Purdue MA 26500 Fall 2022 Midterm II Solutions
Here comes the solution and analysis for Purdue MA 26500 Fall 2022 Midterm II. This second midterm covers topics in Chapter 4 (Vector Spaces) and Chapter 5 (Eigenvalues and Eigenvectors) of the textbook.
Introduction
Purdue Department of Mathematics provides a linear algebra course MA 26500 every semester, which is mandatory for undergraduate students of almost all science and engineering majors.
Textbook and Study Guide
Disclosure: This blog site is reader-supported. When you buy through the affiliate links below, as an Amazon Associate, I earn a tiny commission from qualifying purchases. Thank you.
MA 26500 textbook is Linear Algebra and its Applications (6th Edition) by David C. Lay, Steven R. Lay, and Judi J. McDonald. The authors have also published a student study guide for it, which is available for purchase on Amazon as well.
Exam Information
MA 26500 midterm II covers the topics of Sections 4.1 – 5.7 in the textbook. It is usually scheduled at the beginning of the thirteenth week. The exam format is a combination of multiple-choice questions and short-answer questions. Students are given one hour to finish answering the exam questions.
Based on the knowledge of linear equations and matrix algebra learned in the book chapters 1 and 2, Chapter 4 leads the student to a deep dive into the vector space framework. Chapter 5 introduces the important concepts of eigenvectors and eigenvalues. They are useful throughout pure and applied mathematics. Eigenvalues are also used to study differential equations and continuous dynamical systems, they provide critical information in engineering design,
Reference Links
- Purdue Department of Mathematics Course Achive
- Purdue MA 26500 Spring 2024
- Purdue MA 26500 Exam Archive
Fall 2022 Midterm II Solutions
Problem 1 (10 points)
Let
- A. 25
- B. 17
- C. 9
- D. 1
- E. 0
Problem 1 Solution
Do row reduction as follows:
- Add
times row 1 to row 2 - Add
times row 1 to row 2 - Scale row 2 by
- Add
times row 2 to row 3
So we have 2 pivots, the rank is 2 and the nullity is 3. This results in
The answer is C.
Problem 2 (10 points)
Let
- A.
- B.
- C.
- D.
- E.
Problem 2 Solution
For set
As can be seen, we need 3 pivots to make these column vectors linearly independent. If
The answer is B.
Problem 3 (10 points)
Which of the following statements is always TRUE?
- A. If
for some vector , then is an eigenvalue of . - B. If
is an eigenvector corresponding to eigenvalue 2, then is an eigenvector corresonding to eigenvalue . - C. If
is invertible, then matrix and could have different sets of eigenvalues. - D. If
is an eigenvalue of matrix , then is an eigenvalue of matrix . - E. If
is an eigenvalue of matrix , then matrix is not invertible.
Problem 3 Solution
Per definitions in 5.1 "Eigenvectors and Eigenvalues":
An eigenvector of an
matrix is a nonzero vector such that for some scalar . A scalar is called an eigenvalue of if there is a nontrivial solution of ; such an is called an eigenvector corresponding to .
Statement A is missing the "nonzero" keyword, so it is NOT always TRUE.
For Statement B, given
Statement C involves the definition of Similarity. Denote
This can be proved easily, as seen below
Since , we see that . ■
For Statement D, given
Statement E is FALSE. An eigenvalue
The answer is D.
Problem 4 (10 points)
Let
- A set of polynomials in
satisfying .
- A set of polynomials in
satisfying .
- A set of polynomials in
with integer coefficients.
- A. (i) only
- B. (i) and (ii) only
- C. (i) and (iii) only
- D. (ii) only
- E. (ii) and (iii) only
Problem 4 Solution
Per the definition of Subspace in Section 4.1 "Vector Spaces and Subspaces"
A subspace of a vector space
is a subset of that has three properties:
a. The zero vector ofis in .
b.is closed under vector addition. That is, for each and in , the sum is in .
c.is closed under multiplication by scalars. That is, for each in and each scalar , the vector is in .
So to be qualified as the subspace, the subset should have all the above three properties. Denote the polynomials as
(i) Since
, we have , so .- Obviously, it satisfies the first property as if
for all , is true as well. - Now assume
and are two polynomials in this set and So we have and . Then define a third polynomial It is true that as well. So the set has the second property. - This set does have the third property since
has and it is also in the same set.
This proves that set (i) is a subspace of
.- Obviously, it satisfies the first property as if
(ii) From
, we can deduce that . So any polynomial in this set should satisfy this condition.- Obviously, it satisfies the first property as if
for all , is true as well. - With the same notation of
, and . We have If and , the above ends up with , which is not necessary equal 0. So this polynomial in this set does NOT have the second property.
This proves that set (ii) is NOT a subspace of
.- Obviously, it satisfies the first property as if
(iii) It is easy to tell that this set is NOT a subspace of
. If we do multiplication by floating-point scalars, the new polynomial does not necessarily have an integer coefficient for each term and might not be in the same set.
So the answer is A.
Problem 5 (10 points)
Consider the differential equation
Then the origin is
- A. an attractor
- B. a repeller
- C. a saddle point
- D. a spiral point
- E. none of the above
Problem 5 Solution
First, write the system as a matrix differential equation
Now let's find out the eigenvalues of
Referring to the Complex Eigenvalues discussion at the end of this section, "the origin is called a spiral point of the dynamical system. The rotation is caused by the sine and cosine functions that arise from a complex eigenvalue". Because the complex eigenvalues have a positive real part, the trajectories spiral outward.
So the answer is D.
Refer to the following table for the mapping from
matrix eigenvalues to trajectories:
Eigenvalues Trajectories Repeller/Source Attactor/Sink Saddle Point Spiral (outward) Point Spiral (inward) Point Ellipses (circles if )
Problem 6 (10 points)
Which of the following matrices are diagonalizable over the real numbers?
(ii) (iii) (iv)
- A. (i) and (iii) only
- B. (iii) and (iv) only
- C. (i), (iii) and (iv) only
- D. (i), (ii) and (iii) only
- E. (i), (ii) and (iv) only
Problem 6 Solution
This problem tests our knowledge of Theorem 6 of Section 5.3 "Diagonalization":
An
matrix with distinct eigenvalues is diagonalizable.
So let's find out the eigenvalues for each matrix:
- From the equation
, we can obtain This leads to two roots , .
- From the equation
- Since this is a triangular matrix, the eigenvalue is just 4, with multiplicity 2.
- For the same reason, this
matrix has eigenvalues 1, 5 and 7.
- For the same reason, this
- Use cofactor expansion with
, we have The eigenvalues are 7, 4, and 1.
- Use cofactor expansion with
Now we can see that (i), (iii), and (iv) have distinct eigenvalues, they are diagonalizable matrices.
So the answer is C.
Problem 7 (10 points)
A real
- A.
- B.
- C.
- D.
- E.
Problem 7 Solution
From Section 5.7 "Applications to Differential Equations", we learn that the general solution to a matrix differential equation is
Now use Euler's formula (
So the answer is E.
Problem 8 (10 points)
Let
(2 points) (1) Find
(4 points) (2) Find a basis for the range of
(4 points) (3) Find a basis for the kernel of
Problem 8 Solution
As the mapping rule is
, we can directly write down the transformation as belowIf we denote the 4 entries of a
matrix as , the transformation can be written as So the basis can be the set of three matrices like belowThe kernel (or null space) of such a
is the set of all in vector space such that . Write this as This leads to and . So the original matrix that satified this conditioncan be represented as . This shows that (or ) is the basis for the null space of .
Problem 9 (10 points)
(6 points) (1) Find all the eigenvalues of matrix
(4 points) (2) Find an invertible matrix
Problem 9 Solution
- Apply the equation
, we have So the eigenvalues are 4 an 1. Now to find eigenvector for each eigenvalue, we take the eigenvalue to the system and find the basis vector(s) which would be the eigenvector.- For
, we have the new matrix as This gives with two free variables and . Now in parametric vector form, we can obtain A basis is . - For
, the new matrix is This gives and with one free variable . Again in parametric vector form, we can obtain A basis is .
- For
- From the above solution we can directly write
and below
Problem 10 (10 points)
(4 points) (1) Find the eigenvalues and corresponding eigenvectors of the matrix
(2 points) (2) Find a general solution to the system of differential equations
(4 points) (3) Let
Problem 10 Solution
- To find eigenvalues, write down the determinant as
So the eigenvalues are and . Now follow the same method as Problem 9 solution to get eigenvectors for them.- For
, the new matrix is The eigenvector is . - For
, the new matrix is The eigenvector is .
- For
- The general solution to a matrix differential equation is
So from this, since we already found out the eigenvalues and the corresponding eigenvectors, we can write down - Now apply the initial values of
and , here comes the following equations: This gives and . So .
Summary
Here is the table listing the key knowledge points for each problem in this exam:
Problem # | Points of Knowledge | Book Sections |
---|---|---|
1 | The Rank Theorem | 4.6 "Rank" |
2 | Linear dependence, Invertible Matrix Theorem | 4.3 "Linearly Independent Sets; Bases", 4.6 "Rank" |
3 | Eigenvectors and Eigenvalues | 5.1 "Eigenvectors and Eigenvalues" |
4 | Vector Spaces and Subspaces | 4.1 "Vector Spaces and Subspaces" |
5 | Eigenfunctions of the Differential Equation | 5.7 "Applications to Differential Equations" |
6 | The Diagonalization Theorem, Diagonalizing Matrices | 5.3 "Diagonalization" |
7 | Complex Eigenvalues and Eigenvectors | 5.5 "Complex Eigenvalues" |
8 | Kernel and Range of a Linear Transformation | 4.2 "Null Spaces, Column Spaces, and Linear Transformations" |
9 | Eigenvalues, Basis for Eigenspace, Diagonalizing Matrices | 5.1 "Eigenvectors and Eigenvalues", 5.3 "Diagonalization" |
10 | Eigenvectors and Eigenvalues | 5.1 "Eigenvectors and Eigenvalues", 5.7 "Applications to Differential Equations" |
Gitalking ...