \\ \frac{3}{\sqrt{2}} & 0 & - \frac{3}{\sqrt{2}} \\ 0 & 2 & 0 \end{matrix}] [\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{matrix}]$

Hence the diagonal matrix is $D = [\begin{matrix} 3 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & 2 \end{matrix}]$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

91影视

For each of the matricesAin Exercises 7 through 11, find an orthogonal matrix S and a diagonal matrix D such that $S^{- 1} A S = D$ . Do not use technology.
9. $A = [\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}]$

Short Answer

Step by step solution

Define symmetric matrix

Find the eigenvalues of the given matrix

Find the eigenspaces for λ=2

Find the eigenspaces for λ=-3

Find the eigenspaces for λ=3

Find the orthogonal matrix

Find the diagonal matrix

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Applied Mathematics

Discrete Mathematics

Calculus

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

91影视

For each of the matricesAin Exercises 7 through 11, find an orthogonal matrix S and a diagonal matrix D such that S-1AS=D. Do not use technology.9.A=003020300

Short Answer

Step by step solution

Define symmetric matrix

Find the eigenvalues of the given matrix

Find the eigenspaces for λ=2

Find the eigenspaces for λ=-3

Find the eigenspaces for λ=3

Find the orthogonal matrix

Find the diagonal matrix

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Applied Mathematics

Discrete Mathematics

Calculus

Logic and Functions

Probability and Statistics

Study anywhere. Anytime. Across all devices.

For each of the matricesAin Exercises 7 through 11, find an orthogonal matrix S and a diagonal matrix D such that $S^{- 1} A S = D$ . Do not use technology.
9. $A = [\begin{matrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 3 & 0 & 0 \end{matrix}]$