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 #41 : Eigenvalues And Eigenvectors

A $4 \times 4$ matrix has exactly two distinct eigenvalues: 1 and . Which of the following cannot be its characteristic equation?

Possible Answers:

$\lambda ^{4} - 2 \lambda^{2} + 1 = 0$

$\lambda^{4}+2 \lambda^{3}-2 \lambda - 1 = 0$

$\lambda^{4}-2 \lambda^{3}+2 \lambda - 1 = 0$

Any of the four choices are possible characteristic equations.

$\lambda ^{4}+ 2 \lambda^{2} + 1 = 0$

Correct answer:

$\lambda ^{4}+ 2 \lambda^{2} + 1 = 0$

Explanation:

A $4 \times 4$ matrix has four not necessarily distinct eigenvalues, $\lambda_{1} , \lambda_{2} , \lambda_{3} , \lambda_{4}$ , which are the solutions of its characteristic polynomial equation

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

If there are only two distinct eigenvalues, 1 and , then one of the following three situations happens:

1 has multiplicity 3 and has multiplicity 1, in which case $\lambda_{1} = \lambda_{2} = \lambda_{3}=1 , \lambda_{4} = -1$ ,

and the characteristic equation is

$(\lambda - 1) (\lambda - 1) (\lambda - 1 ) (\lambda +1)=0$

$\lambda^{4}-2 \lambda^{3}+2 \lambda - 1 = 0$

1 has multiplicity 2 and has multiplicity 2, in which case $\lambda_{1} = \lambda_{2} = 1 , \lambda_{3}= \lambda_{4} = -1$ ,

and the characteristic equation is

$(\lambda - 1) (\lambda - 1) (\lambda + 1 ) (\lambda +1)=0$

$\lambda ^{4} - 2 \lambda^{2} + 1 = 0$

1 has multiplicity 1 and has multiplicity 3, in which case $\lambda_{1} = 1 , \lambda_{2} = \lambda_{3}= \lambda_{4} = -1$

and the characteristic equation is

$(\lambda - 1) (\lambda + 1) (\lambda + 1 ) (\lambda +1)=0$

$\lambda^{4}+2 \lambda^{3}-2 \lambda - 1 = 0$

Of the four equations given as choices, only $\lambda ^{4}+ 2 \lambda^{2} + 1 = 0$ cannot be a characteristic equation of the matrix.

Report an Error

Example Question #41 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 3 &-2 \\ - 2 &3 \end{bmatrix}$

Which of the following statements follows from the Cayley-Hamilton Theorem?

Possible Answers:

has two distinct real eigenvalues.

and $A^{2}$ have the same eigenvalues.

Every vector in $\mathbb{R}^{2}$ is an eigenvector of .

None of the other choices gives a correct response.

has one eigenvalue of multiplicity 2.

Correct answer:

None of the other choices gives a correct response.

Explanation:

By the Cayley-Hamilton Theorem, a matrix is a solution of its own characteristic polynomial equation. None of the choices addresses the characteristic equation of .

Report an Error

Example Question #41 : Eigenvalues And Eigenvectors

$A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$ .

Is 2 an eigenvalue of , and if so, what is the dimension of its eigenspace?

Possible Answers:

Yes; the dimension is 3.

Yes; the dimension is 1.

No.

Yes; the dimension is 2.

Correct answer:

Yes; the dimension is 1.

Explanation:

Assume that 2 is an eigenvalue of . Then, if $b = \begin{bmatrix} b_{1} \\ b_{2} \\b_{3} \end{bmatrix}$ is one of its eigenvectors, it follows that

Ab = 2 b , or, equivalently,

(A - 2I) b =O ,

where I,O are the $3 \times 3$ identity and zero matrices, respectively.

