程序代写代做代考 AI Math 215.01, Spring Term 1, 2021 Writing Assignment #3

Math 215.01, Spring Term 1, 2021 Writing Assignment #3
Turn your 􏰁le in on Pweb in pdf format under Writing Assign- ments”. There is no need to put your name on the document since I will grade anonymously, and Pweb will keep track of the authorship of the document.
You will be graded on your writing (use of quanti􏰁ers, state- ment and use of de􏰁nitions, and other mathematical language) as well as the validity and completeness of your mathematical arguments.
1. Let T : R2 ! R2 be an arbitrary injective linear trans- formation and let u~;w~ 2 R2bearbitrary. Show that if w~ 2= Span(u~), then T (w~ ) 2= Span(T (u~)).
De􏰁nition of injective linear transformation: Let P : R2 ! R2 be a linear transformation and let a~;~b 2 R2. If P is injective linear transformation, we have that: if P(a~) = P (~b), then a~ = ~b.
In R2, a linear transformation is a function P : R2 ! R2 that satis􏰁es the following properties : 1.For all ~x;y~ 2 R2, P(~x +y~) = P(~x)+P(y~). 2.For all a 2 R and z~ 2 R2, P (a 􏰄 z~) = a 􏰄 P (z~).
Contrapositive: If T (w~ ) 2 Span(T (u~)), then w~ 2 Span(u~). If the contrapositive statement is true, then the original statement is also true. Let T : R2 ! R2 be an arbi-

trary injective linear transformation. Let u~;w~ 2 R2 be arbitrary. Suppose T (w~ ) 2 Span(T (u~)). Since T (w~ ) 2 Span(T(u~)), we can 􏰁x n 2 R with T(w~) = nT(u~). Since T : R2 ! R2 is a linear transformation and according to the second property of the linear transformation, we have T(w~) = nT(u~) = T(nu~). Since T : R2 ! R2 is an injec- tive linear transformation and according to the de􏰁nition of injective linear transformation, we have w~ = nu~. Since n2R,wecansaythatw~ 2Span(u~). SinceT:R2 !R2 is an arbitrary injective linear transformation and u~; w~ 2 R2 are arbitrary, we can conclude if T (w~ ) 2 Span(T (u~)), then w~ 2 Span(u~). Since we have shown the contrapositive statement is true, we can conclude if w~ 2= Span(u~), then T (w~ ) 2= Span(T (u~)).
,

2. Explain why a 2 􏰅 2 matrix A is invertible if and only if 0 is not an eigenvalue of A.
Hint: Remember what you need to do when you prove an if and only if” statement!
De􏰁nition 3.5.3: An eigenvector of matrix T is a nonzero vector ~v 2 R such that there exists 􏰂 2 R with T~v = 􏰂~v. An eigenvalue of T is a scalar 􏰂 2 R such that there exists a nonzero ~v 2 R2 with T~v = 􏰂~v.
Theorem 3.3.3: Let P : R2 ! R2 be a linear transforma-

xn
. 1. If xm􏰃ny = 0 and at least one of x;y;n;m is nonzero, then
tion,and􏰁xx;y;n;m2RwithP =
there exist nonzero ~v ; w~ 2 R with Null(P ) = Span(~v ) and
range(P) = Span (w~). 2. If all of x;y;n;m equal 0, then Null(P) = R2 and range(P) = f~0g. 3. If xm 􏰃 ny 6= 0, then Null(P ) = f~0g and range(P ) = R2.
Proposition 3.3.4: Let T : R2 ! R2 be a linear transfor- mation. We then have that T is injective if and only if Null(T) = f~0g.
Corollary 3.3.5: Let F : R2 ! R2 be a linear transforma-

x1 x2 y1 y2
tion, and 􏰁x x1;x2;y1;y2 2 R with [F] = 
. The
ym

following are equivalent: 1. x1y2 􏰃 x2y1 6= 0. 2. F is bijec- tive. 3. F is injective. 4. F is surjective.
Proof: If matrix A is invertible, then 0 is not an eigenvalue of A.
Contrapositive: If 0 is an eigenvalue of A, then matrix A is not invertible.
Let A be an arbitrary 2􏰅2 matrix. Let 0 be an eigenvalue of matrix A. Since 0 is an eigenvalue of matrix A and accord- ing to de􏰁nition 3.5.3, we can 􏰁x a nonzero vector ~v 2 R as an eigenvector of A such that A~v = 0~v . Since A~v = 0~v , wehaveA~v =0~v =0. SinceA~v =0and~v isanonzero eigenvector, we have that Null(A) 6= f~0g. Since Null(A) 6= f~0g and according to proposition 3.3.4, we can say that A is not injective. Since A is not injective and according to corollary 3.3.5, A is not bijective which indicates that A is not invertible. Since A is an arbitrary 2 􏰅 2 matrix and 0 is an eigenvalue of A, we have shown if 0 is an eigenvalue of A, then matrix A is not invertible. Since the contrapositive

statement is true, we can conclude if matrix A is invertible, then 0 is not an eigenvalue of A.
Proof: If 0 is not an eigenvalue of A, then matrix A is invertible.
Contrapositive: If matrix A is not invertible, then 0 is an eigenvalue of A.
Let A be an arbitrary singular matrix. Since A is a singular matrix and according to the theorem 3.3.3, item 1 and 2, we can 􏰁x a nonzero vector w~ 2 R2 such that Aw~ = ~0. Suppose w~ is an eigenvector of A. According to de􏰁nition 3.5.3, we can 􏰁x 􏰂 2 R as an eigenvalue of A such that Aw~ = 􏰂w~. Since Aw~ =~0 and Aw~ = 􏰂w~, we have Aw~ = 􏰂w~ = ~0. Since w~ is a nonzero vector, 􏰂 must be 0. Since A is an arbitrary singular matrix, we have shown if matrix A is not invertible, then 0 is an eigenvalue of A. Since the contrapositive statement is true, we can conclude if 0 is not an eigenvalue of A, then matrix A is invertible. Since we have shown both statements are true, we can conclude a 2􏰅2 matrix A is invertible if and only if 0 is not an

eigenvalue of A.

3. Given two 2􏰅2 matrices A and B, write A 􏰆 B to mean that there exists a 2 􏰅 2 invertible matrix P with
􏰃1 B=P AP:
(a) ShowthatA􏰆Aforall2􏰅2matricesA.
Incorporate the answers to the following questions in your answer: First, note that you need to start with an arbitrary 2􏰅2 matrix A|why? Why do you need to 􏰁nd a matrix P that 􏰁ts the de􏰁nition of 􏰆 to complete this problem?
Let A be an arbitrary 2􏰅2 matrix. Let I be a 2􏰅 2 identity matrix. Since I is an identity matrix, I is invertible. Since I is an identity matrix, we have I􏰃1 = I. Now notice A􏰄I = A and I􏰃1􏰄A = A. By substituting A = AI into A = I􏰃1A, we have A = I􏰃1A = I􏰃1(AI) = I􏰃1AI. Since A is an arbitrary 2 􏰅 2 matrix and I is a 2 􏰅 2 invertible matrix, we have shown A 􏰆 A for all 2 􏰅 2 matrices A.
(b) Show that if A and B are 2􏰅2 matrices with A 􏰆 B, then B 􏰆 A.
Incorporate the answers to the following questions in your answer: What is arbitrary in this problem? Here, you already know that if A 􏰆 B, you can choose an invertible matrix, say Q, such that B = Q􏰃1AQ. Why? What might you do with this matrix Q which has been given to you?

Proposition 3.3.18: 1. If P is an invertible 2 􏰅 2 ma- trix, then (P􏰃1)􏰃1) = P. 2. If T and F are both invertible 2 􏰅 2 matrices, then T F is invertible and (TF)􏰃1 = F􏰃1T􏰃1
Let A and B be arbitrary 2􏰅2 matrices with A 􏰆 B. According to the de􏰁nition of 􏰆, we can 􏰁x an invert- ible 2 􏰅 2 matrix Q such that B = Q􏰃1AQ. By mul- tiplying Q on both sides of the equation B = Q􏰃1AQ, we have QB = QQ􏰃1AQ = IAQ = AQ. Then by mul- tiplying Q􏰃1 on both sides of the equation QB = AQ, wehaveQBQ􏰃1 =AQQ􏰃1 =AI=A. SinceQis invertible, we can 􏰁x an invertible 2 􏰅 2 matrix P with P 􏰃1 = Q. By substituting Q = P 􏰃1 into the equa- tion A = QBQ􏰃1, we have A = P􏰃1B(P􏰃1)􏰃1. Ac- cording to proposition 3.3.18, item 1, we have A = P 􏰃1B(P 􏰃1)􏰃1 = P 􏰃1BP . Since A and B are arbitrary 2 􏰅 2 matrices and P is a 2 􏰅 2 invertible matrix, we have shown if A and B are 2􏰅2 matrices with A 􏰆 B, then B 􏰆 A.

(c) Show that if A, B and C are 2 􏰅 2 with both A 􏰆 B and B 􏰆 C, then A 􏰆 C.
Incorporate the answers to the following questions in your answer: What is arbitrary in this problem? Once again, you can choose two special matrices|why?
Proposition 3.3.18: 1. If P is an invertible 2 􏰅 2 ma- trix, then (P􏰃1)􏰃1) = P. 2. If T and F are both invertible 2 􏰅 2 matrices, then T F is invertible and (TF)􏰃1 = F􏰃1T􏰃1
Let A, B, and C be arbitrary 2 􏰅 2 matrices with A 􏰆 B and B 􏰆 C. According to the de􏰁nition of 􏰆, we can 􏰁x 2 􏰅 2 invertible matrices Q and P such that B = P􏰃1AP and C = Q􏰃1BQ. By substi- tuting B = P􏰃1AP into equation C = Q􏰃1BQ, we have C = Q􏰃1(P 􏰃1AP )Q = (Q􏰃1P 􏰃1)A(P Q). Ac- cording to proposition 3.3.18, item 2, we have C = (Q􏰃1P 􏰃1)A(P Q) = (P Q)􏰃1A(P Q). Since P and Q are invertible matrices and according to proposition 3.3.18, item 2, we know that P Q is an invertible 2 􏰅 2 matrix. Since P Q is an invertible 2 􏰅 2 matrix, we can 􏰁x an invertible 2 􏰅 2 matrix Z such that Z = PQ. By substituting Z into equation C = (P Q)􏰃1A(P Q),

we have C = Z􏰃1AZ. Since A, B, and C are arbitrary 2 􏰅 2 matrices and Z is a 2 􏰅 2 invertible matrix, we can conclude if A, B and C are 2 􏰅 2 with both A 􏰆 B and B 􏰆 C, then A 􏰆 C.
Cultural Aside: Using the problem above along with our work in class, it follows that A 􏰆 B if and only if A and B are both representations of a common lin- ear transformation T : R2 ! R2, but with respect to possibly di􏰀erent coordinates. In this problem, you are proving that 􏰆 is something called an equivalence rela- tion, a concept that you will see repeatedly throughout your mathematical journey.

4. Comment on working with partner(s): We completed the problem set individually and checked our work together. Although the methods we used to prove the problems were di􏰀erent, we ultimately obtained similar conclusions. We believe most of our proofs are valid.