Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #521 : Linear Algebra

$A= \begin{bmatrix} 30 & a & a &a \\ a & -12 & a & a \\ a & a & 13 & a \\ a & a & a & 22 \end{bmatrix}$ where $a \in \mathbb{R}$ and is a diagonally dominant matrix. Note that every nondiagonal element is the same value

Give a necessary and sufficient condition for the four Gershgorin discs of to be mutually disjoint.

Possible Answers:

- 3 <a < 3

$- \frac{8}{3} <a < \frac{8}{3}$

$- \frac{2}{3} <a < \frac{2}{3}$

$- \frac{4}{3} <a < \frac{4}{3}$

$- \frac{1}{3} <a < \frac{1}{3}$

Correct answer:

$- \frac{4}{3} <a < \frac{4}{3}$

Explanation:

A Gershgorin disc (or circle) for a diagonally dominant matrix is a closed circle in the complex plane with one of the diagonal entries as a center. The radius of that particular Gershgorin circle is the sum of the absolute values in the same row as this diagonal element.

The four Gershgorin circles of have their centers at , 13, 22, and 30 on the complex plane (and, specifically, on the real axis). The radius of each of the circles is |3a | ; for the circles to be mutually disjoint, the distance between their centers must be greater than |6a | .

The least distance between any two diagonal elements is 30 - 22 = 8 , so

|6a |< 8 ,

$|a| < \frac{4}{3}$ .

$- \frac{4}{3} <a < \frac{4}{3}$ is a necessary and sufficient condition for no two of the Gershgorin discs to intersect.

Report an Error

Example Question #64 : Eigenvalues And Eigenvectors

A $4 \times 4$ matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Calculate $\det A$ .

Possible Answers:

Correct answer:

Explanation:

The determinant of a matrix is equal to the product of its eigenvalues, so

$\det A = ( 1+i )( 1-i) (-1+i) ( -1-i )$

$= (1^{2}+ 1^{2})\left [ (-1)^{2}+ 1^{2} \right ]$

$= 2 \cdot 2$

= 4

Report an Error

Example Question #65 : Eigenvalues And Eigenvectors

A $4 \times 4$ matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Give the trace of .

Possible Answers:

Correct answer:

Explanation:

The trace of a matrix is equal to the sum of its eigenvalues, so

$\textup{Tr}(A) = (1+i) + (1-i) + (-1+i) + (-1-i) = 0$

Report an Error

Example Question #62 : Eigenvalues And Eigenvectors

The characteristic equation of a matrix is

$\lambda^{5}+ 7\lambda^{3} - 6\lambda^{2}+ 128 = 0$

Give the trace of .

Possible Answers:

$\textup{Tr}(A) = 128$

$\textup{Tr}(A) =- 7$

$\textup{Tr}(A) = 7$

$\textup{Tr}(A) =- 128$

$\textup{Tr}(A) = 0$

Correct answer:

$\textup{Tr}(A) = 0$

Explanation:

The trace of a matrix is equal to the sum of the eigenvalues of the matrix. Since the characteristic equation has degree five, the matrix has five (not necessarily distinct) eigenvalues. If $\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}, \lambda_{5}$ are the eigenvalues of the matrix, then its characteristic equation is

$(\lambda - \lambda_{1}) (\lambda - \lambda_{2})(\lambda - \lambda_{3})(\lambda - \lambda_{4})(\lambda - \lambda_{5})= 0$

When expanded, the coefficient of its degree-four term is $- (\lambda_{1}+ \lambda_{2} +\lambda_{3} +\lambda_{4} +\lambda_{5})$ . The degree-four term in the characteristic polynomial is "missing", so the implied coefficient is 0.

It follows that the sum of the eigenvalues is 0, and that, consequently, $\textup{Tr}(A) = 0$ .

Report an Error

Example Question #67 : Eigenvalues And Eigenvectors

The characteristic equation of a matrix is

$\lambda^{5}+ 7\lambda^{3} - 6\lambda^{2}+ 128 = 0$

Give the determinant of .

