# 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 quantiers, state- ment and use of denitions, 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~)).

Denition 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 satises 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 denition 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!

Denition 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 xmny = 0 and at least one of x;y;n;m is nonzero, then

tion,andxx;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 22 matrix. Let 0 be an eigenvalue of matrix A. Since 0 is an eigenvalue of matrix A and accord- ing to denition 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 denition 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 22 matrix A is invertible if and only if 0 is not an

eigenvalue of A.

3. Given two 22 matrices A and B, write A B to mean that there exists a 2 2 invertible matrix P with

1 B=P AP:

(a) ShowthatAAforall22matricesA.

Incorporate the answers to the following questions in your answer: First, note that you need to start with an arbitrary 22 matrix A|why? Why do you need to nd a matrix P that ts the denition of to complete this problem?

Let A be an arbitrary 22 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 I1 = I. Now notice AI = A and I1A = A. By substituting A = AI into A = I1A, we have A = I1A = I1(AI) = I1AI. 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 22 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 = Q1AQ. 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 (P1)1) = P. 2. If T and F are both invertible 2 2 matrices, then T F is invertible and (TF)1 = F1T1

Let A and B be arbitrary 22 matrices with A B. According to the denition of , we can x an invert- ible 2 2 matrix Q such that B = Q1AQ. By mul- tiplying Q on both sides of the equation B = Q1AQ, we have QB = QQ1AQ = IAQ = AQ. Then by mul- tiplying Q1 on both sides of the equation QB = AQ, wehaveQBQ1 =AQQ1 =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 = QBQ1, we have A = P1B(P1)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 22 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 (P1)1) = P. 2. If T and F are both invertible 2 2 matrices, then T F is invertible and (TF)1 = F1T1

Let A, B, and C be arbitrary 2 2 matrices with A B and B C. According to the denition of , we can x 2 2 invertible matrices Q and P such that B = P1AP and C = Q1BQ. By substi- tuting B = P1AP into equation C = Q1BQ, we have C = Q1(P 1AP )Q = (Q1P 1)A(P Q). Ac- cording to proposition 3.3.18, item 2, we have C = (Q1P 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 = Z1AZ. 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 dierent 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 dierent, we ultimately obtained similar conclusions. We believe most of our proofs are valid.