Preparation problems for the discussion sections on November 11th and 13th1. Let A =1 00 11 1.a. Find the QR decomposition of A: write A = QR where Q is a matrix with orthonormalcolumns and R is an upper triangular matrix.b. Let b =001. Use the QR decomposition of A to find the least squares solution ofAbx = b (by solving Rbx = QTb).Solution:a. We start with columns of A(= [v1v2]) and we use Gram-Schmidt to find columns ofQ(= [q1q2]):q1=v1kv1k=101k101k=1√201√2and,q2=v2− (q1· v2)q1kv2− (q1· v2)q1k=011− (1√201√2·011)1√201√2k011− (1√201√2·011)1√201√2k=−12112k−12112k=r23−12112=−1√62√61√6Hence,Q =1√2−1√602√61√21√6We have:R = QTA =1√201√2−1√62√61√61 00 11 1="√21√203√6#b. We have to solve Rbx = QTb:"√21√203√6#ab=1√201√2−1√62√61√6001=1√21√6Therefore,bx =1313.2. a. Compare det1 23 4and the “row flipped” determinant det3 41 2.1b. If A =0 0 0 0 10 0 0 1 00 0 1 0 00 1 0 0 01 0 0 0 0, what is det(A)?c. If A =1 1 42 2 53 3 6, what is det(A)?d. If A =1 4 52 5 73 6 9, what is det(A)?e. If A, B are 3 × 3 matrices with det(A) = 2, det(B) = −1, calculate(i) det(BAT),(ii) det(BAB−1),(iii) det(A−1).f. If A =1 2 33 2 11 1 3, find det(A) by expanding along the last column.Solution:a. We have:det(1 23 4) = 1.4 − 2.3 = −2and,det(3 41 2) = 3.2 − 4.1 = 2So, det(1 23 4) = −det(3 41 2). This agrees with the fact that we know that theinterchange of two rows changes the sign of the determinant.b. We transform A into an upper triangular matrix using row operations:A =0 0 0 0 10 0 0 1 00 0 1 0 00 1 0 0 01 0 0 0 0R1↔R5,R2↔R4−−−−−−−−−→1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1Since we swap rows twice, we have:det(A) = −(−det(1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1)) = 1c. We transform A into an upper triangular matrix using row operations:A =1 1 42 2 53 3 6R2→R2−2R1,R3→R3−3R1−−−−−−−−−−−−−−−→1 1 40 0 −30 0 −62Since the row operations that we used do not change the value of the determinant,we have:det(A) = det(1 1 40 0 −30 0 −6) = 1.0.(−6) = 0d. We transform A into an upper triangular matrix using row operations:A =1 4 52 5 73 6 9R2→R2−2R1,R3→R3−3R1−−−−−−−−−−−−−−−→1 4 50 −3 −30 −6 −6R3→R3−2R2−−−−−−−→1 4 50 −3 −30 0 0Since the row operations that we used do not change the value of the determinant,we have:det(A) = det(1 4 50 −3 −30 0 0) = 0e. (i) det(BAT) = det(B) det(AT) = det(B) det(A) = −2,(ii) det(BAB−1) = det(B) det(A) det(B−1) = det(B) det(A)1det(B)= det(A) = 2,(iii) det(A−1) =1det(A)=12.f. We have:det(A) = 3 det(3 21 1) − 1 det(1 21 1) + 3 det(1 23 2) = 3.1 − 1.(−1) + 3.(−4) = −83. a. Someone tells you that det is linear, so det(3A) = 3 det(A). What do you answer?(What about det(31 00 1)? If A is a 3 × 3 matrix, and det(A) = 2 what is det(3A)?)b. Somebody tells you that the matrixA =1 2 −2 02 3 −4 0−1 −2 0 00 2 5 0is invertible. What do you say?c. LetA =1 2 −2 02 3 −4 1−1 −2 0 20 2 5 3.Calculate det(A). Is A invertible?d. Let A be a 3 × 3 matrix so that A11−1= 0. What is det(A).Solution:a. In general, if A is a n × n matrix then det(3A) = 3ndet(A). In particular, if A is a3 × 3 matrix, and det(A) = 2 then det(3A) = 33det(A) = 27.2 = 54.Hence, if det(3A) = 3 det(A), then either n = 1 (i.e., A is a 1 ×1 matrix) or det A = 0.Otherwise, the claim that det(3A) = 3 det(A) is false.b. Since A has a column of zeros, det(A) = 0. In other words, A is not invertible. Weshould tell the person to review their linear algebra.3c. We transform A into an upper triangular matrix using row operations:A =1 2 −2 02 3 −4 1−1 −2 0 20 2 5 3R2→R2−2R1 , R3→R3+R1−−−−−−−−−−−−−−−−−→1 2 −2 00 −1 0 10 0 −2 20 2 5 3R4→R4+2R3,R4→R4+5/2R1−−−−−−−−−−−−−−−−−→1 2 −2 00 −1 0 10 0 −2 20 0 0 10Since the row operations that we used do not change the value of the determinant,we have:det(A) = det(1 2 −2 00 −1 0 10 0 −2 20 0 0 10) = 20A is invertible since det(A) 6= 0.d. Since Ax = 0 has a non-zero solution, A is not invertible, i.e., det(A) = 0.4. Reading through your favorite linear algebra textbook, you find the following interestingstatement: if the columns of A are independent, then the orthogonal projection onto ColA hasprojection matrix A(ATA)−1AT.a. How does this formula simplify in the case when A has orthonormal columns?b. Let Q =1 00350 −45. What is the projection matrix corresponding to the orthogonalprojection onto Col(Q)?c. Let Q =1 0 0035450 −4535. What is the projection matrix corresponding to the orthogonalprojection onto Col(Q)? Explain why your answer is not surprising.d. (optional) Can you explain the formula A(ATA)−1ATfor the projection matrix usingthe normal equations for least squares?Solution:a. If A has orthonormal columns then ATA = I. So the projection matrix is:A(ATA)−1AT= AATb. Since Q has orthonormal columns, the projection matrix is:QQT=1 00350 −451 0 0035−45=1 0 00925−12250 −12251625c. Q has orthonormal columns and is square, so is orthogonal and satisfies Q−1= QT.Therefore, the projection matrix QQTis equal to I.Explanation: since the columns of Q are linearly independent and Q has 3 columns,the columns of Q form a basis for R3. In other words, Col(Q) = R3and projection ofeach vector in R3onto Col(Q) is itself, i.e., the projection matrix is I.5. True or False? Justify your answers!a. Let Q be a 3 × 3 orthogonal matrix. Then det(Q) = 1.4b. If det(A) =det(B) = 0 then det(A + B) = 0.c. We say
View Full Document
Unlocking...