$A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}$ , so

$A - 2 I = \begin{bmatrix} 2-2 & 3 & 6 \\ 0 & 4-2 & 4 \\ 2 & 0 & 4-2 \end{bmatrix}$

$= \begin{bmatrix} 0& 3 & 6 \\ 0 & 2 & 4 \\ 2 & 0 & 2\end{bmatrix}$

Changing to reduced row echelon form:

$R1 \leftrightarrow R3$

$\begin{bmatrix}2 & 0 & 2 \\ 0 & 2 & 4 \\ 0& 3 & 6 \end{bmatrix}$

$\frac{1}{2}R1 \rightarrow R1$

$\begin{bmatrix}1 & 0 & 1 \\ 0 & 2 & 4 \\ 0& 3 & 6 \end{bmatrix}$

$\frac{1}{2}R2 \rightarrow R2$

$\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2\\ 0& 3 & 6 \end{bmatrix}$

$-3R2+ R3 \rightarrow R3$

$\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2\\ 0& 0 & 0 \end{bmatrix}$

The matrix is in reduced row echelon form. There is one column which does not contain leading 1s, so 2 is indeed an eigenvalue. The eigenspace is the set of all vectors (x, y, z) such that x= -z , y = -2z - that is, the set of vectors t(-1,- 2, 1) . The eigenspace has dimension 1.

Report an Error

Example Question #41 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} a & 4 & -2 \\ 1 & 3 & 0 \\ - 6 & 4 & a \end{bmatrix}$

Evaluate so that the sum of the eigenvalues of is 10.

Possible Answers:

a = -6.5

a = 7

a = 3.5

The sum of the eigenvalues of is 10 regardless of the value of .

a = -13

Correct answer:

a = 3.5

Explanation:

The sum of the eigenvalues of a square matrix is equal to its trace, the sum of its diagonal elements. Examine these elements, which are in red below:

$A = \begin{bmatrix} {\color{Red}\textbf{a} } & 4 & -2 \\ 1 & {\color{Red}\textbf{3}} & 0 \\ - 6 & 4 & {\color{Red}\textbf{a}} \end{bmatrix}$

Set the trace equal to 10 and solve for :

$\textup{Tr} (A ) = 10$

a+ a + 3 = 10

2 a+ 3 = 10

2a = 7

a = 3.5

Report an Error

Example Question #41 : Eigenvalues And Eigenvectors

$A= \begin{bmatrix} 0.1 & 0.4 & 0. 7 \\ 0. 5 & 0. 2 & 0.3 \\ 0.4 &0.4 & 0 \end{bmatrix}$

True or false: 1 is an eigenvalue of .

Possible Answers:

True

False

Correct answer:

True

Explanation:

Examine the columns of :

$A= \begin{bmatrix} {\color{Red} 0.1} & {\color{DarkGreen} 0.4} & {\color{Blue} 0. 7} \\ {\color{Red} 0. 5} & {\color{DarkGreen} 0. 2} &{\color{Blue} 0.3} \\ {\color{Red} 0.4 }&{\color{DarkGreen} 0.4} & {\color{Blue} 0} \end{bmatrix}$

The entries of each column add up to 1:

Column 1: 0.1 + 0.5 + 0.4 = 1

Column 2: 0.4+ 0.2 + 0.4 = 1

Column 3: 0.7+0.3 + 0 = 1

It follows that could be a stochastic matrix for a state system; one of the properties of such a matrix is that one of its eigenvalues must be 1.

Report an Error

Example Question #41 : Eigenvalues And Eigenvectors

A $4 \times 4$ real matrix has as two of its eigenvalues and 2-3i . Give its characteristic equation.

Possible Answers:

$\lambda ^{4}+\lambda ^{2}-36 \lambda +52 = 0$

$\lambda ^{4}-4 \lambda^{3}+9 \lambda^{2}+16 \lambda-52 = 0$

Insufficient information is provided to answer the question.

$\lambda ^{4}+\lambda ^{2}+36 \lambda +52 = 0$

$\lambda ^{4}+4 \lambda ^{3}+9 \lambda ^{2}-16 \lambda -52 = 0$

Correct answer:

Insufficient information is provided to answer the question.

Explanation:

A $4 \times 4$ matrix will have four eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since 2-3i is a zero, so it 2+3i .

We now know three of the zeroes, and since only one eigenvalue is unknown, it must be a real value. However, there is no restriction on this zero. Therefore, we have no way of determining the fourth zero - and, consequently, no way of figuring out the characteristic equation with any certainty.

Report an Error

Example Question #42 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 7 & a \\ -3 & b \end{bmatrix}$ for some $a,b \in \mathbb{C}$ .

has as its eigenvalues $4+ \sqrt{3}$ and $4 - \sqrt{3}$ . Which of the following is equal to ?

Possible Answers:

a = -2

a= -4

a= 2

a= 4

a=0

Correct answer:

a= 2

Explanation:

The sum of the eigenvalues of a matrix is equal to the trace of a matrix - the sum of its diagonal elements; the product of a matrix is equal to its determinant.

Therefore, to find , it is necessary to first find . The trace of is equal to diagonal sum b+7 , so set that equal to the sum of the given eigenvalues and solve:

$b+7 = (4+ \sqrt{3}) + (4- \sqrt{3})$

b+7 = 8

b =1

The matrix is therefore

$A = \begin{bmatrix} 7 & a \\ -3 & 1 \end{bmatrix}$

can be found by setting the determinant - the upper-left to lower-right product minus the upper-right to lower-left product - equal to the product of the eigenvalues:

$7(1)-a(-3) = (4+ \sqrt{3}) (4- \sqrt{3})$

3a+ 7 = 13

3a = 6

a= 2 ,

the correct choice.

Report an Error

Example Question #41 : Eigenvalues And Eigenvectors

$A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$

Which of the following holds for the eigenvectors of ?

Possible Answers:

All eigenvectors of are scalar multiples of $\begin{bmatrix} 3 \\ 2 \\ 1 \end{bmatrix}$ .

All eigenvectors of are scalar multiples of $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ .

All eigenvectors of are scalar multiples of $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ .

Every vector in $\mathbb{R}^{3}$ is an eigenvector of .

All eigenvectors of are scalar multiples of $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ .

Correct answer:

All eigenvectors of are scalar multiples of $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ .

Explanation:

is a lower triangular matrix - all elements above its main diagonal are equal to 0. Such a matrix has as its eigenvalues its diagonal elements. All three diagonal elements in are 1, so 1 is the only eigenvalue of .

To find the eigenvector(s) of corresponding to $\lambda = 1$ , first, find the matrix

$A - \lambda I = A - I$ .

$A-I = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}- \begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\ 0&0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 3 & 2 & 0 \end{bmatrix}$

Perform the Gauss-Jordan elimination process to get this matrix in reduced row-echelon form:

$R1 \leftrightarrow R2$

$R2 \leftrightarrow R3$

$\begin{bmatrix} 2 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2} R1 \rightarrow R1$

$\begin{bmatrix} 1 & 0 & 0 \\ 3 & 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$-3R1 + R2 \rightarrow R2$

$\begin{bmatrix} 1 & 0 & 0 \\ 0& 2 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$\frac{1}{2} R2 \rightarrow R2$

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

This matrix is interpreted as

x = 0 , y = 0 , arbitrary.

The eigenvectors must all take the form

$\begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}$

for some scalar - equivalently, all eigenvectors of are scalar multiples of

$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ .

Report an Error

Example Question #51 : Eigenvalues And Eigenvectors

A $3 \times 3$ real matrix has as two of its eigenvalues 2 and 1 +5 i . Give its characteristic equation.

Possible Answers:

$\lambda ^{3} -28 \lambda+48 = 0$

Insufficient information is provided to answer the question.

$\lambda ^{3} -4\lambda ^{2} +30\lambda -52 = 0$

$\lambda ^{3} -4\lambda ^{2} -20\lambda+48 = 0$

$\lambda ^{3} +22 \lambda -52 = 0$

Correct answer:

$\lambda ^{3} -4\lambda ^{2} +30\lambda -52 = 0$

Explanation:

A $3 \times 3$ matrix will have three (not necessarily distinct) eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since 1 +5 i is a zero of this equation, so is its complex conjugate, 1 -5 i .

The characteristic equation of a $3 \times 3$ equation sets a degree-3 polynomial equal to 0. Since 2, 1 +5 i , and 1 -5 i are its zeroes, this polynomial is

$(\lambda -2) [\lambda -(1+5i)][\lambda -(1-5i)]$

$=(\lambda -2) [(\lambda -1)-5i] [(\lambda -1)+5i]$

$=(\lambda -2) [(\lambda -1)^{2}- (5i)^{2}]$

$=(\lambda -2) (\lambda ^{2} -2\lambda +1 +25)$

$=(\lambda -2) (\lambda ^{2} -2\lambda +26)$

$= \lambda ^{3} -4\lambda ^{2} +30\lambda -52$ ,

The characteristic equation is

$\lambda ^{3} -4\lambda ^{2} +30\lambda -52 = 0$ .

Report an Error

Example Question #52 : Eigenvalues And Eigenvectors

is a $3 \times 3$ matrix; $\lambda$ is an eigenvalue of , with eigenspace of dimension 2. $\textbf{v}_{1}= \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$ and $\textbf{v}_{2} =\begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix}$ are two eigenvectors of corresponding to $\lambda$ .

Does exist so that $\textbf{u} = \begin{bmatrix} 1 \\ 0 \\ x \end{bmatrix}$ is also an eigenvector of corresponding to $\lambda$ ? If so, what is ?

Possible Answers:

x= 5

x = -5

x = 5

No such exists.

x= 3

Correct answer:

x= 3

Explanation:

The two eigenvectors given, $\textbf{v}_{1}$ and $\textbf{v}_{2}$ , are linearly independent, since they are not scalar multiples of each other; therefore, they form a basis of the 2-dimensional eigenspace of $\lambda$ . $\textbf{u}$ is an eigenvector corresponding to $\lambda$ if and only if it is a linear combination of the two given basis vectors, or, equivalently, if there exist a,b so that

$a \textbf{v}_{1} +b \textbf{v}_{2} = \textbf{u}$

Rewrite this as follows:

$a \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} +b \begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ x \end{bmatrix}$

This is equivalent to a system of three linear equations in two variables:

a-b = 1

-2a+b = 0

a-2b = x

Solve the system by first, rewriting the second equation as

b= 2a

The first equation becomes

a-2a = 1

-a = 1

a = -1

Substitute in the first equation:

-1 -b = 1

-b = 2

b = -2

Now substitute for and in the third equation:

-1-2 (-2) = x

x= 3 , the correct response.

Report an Error

View Linear Algebra Tutors

Glen
Certified Tutor

Gonzaga University, Bachelor of Science, Computer Science. Bowling Green State University-Main Campus, Master of Science, Com...

View Linear Algebra Tutors

Mohammad
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Mathematics. Rutgers University-New Brunswick, Master of Science, Stat...

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

SSAT Tutors in Atlanta, GRE Tutors in Atlanta, Reading Tutors in Philadelphia, ACT Tutors in Denver, Computer Science Tutors in Denver, ACT Tutors in Chicago, English Tutors in Dallas Fort Worth, Spanish Tutors in Boston, SAT Tutors in Houston, French Tutors in Chicago

Popular Courses & Classes

Spanish Courses & Classes in Chicago, GRE Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in New York City, GMAT Courses & Classes in Atlanta, SSAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Los Angeles

Popular Test Prep

ACT Test Prep in Houston, GMAT Test Prep in Seattle, SSAT Test Prep in Phoenix, SSAT Test Prep in Atlanta, ISEE Test Prep in Houston, SAT Test Prep in San Diego, LSAT Test Prep in Philadelphia, ISEE Test Prep in San Diego, LSAT Test Prep in Miami, GMAT Test Prep in New York City

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 #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #42 : Eigenvalues And Eigenvectors

Example Question #41 : Eigenvalues And Eigenvectors

Example Question #51 : Eigenvalues And Eigenvectors

Example Question #52 : Eigenvalues And Eigenvectors

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: