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

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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,

Fall 2022 Midterm II Solutions

Problem 1 (10 points)

Let A=[102011242123732] Let a be the rank of A and b be the nullity of A, find 5b3a

  • A. 25
  • B. 17
  • C. 9
  • D. 1
  • E. 0

Problem 1 Solution

Do row reduction as follows:

  1. Add 1 times row 1 to row 2
  2. Add 2 times row 1 to row 2
  3. Scale row 2 by 12
  4. Add 3 times row 2 to row 3

[102011242123732][102010222003330][102010111000000]

So we have 2 pivots, the rank is 2 and the nullity is 3. This results in 5b3a=5332=9.

The answer is C.

Problem 2 (10 points)

Let uu=[201], vv=[310], and ww=[11c] where c is a real number. The set {uu,vv,ww} is a basis for R3 provided that c is not equal

  • A. 2
  • B. 2
  • C. 3
  • D. 3
  • E. 1

Problem 2 Solution

For set {uu,vv,ww} to be a basis for R3, the three vectors should be linearly independent. Let's create a matrix with these vectors as columns, then do row reduction like below [23101110c][10c011231][10c0110312c][10c0110042c]

As can be seen, we need 3 pivots to make these column vectors linearly independent. If c is 2, the last row above has all-zero entries, there would be only 2 pivots. So C cannot be 2 for these three vectors to be linearly independent.

The answer is B.

Problem 3 (10 points)

Which of the following statements is always TRUE?

  • A. If Axx=λxx for some vector xx, then λ is an eigenvalue of A.
  • B. If vv is an eigenvector corresponding to eigenvalue 2, then vv is an eigenvector corresonding to eigenvalue 2.
  • C. If B is invertible, then matrix A and B1AB could have different sets of eigenvalues.
  • D. If λ is an eigenvalue of matrix A, then λ2 is an eigenvalue of matrix A2.
  • E. If 5 is an eigenvalue of matrix B, then matrix B5I is not invertible.

Problem 3 Solution

Per definitions in 5.1 "Eigenvectors and Eigenvalues":

An eigenvector of an n×n matrix A is a nonzero vector xx such that Axx=λxx for some scalar λ. A scalar λ is called an eigenvalue of A if there is a nontrivial solution xx of Axx=λxx; such an xx is called an eigenvector corresponding to λ.

Statement A is missing the "nonzero" keyword, so it is NOT always TRUE.

For Statement B, given Avv=2vv, we can obtain A(vv)=2(vv). The eigenvalue is still 2, not 2. This statement is FALSE.

Statement C involves the definition of Similarity. Denote P=B1AB, we have BPB1=BB1ABB1=A So A and P are similar. Similar matrices have the same eigenvalues (Theorem 4 in Section 5.2 "The Characteristic Equation"). Statement C is FALSE

This can be proved easily, as seen below det(AλI)=det(BPB1λI)=det(BPB1λBB1)=det(B)det(PλI)det(B1)=det(B)det(B1)det(PλI) Since det(B)det(B1)=det(BB1)=detI=1, we see that det(AλI)=det(PλI). ■

For Statement D, given Axx=λxx, we can do the following deduction A2xx=AAxx=Aλxx=λAxx=λ2xx So it is always TRUE that λ2 is an eigenvalue of matrix A2.

Statement E is FALSE. An eigenvalue 5 means matrix B(5)I is not invertible since det(B(5)I)=det(B+5I)=0. But the statement refers to a different matrix B5I.

The answer is D.

Problem 4 (10 points)

Let P3 be the vector space of all polynomials of degree at most 3. Which of the following subsets are subspaces of P3?

  1. A set of polynomials in P3 satisfying p(0)=p(1).
  2. A set of polynomials in P3 satisfying p(0)p(1)=0.
  3. A set of polynomials in P3 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 V is a subset H of V that has three properties:
a. The zero vector of V is in H.
b. H is closed under vector addition. That is, for each uu and vv in H, the sum uu+vv is in H.
c. H is closed under multiplication by scalars. That is, for each uu in H and each scalar c, the vector cuu is in H.

So to be qualified as the subspace, the subset should have all the above three properties. Denote the polynomials as p(x)=a0+a1x+a2x2+a3x3.

  • (i) Since p(0)=p(1), we have a0=a0+a1+a2+a3, so a1+a2+a3=0.

    • Obviously, it satisfies the first property as if ai=0 for all i, a1+a2+a3=0 is true as well.
    • Now assume p1(x) and p2(x) are two polynomials in this set and p1(x)=a0+a1x+a2x2+a3x3p2(x)=b0+b1x+b2x2+b3x3 So we have a1+a2+a3=0 and b1+b2+b3=0. Then define a third polynomial p3(x)=p1(x)+p2(x)=(a0+b0)+(a1+b1)x+(a2+b2)x2+(a3+b3)x3=c0+c1x+c2x2+c3x3 It is true that c1+c2+c3=0 as well. So the set has the second property.
    • This set does have the third property since cp(x) has ca1+ca2+ca3=0 and it is also in the same set.

    This proves that set (i) is a subspace of P3.

  • (ii) From p(0)p(1)=0, we can deduce that a0(a0+a1+a2+a3)=0. So any polynomial in this set should satisfy this condition.

    • Obviously, it satisfies the first property as if ai=0 for all i, a0(a0+a1+a2+a3)=0 is true as well.
    • With the same notation of p1(x), p2(x) and p3(x). We have c0(c0+c1+c2+c3)=(a0+b0)(a0+b0+a1+b1+a2+b2+a3+b3)=(a0+b0)((a0+a1+a2+a3)+(b0+b1++b2+b3))=a0(a0+a1+a2+a3)+a0(b0+b1++b2+b3)+b0(a0+a1+a2+a3)+b0(b0+b1++b2+b3)=a0(b0+b1+b2+b3)+b0(a0+a1+a2+a3) If a0=0 and b00, the above ends up with b0(a1+a2+a3), 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 P3.

  • (iii) It is easy to tell that this set is NOT a subspace of P3. 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 [x(t)y(t)]=[1322][x(t)y(t)].

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 xx(t)=Axx(t). We learn from Section 5.7 "Applications to Differential Equations" that each eigenvalue–eigenvector pair provides a solution.

Now let's find out the eigenvalues of A. From det(AλI)=0, we have |1λ322λ|=λ23λ+8=0 This only gives two complex numbers as eigenvalues λ=3±23i2

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 2×2 matrix eigenvalues to trajectories:

Eigenvalues Trajectories
λ1>0,λ2>0 Repeller/Source
λ1<0,λ2<0 Attactor/Sink
λ1<0,λ2>0 Saddle Point
λ=a±bi,a>0 Spiral (outward) Point
λ=a±bi,a<0 Spiral (inward) Point
λ=±bi Ellipses (circles if b=1)

Problem 6 (10 points)

Which of the following matrices are diagonalizable over the real numbers?

  1. [2536] (ii) [4104] (iii) [113052007] (iv) [711022013]
  • 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 n×n matrix with n distinct eigenvalues is diagonalizable.

So let's find out the eigenvalues for each matrix:

    1. From the equation detAλI=0, we can obtain |2λ536λ|=(λ2)(λ+6)+15=(λ+1)λ+3)=0 This leads to two roots λ1=1, λ2=3.
    1. Since this is a triangular matrix, the eigenvalue is just 4, with multiplicity 2.
    1. For the same reason, this 3×3 matrix has eigenvalues 1, 5 and 7.
    1. Use cofactor expansion with C1,1, we have |7λ1102λ2013λ|=(7λ)(1)1+1|2λ213λ|=(7λ)(λ25λ+62)=(7λ)(λ4)(λ1) The eigenvalues are 7, 4, and 1.

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 2×2 matrix A has an eigenvalue λ1=2+i with corresponding eigenvector vv1=[3i4+i]. Which of the following is the general REAL solution to the system of differential equations xx(t)=Axx(t)

  • A. c1e2t[3costsint4cost+sint]+c2e2t[3sint+cost4sintcost]
  • B. c1e2t[3cost+sint4costsint]+c2e2t[3sintcost4sintcost]
  • C. c1e2t[3costsint4cost+sint]+c2e2t[3sintcost4sintcost]
  • D. c1e2t[3cost+sint4costsint]+c2e2t[3sint+cost4sintcost]
  • E. c1e2t[3cost+sint4costsint]+c2e2t[3sintcost4sint+cost]

Problem 7 Solution

From Section 5.7 "Applications to Differential Equations", we learn that the general solution to a matrix differential equation is xx(t)=c1vv1eλ1t+c2vv2eλ2t For a real matrix, complex eigenvalues and associated eigenvectors come in conjugate pairs. Hence we know that λ2=2i and vv2=[3+i4i]. However, we do not need these two to find our solution here. The real and imaginary parts of vv1eλ1t are (real) solutions of xx(t)=Axx(t), because they are linear combinations of vv1eλ1t and vv2eλ2t. (See the proof in "Complex Eigenvalues" of Section 5.7)

Now use Euler's formula (eix=cosx+isinx), we have vv1eλ1t=e(2+i)t[3i4+i]=e2t(cost+isint)[3i4+i]=e2t[(3cost+sint)+(3sintcost)i(4costsint)+(4sint+cost)i] The general REAL solution is the linear combination of the REAL and IMAGINARY parts of the result above, it is c1e2t[3cost+sint4costsint]+c2e2t[3sintcost4sint+cost]

So the answer is E.

Problem 8 (10 points)

Let T:M2×2M2×2 be a linear map defined as AA+AT.

(2 points) (1) Find T([1234])

(4 points) (2) Find a basis for the range of T.

(4 points) (3) Find a basis for the kernel of T.

Problem 8 Solution

  1. As the mapping rule is AA+AT, we can directly write down the transformation as below T([1234])=[1234]+[1234]T=[2558]

  2. If we denote the 4 entries of a 2×2 matrix as [abcd], the transformation can be written as T([abcd])=[abcd]+[abcd]T=[2ab+cb+c2d]=2a[1000]+(b+c)[0110]+2d[0001] So the basis can be the set of three 3×3 matrices like below {[1000],[0110],[0001]}

  3. The kernel (or null space) of such a T is the set of all uu in vector space V such that T(uu)=00. Write this as T([abcd])=[2ab+cb+c2d]=[0000] This leads to a=d=0 and c=b. So the original matrix A that satified this conditioncan be represented as c[0110]. This shows that [0110] (or [0110]) is the basis for the null space of T.

Problem 9 (10 points)

(6 points) (1) Find all the eigenvalues of matrix A=[400121123], and find a basis for the eigenspace corresponding to each of the eigenvalues.

(4 points) (2) Find an invertible matrix P and a diagonal matrix D such that [400121123]=PDP1

Problem 9 Solution

  1. Apply the equation detAλI=0, we have |4λ0012λ1123λ|=(4λ)|2λ123λ|=(λ4)2(λ1)=0 So the eigenvalues are 4 an 1. Now to find eigenvector for each eigenvalue, we take the eigenvalue to the system (AλI)xx=00 and find the basis vector(s) which would be the eigenvector.
    • For λ1=λ2=4, we have the new matrix as [000121121][000121000] This gives x12x2+x3=0 with two free variables x2 and x3. Now in parametric vector form, we can obtain [x1x2x3]=[2x2x3x2x3]=x2[210]+x3[101] A basis is {[210],[101]}.
    • For λ3=1, the new matrix is [300111122][100011022][100011000] This gives x1=0 and x2=x3 with one free variable x3. Again in parametric vector form, we can obtain [x1x2x3]=[0x3x3]=x3[011] A basis is {[011]}.
  2. From the above solution we can directly write P and D below P=[210101011]D=[400040001]

Problem 10 (10 points)

(4 points) (1) Find the eigenvalues and corresponding eigenvectors of the matrix [5142]

(2 points) (2) Find a general solution to the system of differential equations [x(t)y(t)]=[5142][x(t)y(t)]

(4 points) (3) Let [x(t)y(t)] be a particular soilution to the initial value problem [x(t)y(t)]=[5142][x(t)y(t)],[x(0)y(0)]=[37]. Find x(1)+y(1).

Problem 10 Solution

  1. To find eigenvalues, write down the determinant as |5λ142λ|=(λ+6)(λ+1)=0 So the eigenvalues are λ1=6 and λ2=1. Now follow the same method as Problem 9 solution to get eigenvectors for them.
    • For λ1=6, the new matrix is [1144][1100] The eigenvector is [11].
    • For λ1=1, the new matrix is [4141][4100] The eigenvector is [14].
  2. The general solution to a matrix differential equation is xx(t)=c1vv1eλ1t+c2vv2eλ2t So from this, since we already found out the eigenvalues and the corresponding eigenvectors, we can write down [x(t)y(t)]=c1[11]e6t+c2[14]et
  3. Now apply the initial values of x(0) and y(0), here comes the following equations: c1+c2=3c1+4c2=7 This gives c1=1 and c2=2. So x(1)+y(1)=e6+2e1e6+8e1=10e1.

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"