Possible Answers:

$\det A =- 7$

$\det A = 128$

$\det A = 0$

$\det A = -128$

$\det A = 7$

Correct answer:

$\det A = -128$

Explanation:

The determinant of a matrix is equal to the product of the eigenvalues of the matrix. Since the characteristic equation has degree five, the matrix has five (not necessarily distinct) eigenvalues. If $\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}, \lambda_{5}$ are the eigenvalues of the matrix, then its characteristic equation is

$(\lambda - \lambda_{1}) (\lambda - \lambda_{2})(\lambda - \lambda_{3})(\lambda - \lambda_{4})(\lambda - \lambda_{5})= 0$

When expanded, this polynomial has constant term $- \lambda_{1} \lambda_{2} \lambda_{3} \lambda_{4} \lambda_{5}$ . It follows that the product of the eigenvalues is , and that, consequently, $\det A = -128$ .

Report an Error

Example Question #68 : Eigenvalues And Eigenvectors

The characteristic equation of a matrix is

$\lambda ^{4}+ 5\lambda ^{3}- 4 \lambda - 200 = 0$

Which statement must be true?

Possible Answers:

$\det A = -5$

None of the other statements need be true.

$\det A = 200$

$\det A = -200$

$\det A = 5$

Correct answer:

$\det A = -200$

Explanation:

The determinant of a matrix is equal to the product of the eigenvalues of the matrix. Since the characteristic equation has degree four, the matrix has four (not necessarily distinct) eigenvalues. If $\lambda_{1}, \lambda_{2}, \lambda_{3}, \lambda_{4}$ are the eigenvalues of the matrix, then its characteristic equation is

$(\lambda - \lambda_{1}) (\lambda - \lambda_{2})(\lambda - \lambda_{3})(\lambda - \lambda_{4})= 0$

When expanded, this polynomial has constant term $+ \lambda_{1} \lambda_{2} \lambda_{3} \lambda_{4}$ . It follows that the product of the eigenvalues is , and that, consequently, $\det A = -200$ .

Report an Error

Example Question #69 : Eigenvalues And Eigenvectors

Consider the diagonally dominant matrix

$A= \begin{bmatrix} 12 & 1 & -2 & 3 \\ 1 & 20 & - 2 & -4 \\ 1 & 2 & 16 & 1 \\ 1 & -2 & 3 & 14 \end{bmatrix}$

Give the radius of the Gershgorin disc with center 20.

Possible Answers:

Correct answer:

Explanation:

A Gershgorin disc for a diagonally dominant matrix is a circle in the complex plane with one of the diagonal entries as a center. The radius of the disc corresponding to that diagonal entry is the sum of the absolute values of the entries in the same row.

Observe the elements on the second row, whose diagonal element is 20.

$A= \begin{bmatrix} 12 & 1 & -2 & 3 \\ {\color{DarkOrange}\textbf{1 }}& \textbf{20 }& {\color{DarkOrange}\textbf{-2}}& {\color{DarkOrange} \textbf{-4}} \\ 1 & 2 & 16 & 1 \\ 1 & -2 & 3 & 14 \end{bmatrix}$

The sum of the absolute values other elements in that row - and the radius of the Gershgorin disc with its center at 20 - is

r= |1| + |-2| + |-4| = 7 .

Report an Error

Example Question #70 : Eigenvalues And Eigenvectors

A matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Give the set of eigenvalues of $A^{-1}$ .

Possible Answers:

$\left \{ \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right \}$

$\begin{Bmatrix} 2+2i, 2-2i, -2+2i, -2-2i \end{Bmatrix}$

$\left \{ \sqrt{2} + i \sqrt{2}, \sqrt{2} - i \sqrt{2}, -\sqrt{2} + i \sqrt{2},- \sqrt{2} - i \sqrt{2} \right \}$

$\left \{ \frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2}- \frac{\sqrt{2}}{2}i , -\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i, -\frac{\sqrt{2}}{2}- \frac{\sqrt{2}}{2}i , \right \}$

$\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$

Correct answer:

$\left \{ \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right \}$

Explanation:

If is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . The eigenvalues of are $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ , so the eigenvalues of these numbers are the reciprocals of these. The reciprocal of 1 + i is

$\frac{1}{1 + i} = \frac{1(1-i)}{(1 + i)(1-i)} = \frac{1 - i}{2} = \frac{1}{2}- \frac{1}{2}i$ .

Similarly,

$\frac{1}{1 - i} = \frac{1}{2}+ \frac{1}{2}i$

$\frac{1}{-1 + i} =- \frac{1}{2}- \frac{1}{2}i$

$\frac{1}{-1 - i} =- \frac{1}{2}+ \frac{1}{2}i$

The set of eigenvalues of $A^{-1}$ is $\left \{ \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right \}$ .

Report an Error

Example Question #451 : Operations And Properties

is an involutory matrix.

True, false, or indeterminate: 0 is an eigenvalue of .

Possible Answers:

True

Indeterminate

False

Correct answer:

False

Explanation:

An eigenvalue of an involutory matrix must be either 1 or . This can be seen as follows:

Let $\lambda$ be an eigenvalue of involutory matrix . Then for some eigenvector $\textbf{v}$ ,

$A\textbf{v} = \lambda \textbf{v}$

Premultiply both sides by :

$AA\textbf{v} =A \lambda \textbf{v}$

$A^{2}\textbf{v} = \lambda A \textbf{v}$

By definition, an involutory matrix has as its square, so

$I \textbf{v} = \lambda A \textbf{v}$

$\textbf{v} = \lambda A \textbf{v}$

$\frac{1}{ \lambda }\textbf{v} =A \textbf{v}$

By transitivity,

$\lambda \textbf{v} = \frac{1}{ \lambda }\textbf{v}$

Thus, $\lambda = \frac{1}{\lambda}$ , or $\lambda^{2} = 1$

It follows that $\lambda = \pm 1$ . The statement is false.

Report an Error

Example Question #452 : Operations And Properties

The trace of a singular $2 \times 2$ matrix is 12. Give its set of eigenvalues.

Possible Answers:

$\left \{ -6, 6\right \}$

$\left \{ 12\right \}$

Insufficient information is given to answer the question.

$\left \{ 0, 6\right \}$

$\left \{ 0, 12\right \}$

Correct answer:

$\left \{ 0, 12\right \}$

Explanation:

, being a singular matrix, must have 0 as an eigenvalue. Let $\lambda$ be its other eigenvalue..

The trace of a matrix is equal to the sum of its eigenvalues, so

$\lambda+ 0 = \textup{Tr}(A)= 12$ ,

and

$\lambda = 12$

The set of eigenvalues of is $\left \{ 0, 12\right \}$ .

Report an Error

View Linear Algebra Tutors

Yosef
Certified Tutor

Cooper Union for the Advancement of Science and Art, Bachelor of Engineering, Mechanical Engineering. Rensselaer Polytechnic ...

View Linear Algebra Tutors

Timothy
Certified Tutor

Bethune-Cookman University, Bachelor of Science, Mathematics. University of Phoenix, Master of Science, Computer and Informat...

View Linear Algebra Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Spanish Tutors in Atlanta, Spanish Tutors in Boston, English Tutors in Washington DC, English Tutors in Los Angeles, SSAT Tutors in Boston, Algebra Tutors in Seattle, English Tutors in San Diego, Biology Tutors in New York City, Spanish Tutors in Philadelphia, ISEE Tutors in Seattle

Popular Courses & Classes

MCAT Courses & Classes in Boston, MCAT Courses & Classes in Miami, ACT Courses & Classes in Denver, GMAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Chicago, SSAT Courses & Classes in Chicago, GRE Courses & Classes in Miami, GRE Courses & Classes in Houston, SSAT Courses & Classes in Los Angeles, ACT Courses & Classes in Chicago

Popular Test Prep

LSAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Atlanta, GRE Test Prep in Denver, GMAT Test Prep in New York City, GRE Test Prep in Los Angeles, SSAT Test Prep in Phoenix, GMAT Test Prep in San Diego, ISEE Test Prep in San Diego, MCAT Test Prep in Dallas Fort Worth, SAT Test Prep in Houston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #521 : Linear Algebra

Example Question #64 : Eigenvalues And Eigenvectors

Example Question #65 : Eigenvalues And Eigenvectors

Example Question #62 : Eigenvalues And Eigenvectors

Example Question #67 : Eigenvalues And Eigenvectors

Example Question #68 : Eigenvalues And Eigenvectors

Example Question #69 : Eigenvalues And Eigenvectors

Example Question #70 : Eigenvalues And Eigenvectors

Example Question #451 : Operations And Properties

Example Question #452 : Operations And Properties

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: